{"id":3251,"date":"2017-09-27T22:36:55","date_gmt":"2017-09-27T22:36:55","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3251"},"modified":"2024-09-04T14:32:25","modified_gmt":"2024-09-04T14:32:25","slug":"resolucoes-dinamica-2014-2013","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/mecanica\/vestibulares-recentes-por-assunto-2014-e-2013\/resolucoes-dinamica-2014-2013\/","title":{"rendered":"Resolu\u00e7\u00f5es Din\u00e2mica – 2014 – 2013"},"content":{"rendered":"

RESOLU\u00c7\u00d5ES \u2013 2014 – 2013<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>A<\/b><\/span><\/span>s tr\u00eas leis de Newton<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se ele est\u00e1 com\u00a0<\/b><\/span><\/span><\/span>velocidade constante<\/b><\/span><\/span><\/span>, independente da inclina\u00e7\u00e3o, ele est\u00e1 em\u00a0<\/b><\/span><\/span><\/span>MRU na vertical e na horizontal<\/b><\/span><\/span><\/span>e, pelo princ\u00edpio da in\u00e9rcia, a\u00a0<\/b><\/span><\/span><\/span>resultante das for\u00e7as que agem sobre ele \u00e9 nula (equil\u00edbrio din\u00e2mico) —\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A for\u00e7a resultante sobre o bloco na dire\u00e7\u00e3o do movimento \u00e9 a for\u00e7a de atrito F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcN=\u03bcP= \u03bcmg= 0,4.m.10 —\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=4m\u00a0<\/b><\/span><\/span><\/span>— segunda lei de Newton —\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=ma\u00a0<\/b><\/span><\/span><\/span>— 4m=ma —\u00a0<\/b><\/span><\/span><\/span>a=4m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/span>— Torricelli —\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.a.d\u00a0<\/b><\/span><\/span><\/span>— 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 20<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.(-4)d — d=400\/8 —<\/b><\/span><\/span><\/span>d=50m<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

R- E<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como ele se move com\u00a0<\/b><\/span><\/span><\/span>velocidade constante na mesma altitude,\u00a0<\/b><\/span><\/span><\/span>na horizontal\u00a0<\/b><\/span><\/span><\/span>ele est\u00e1 em\u00a0<\/b><\/span><\/span><\/span>equil\u00edbrio din\u00e2mico (MRU)<\/b><\/span><\/span><\/span>\u00a0e, assim a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a resultante\u00a0<\/b><\/span><\/span><\/span>na\u00a0<\/b><\/span><\/span><\/span>horizonta<\/b><\/span><\/span><\/span>l \u00e9\u00a0<\/b><\/span><\/span><\/span>nula\u00a0<\/b><\/span><\/span><\/span>(as for\u00e7as horizontais se anulam).<\/b><\/span><\/span><\/span><\/p>\n

Como ele permanece na\u00a0<\/b><\/span><\/span><\/span>mesma altitude<\/b><\/span><\/span><\/span>\u00a0(n\u00e3o sobe nem desce) a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a resultante na vertical \u00e9 nula\u00a0<\/b><\/span><\/span><\/span>(as for\u00e7as verticais se anulam).<\/b><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Desprezando-se a resist\u00eancia do ar<\/b><\/span><\/span><\/span><\/span>, qualquer corpo que estiver no espa\u00e7o ap\u00f3s lan\u00e7amento<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\"\"<\/p>\n

verical,\u00a0<\/b><\/span><\/span><\/span><\/span>lan\u00e7amento obl\u00edquo<\/b><\/span><\/span><\/span><\/span>,queda livre, etc, ter\u00e1 como\u00a0<\/b><\/span><\/span><\/span><\/span>for\u00e7a resultante a for\u00e7a peso\u00a0<\/b><\/span><\/span><\/span><\/span>que tem sempre\u00a0<\/b><\/span><\/span><\/span><\/span>dire\u00e7\u00e3o vertival e sentido para baixo<\/b><\/span><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- A<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0Na vertical:<\/b><\/span><\/span><\/span>\u00a0Como cada dobradi\u00e7a sustenta o peso do port\u00e3o\u00a0<\/b><\/span><\/span><\/span>P=800N (vertical e para baixo)\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>cada uma suporta\u00a0<\/b><\/span><\/span><\/span>metade desse peso<\/b><\/span><\/span><\/span>, cada\u00a0<\/b><\/span><\/span><\/span>dobradi\u00e7a receber\u00e1 uma for\u00e7a vertical e para cima de P\u2019=800\/2<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

— P\u2019=400N<\/b><\/span><\/span><\/span>\u00a0(figura acima). Assim, na\u00a0<\/b><\/span><\/span><\/span>vertical o port\u00e3o est\u00e1 em equil\u00edbrio<\/b><\/span><\/span><\/span>, pois 2P\u2019=P.<\/b><\/span><\/span><\/span><\/p>\n

Na horizontal<\/b><\/span><\/span><\/span>: na\u00a0<\/b><\/span><\/span><\/span>dobradi\u00e7a superior\u00a0<\/b><\/span><\/span><\/span>deve surgir uma\u00a0<\/b><\/span><\/span><\/span>for\u00e7a horizontal e para a esquerda\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0impedindo que o port\u00e3o gire no sentido hor\u00e1rio e\u00a0<\/b><\/span><\/span><\/span>na inferior uma for\u00e7a horizontal e para a direita\u00a0\"\"<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

tamb\u00e9m impedindo que ele gire no sentido hor\u00e1rio (figura acima).<\/b><\/span><\/span><\/span><\/p>\n

Colocando o\u00a0<\/b><\/span><\/span><\/span>polo 0<\/b><\/span><\/span><\/span>\u00a0(eixo de rota\u00e7\u00e3o) na\u00a0<\/b><\/span><\/span><\/span>dobradi\u00e7a inferior\u00a0<\/b><\/span><\/span><\/span>vamos calcular o\u00a0<\/b><\/span><\/span><\/span>momento de cada for\u00e7a em rela\u00e7\u00e3o a 0,\u00a0<\/b><\/span><\/span><\/span>estabelecendo o sentido hor\u00e1rio de rota\u00e7\u00e3o como positivo:<\/b><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span><\/span>Ns<\/b><\/span><\/span><\/span><\/sub><\/span>= – N<\/b><\/span><\/span><\/span><\/span>s<\/b><\/span><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span><\/span>= – N<\/b><\/span><\/span><\/span><\/span>s<\/b><\/span><\/span><\/span><\/sub><\/span>.1,6\u00a0<\/b><\/span><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span><\/span>M<\/b><\/span><\/span><\/span><\/span>NS<\/b><\/span><\/span><\/span><\/sub><\/span>= – 1,6N<\/b><\/span><\/span><\/span><\/span>s<\/b><\/span><\/span><\/span><\/sub><\/span><\/p>\n

\"\"<\/p>\n

M<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>= + P.d<\/b><\/span><\/span><\/span><\/span>= + 800.0,8\u00a0<\/b><\/span><\/span><\/span><\/span>— M<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>= + 640 N.m<\/b><\/span><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span><\/span>ni<\/b><\/span><\/span><\/span><\/sub><\/span>= N<\/b><\/span><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span><\/span>=N<\/b><\/span><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span><\/sub><\/span>.0\u00a0<\/b><\/span><\/span><\/span><\/span>— M<\/b><\/span><\/span><\/span><\/span>Ni<\/b><\/span><\/span><\/span><\/sub><\/span>=0<\/b><\/span><\/span><\/span><\/span><\/p>\n

No equil\u00edbrio de rota\u00e7\u00e3o a soma dos momentos de cada for\u00e7a deve ser nula:<\/b><\/span><\/span><\/span><\/p>\n

– 1,6N<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 640 = 0 — N<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=640\/1,6 — N<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=400N — como na horizontal as for\u00e7as devem se anular para que haja equil\u00edbrio, N<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=N<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=400N.<\/b><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na\u00a0<\/b><\/span><\/span><\/span>sequ\u00eancia da figura abaixo\u00a0<\/b><\/span><\/span><\/span>que a for\u00e7a aplicada no tubo 1 tem dire\u00e7\u00e3o paralela a do tubo 2 formando tamb\u00e9m um \u00e2ngulo de 60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0com a horizontal e, como n\u00e3o existe atrito essa\u00a0<\/b><\/span><\/span><\/span>for\u00e7a\u00a0\"\"<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00e9 transmitida at\u00e9 ao aspirador formando um \u00e2ngulo de\u00a0<\/b><\/span><\/span><\/span>60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/span>com a horizontal<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Essa for\u00e7a foi\u00a0<\/b><\/span><\/span><\/span>decomposta em duas parcelas\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span>Fcos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=4.0,5=<\/b><\/span><\/span><\/span>2N<\/b><\/span><\/span><\/span>\u00a0e em\u00a0\"\"<\/b><\/span><\/span><\/span>. Como o aspirador\u00a0<\/b><\/span><\/span><\/span>n\u00e3o<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>se move na vertical a for\u00e7a peso\u00a0\"\"\u00a0anula as for\u00e7as\u00a0\"\"\u00a0e a normal\u00a0\"\"<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span>\u00a0Assim, a \u00fanica for\u00e7a respons\u00e1vel pelo movimento do aspirador \u00e9\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=2N\u00a0<\/b><\/span><\/span><\/span>que \u00e9 a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a resultante\u00a0<\/b><\/span><\/span><\/span>sobre o mesmo.<\/b><\/span><\/span><\/span><\/p>\n

Aplicando a segunda lei de Newton\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=ma\u00a0<\/b><\/span><\/span><\/span>— 2=2.a —\u00a0<\/b><\/span><\/span><\/span>a=1m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span><\/p>\n

Observe a an\u00e1lise de cada trecho:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Resumindo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- E<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span> <\/b>Pelo enunciado\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/sub><\/span>=2F<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>— durante o movimento com\u00a0<\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o de a=0,2m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0a for\u00e7a de atrito F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0=240N e \u00e9 constante.<\/b><\/span><\/span><\/span><\/p>\n

Aplicando a segunda lei de Newton\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=ma\u00a0<\/b><\/span><\/span><\/span>— F<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>\u00a0+ F<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0= ma — F<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 2F<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0= ma —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

3F<\/b><\/span><\/span><\/span><\/span>D<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0\u2013 F<\/b><\/span><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0=ma — 3F<\/b><\/span><\/span><\/span><\/span>D<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0\u2013 240 = 120.0,2 — F<\/b><\/span><\/span><\/span><\/span>D<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0= (24 + 240)\/3 —\u00a0<\/b><\/span><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span><\/span>D<\/b><\/span><\/span><\/span><\/sub><\/span>=88N.<\/b><\/span><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/sub><\/span>=2F<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>=2.88<\/b><\/span><\/span><\/span>=176N.<\/b><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0Peso<\/b><\/span><\/span><\/span>\u00a0do paciente —\u00a0<\/b><\/span><\/span><\/span>P=<\/b><\/span><\/span><\/span>m.g=50.10=<\/b><\/span><\/span><\/span>500N<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>parcela horizontal da tra\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=Tcos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=T\u221a3\/2 —\u00a0<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=0,87T.<\/b><\/span><\/span><\/span><\/p>\n

Observe na figura que ele n\u00e3o deve se deslocar horizontalmente ent\u00e3o a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a de atrito deve anular a parcela horizontal da tra\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcN<\/b><\/span><\/span><\/span>=\u03bcP=0,26.500 —\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=130N\u00a0<\/b><\/span><\/span><\/span>— F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0= T<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\u00a0— 130 = 0,87T — T=<\/b><\/span><\/span><\/span><\/p>\n

130\/0,87 —\u00a0<\/b><\/span><\/span><\/span>T=148,5N<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

At\u00e9 T=149,5N e\u00a0<\/b><\/span><\/span><\/span>m=14,95kg ele n\u00e3o se desloca e assim, a massa m\u00e1xima deve ser de m=13kg.<\/b><\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A for\u00e7a resultante<\/b><\/span><\/span><\/span>\u00a0na\u00a0<\/b><\/span><\/span><\/span>dire\u00e7\u00e3o do movimento \u00e9 a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a de atrito<\/b><\/span><\/span><\/span>,\u00a0<\/b><\/span><\/span><\/span>constante<\/b><\/span><\/span><\/span>, contr\u00e1ria ao mesmo que produz uma\u00a0<\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o de retardamento tamb\u00e9m constante<\/b><\/span><\/span><\/span>\u00a0de intensidade F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=ma —\u00a0<\/b><\/span><\/span><\/span>V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0– at\u00a0<\/b><\/span><\/span><\/span>— fun\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>de primeiro grau (reta decrescente), velocidade diminuindo at\u00e9 parar, quando V=0.<\/b><\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se voc\u00ea n\u00e3o domina a teoria, ela est\u00e1 detalhada a seguir:<\/b><\/span><\/span><\/span><\/p>\n

For\u00e7a de atrito est\u00e1tico (\"\") e for\u00e7a de atrito din\u00e2mico(\"\").<\/b><\/span><\/span><\/span><\/p>\n

Considere um bloco de massa m sujeito a uma for\u00e7a externa\u00a0\"\"de intensidade vari\u00e1vel<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Quando o corpo estiver em repouso e n\u00e3o houver for\u00e7a externa, F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=0.\u00a0 Com o corpo permanecendo em repouso, aumentando F, a intensidade da for\u00e7a de atrito est\u00e1tico tamb\u00e9m aumenta (figuras acima), at\u00e9 que o corpo fique na imin\u00eancia de movimento (figura abaixo).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Quando o corpo est\u00e1 na imin\u00eancia de movimento a intensidade da for\u00e7a de atrito est\u00e1tico \u00e9 m\u00e1xima. Uma for\u00e7a de intensidade maior que F<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0faz com que o corpo entre em movimento e, a partir da\u00ed a for\u00e7a de atrito \u00e9 din\u00e2mica ou cin\u00e9tica.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Depois que o bloco entra em movimento a for\u00e7a de atrito \u00e9 denominada for\u00e7a de atrito din\u00e2mica, e \u00e9 sempre a mesma, \u00a0independente da velocidade A intensidade da for\u00e7a de atrito din\u00e2mica \u00e9 ligeiramente menor que a intensidade da for\u00e7a de atrito est\u00e1tica m\u00e1xima.<\/b><\/span><\/span><\/span><\/p>\n

Suponha, nos exemplos acima, que se F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0tiver intensidade de 3N, F<\/b><\/span><\/span><\/span>at1<\/b><\/span><\/span><\/sub><\/span>\u00a0tamb\u00e9m ter\u00e1 intensidade de 3N e o corpo continua em repouso.. Se F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=5N, F<\/b><\/span><\/span><\/span>at2<\/b><\/span><\/span><\/sub><\/span>=5N e o corpo continua em repouso. Se F<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=7N, a m\u00e1xima intensidade de atrito est\u00e1tico, que corresponde \u00e0 imin\u00eancia de movimento, \u00e9 dada por:<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>.N<\/b><\/span><\/span><\/span><\/p>\n

Onde,\u00a0F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 for\u00e7a de atrito est\u00e1tico m\u00e1ximo ou simplesmente for\u00e7a de atrito est\u00e1tico, m<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u00e9 coeficiente de atrito est\u00e1tico e N a intensidade da for\u00e7a normal, que o bloco troca com o plano.<\/b><\/span><\/span><\/span><\/p>\n

Portanto o bloco s\u00f3 entrar\u00e1 em movimento com qualquer for\u00e7a que seja maior que 7N. Assim, se aplicarmos por exemplo, 8N o bloco entrar\u00e1 em movimento e agora a for\u00e7a de atrito ser\u00e1 din\u00e2mica e dada por:<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>atd<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>.N<\/b><\/span><\/span><\/span><\/p>\n

Onde,\u00a0F<\/b><\/span><\/span><\/span>atd<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 for\u00e7a de atrito din\u00e2mico, m<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 coeficiente de atrito din\u00e2mico e N a intensidade da for\u00e7a normal, que o bloco troca com o plano.<\/b><\/span><\/span><\/span><\/p>\n

Verifica-se experimentalmente que F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>>F<\/b><\/span><\/span><\/span>atd<\/b><\/span><\/span><\/sub><\/span>\u00a0e que m<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>>m<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

O gr\u00e1fico abaixo representa todo o processo explicado acima.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0<\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Na figura 1 foram colocados os m\u00f3dulos de todas as for\u00e7as que agem sobre cada bloco —\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>. P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0e P<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>que correspondem aos pesos de cada bloco —\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>el<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 for\u00e7a el\u00e1stica aplicada pela mola sobre os blocos C e B —\u00a0<\/b><\/span><\/span><\/span>Fat\u00a0<\/b><\/span><\/span><\/span>– for\u00e7a de atrito sobre o bloco B, contr\u00e1ria ao movimento —\u00a0<\/b><\/span><\/span><\/span>T\u00a0<\/b><\/span><\/span><\/span>\u2013 for\u00e7a de tra\u00e7\u00e3o trocada entre os blocos A e B —\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>NB<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 parcela do peso perpendicular ao plano inclinado que o comprime contra o mesmo —\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>PB<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>\u2013 parcela do peso paralela ao plano respons\u00e1vel pela descida do bloco —\u00a0<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span>\u00a0\u2013 acelera\u00e7\u00e3o de cada bloco no sentido hor\u00e1rio — na figura 2 foram colocadas apenas as for\u00e7as que agem na dire\u00e7\u00e3o do movimento de cada bloco:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span><\/span>pB<\/b><\/span><\/span><\/sub><\/span>=Psen37<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=mgsen37<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.10.0,6=<\/b><\/span><\/span><\/span>6m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bc.N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>= \u03bcP<\/b><\/span><\/span><\/span>NB<\/b><\/span><\/span><\/sub><\/span>= \u03bcm<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.10.cos37<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=0,4.m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.10.0,8<\/b><\/span><\/span><\/span>=3,2m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>el<\/b><\/span><\/span><\/sub><\/span>=k.x<\/b><\/span><\/span><\/span>=1240.0,02<\/b><\/span><\/span><\/span>=24,8.<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>.a\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>T + P<\/b><\/span><\/span><\/span>pB<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 (F<\/b><\/span><\/span><\/span>el<\/b><\/span><\/span><\/sub><\/span>\u00a0+ F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>) = m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.a\u00a0<\/b><\/span><\/span><\/span>— 22,8 + 6m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 24,8 \u2013 3,2m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.2,4 — 6m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 5,6m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= 2 — m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=2\/0,4 —\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=5kg<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

R- E.<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0Princ\u00edpio\u00a0da\u00a0In\u00e9rcia\u00a0(Primeira Lei de Newton)<\/b><\/span><\/span><\/span><\/p>\n

\u201c<\/span>Todo corpo que esteja em\u00a0<\/b><\/span><\/span><\/span>repouso ou em movimento retil\u00edneo e uniforme\u00a0<\/b><\/span><\/span><\/span>(movendo-se em trajet\u00f3ria reta com velocidade\u00a0 vetorial constante),\u00a0<\/b><\/span><\/span><\/span>tende a continuar nestes estados se a for\u00e7a resultante que age sobre ele for nula\u201d.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- A<\/b><\/span><\/span><\/span><\/p>\n

For\u00e7as no MCU<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Cada garota encontra-se em movimento circular uniforme com a mesma velocidade angular\u00a0<\/b><\/span><\/span><\/span>(W),\u00a0<\/b><\/span><\/span><\/span>pois \u201cvarrem\u201d o mesmo \u00e2ngulo no mesmo tempo — em todo movimento circular surge sempre uma for\u00e7a resultante denominada for\u00e7a resultante centr\u00edpeta\u00a0\"\"\u00a0com as seguintes caracter\u00edsticas:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Essa for\u00e7a resultante centr\u00edpeta sobre a garota 1 tem intensidade F<\/b><\/span><\/span><\/span>c1<\/b><\/span><\/span><\/sub><\/span>=T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=m.W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.2R —\u00a0<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2mW<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>R (I)\u00a0<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span><\/p>\n

sobre a garota 2 — F<\/b><\/span><\/span><\/span>c2<\/b><\/span><\/span><\/sub><\/span>=T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=mW<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>R —\u00a0<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= mW<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>R (II)\u00a0<\/b><\/span><\/span><\/span>— veja figuras abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

substituindo (I) em (II) — T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 2mW<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>R = mW<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>R —\u00a0<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=3mW<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>R e T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2mW<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>R (I)\u00a0<\/b><\/span><\/span><\/span>— dados —\u00a0<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=120N —<\/b><\/span><\/span><\/span>120=2mw<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>R —<\/b><\/span><\/span><\/span>mw<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>R<\/b><\/span><\/span><\/span>=120\/20<\/b><\/span><\/span><\/span>=60\u00a0<\/b><\/span><\/span><\/span>— T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=3.60 — T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=180N —\u00a0<\/b><\/span><\/span><\/span>R- E.<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Na dire\u00e7\u00e3o radial<\/b><\/span><\/span><\/span>— o exerc\u00edcio afirma quea pessoa se solta do cip\u00f3 no instante em que sua velocidade fica novamente igual a zero, ponto S da figura — se nesse ponto a\u00a0<\/b><\/span><\/span><\/span>velocidade \u00e9 nula V<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>=0,\u00a0<\/b><\/span><\/span><\/span>nele a for\u00e7a resultante centr\u00edpeta tamb\u00e9m \u00e9 nula, pois F<\/b><\/span><\/span><\/span>CS<\/b><\/span><\/span><\/sub><\/span>=m.(V<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=m.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0 —\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>CS<\/b><\/span><\/span><\/sub><\/span>=0\u00a0<\/b><\/span><\/span><\/span>— ent\u00e3o a acelera\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

centr\u00edpeta em S ser\u00e1 tamb\u00e9m nula, pois F<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=m.a<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=0 —\u00a0<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=0\u00a0<\/b><\/span><\/span><\/span>e isso ocorre porque a tens\u00e3o\u00a0<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>\u00a0anula a componente oposta do peso\u00a0<\/b><\/span><\/span><\/span>P\u2019=Pcos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>Na dire\u00e7\u00e3o tangencial<\/b><\/span><\/span><\/span>— a for\u00e7a resultante<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0sobre o corpo em\u00a0<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span>\u00a0(<\/b><\/span><\/span><\/span>respons\u00e1vel pelo movimento na<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

dire\u00e7\u00e3operpendicular ao cip\u00f3<\/b><\/span><\/span><\/span>) \u00e9 a parcela tangencial do peso P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=Psen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0— F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=ma — F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0—<\/b><\/span><\/span><\/span><\/p>\n

ma=mgsen30 — a=gsen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=10.(1\/2) —\u00a0<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=5m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>— essa acelera\u00e7\u00e3o \u00e9 a tangencial e forma um \u00e2ngulo de 60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0com a vertical —<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Para ir de A at\u00e9 B o motociclista percorre meio arco de circunfer\u00eancia de raio R, ou seja, percorre \u0394S=2\u03c0R\/2— \u0394S=\u03c0R — sua menor velocidade m\u00e9dia ser\u00e1 V=\u0394S\/\u0394t, onde \u0394t \u00e9 o intervalo de tempo que ele demora para percorrer \u0394S — V=\u03c0R\/\u0394t— note que essa \u00e9 velocidade m\u00e9dia e, portanto, \u00e9 a mesma em A e em B — \u0394t=\u03c0R\/V (I) — sendo o objeto abandonado (V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0) em A ele cai em queda livre com acelera\u00e7\u00e3o igual \u00e0 da gravidade g, percorrendo a dist\u00e2ncia vertical 2R=\u0394S — equa\u00e7\u00e3o hor\u00e1ria do espa\u00e7o de uma queda livre — \u0394S=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u0394t + g.(\u0394t)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 — 2R=0 + g.(\u0394t)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 (II) — observe que o tempo que o objeto demora para chegar em B em queda livre, \u00e9 o mesmo que o motociclista demorou tamb\u00e9m para chegar em B e, assim, voc\u00ea pode substituir (I) em (II) — 2R=g.(\u03c0R\/V)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 — 4R=g.(\u03c0R\/V)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0— 4R=g.\u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0— 4=g.\u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.R\/V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0— 4V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=g.\u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.R — V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>= g.\u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.R\/4 (III) — mas, no ponto A as for\u00e7as que agem sobre o motociclista s\u00e3o seu peso P (vertical e para baixo) e a for\u00e7a normal N que ele troca com a superf\u00edcie superior do globo<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(tamb\u00e9m vertical e para baixo)— em todo movimento circular existe sempre uma for\u00e7a resultante centr\u00edpeta dirigida para o centro da circunfer\u00eancia, de intensidade F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R — no ponto A F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=P + N — mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=N + P (IV) — substituindo (III) em (IV) — m.(g.\u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.R\/4)\/R=N + P — m. g.\u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/4=N + P — mg=P — P.\u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/4=N + P — \u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=10 —<\/b><\/span><\/span><\/span><\/p>\n

10P\/4=N + P — 2,5P=N + P — 2,5P \u2013 P=N — 1,5P=N —\u00a0<\/b><\/span><\/span><\/span>N\/P=1,5.<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a)\u00a0<\/b><\/span><\/span><\/span>Energia mec\u00e2nica no ponto I\u00a0<\/b><\/span><\/span><\/span>— E<\/b><\/span><\/span><\/span>mI<\/b><\/span><\/span><\/sub><\/span>=mgh + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2= mgh + m0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mI<\/b><\/span><\/span><\/sub><\/span>=mgh<\/b><\/span><\/span><\/span><\/p>\n

Energia mec\u00e2nica no ponto II\u00a0<\/b><\/span><\/span><\/span>— E<\/b><\/span><\/span><\/span>mII<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span>\u00a0mgh + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2= mg0 + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mII<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span><\/p>\n

Pelo\u00a0<\/b><\/span><\/span><\/span>teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mI<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mII<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>— mgh=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 —\u00a0<\/b><\/span><\/span><\/span>V=\u221a(2gh)<\/b><\/span><\/span><\/span><\/p>\n

b) No\u00a0<\/b><\/span><\/span><\/span>ponto II\u00a0<\/b><\/span><\/span><\/span>as for\u00e7as que agem sobre o corpo s\u00e3o seu\u00a0<\/b><\/span><\/span><\/span>peso<\/b><\/span><\/span><\/span>\u00a0(vertical e para baixo) e for\u00e7a de\u00a0<\/b><\/span><\/span><\/span>tra\u00e7\u00e3o no fio\u00a0<\/b><\/span><\/span><\/span>(vertical e para cima) — em todo corpo em movimento circular surge uma for\u00e7a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

resultante centr\u00edpeta\u00a0<\/b><\/span><\/span><\/span>de dire\u00e7\u00e3o radial, dirigida para o centro da circunfer\u00eancia (no caso, para cima) e de intensidade\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/d\u00a0<\/b><\/span><\/span><\/span>— mas,\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=T \u2013 P\u00a0<\/b><\/span><\/span><\/span>— mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/d = T \u2013 mg — m.(\u221a(2gh)<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>\/d = T \u2013 mg —<\/b><\/span><\/span><\/span><\/p>\n

m.2gh\/d = T \u2013 mg — T=mg + 2mgh\/d —\u00a0<\/b><\/span><\/span><\/span>T=mg(1 + 2h\/d)<\/b><\/span><\/span><\/span><\/p>\n

Trabalho- Energia<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como o entulho de massa\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>\u00a0est\u00e1\u00a0<\/b><\/span><\/span><\/span>descendo com velocidade constante\u00a0<\/b><\/span><\/span><\/span>ele est\u00e1 em\u00a0<\/b><\/span><\/span><\/span>MR( (equil\u00edbrio din\u00e2mico<\/b><\/span><\/span><\/span>) e a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a resultante sobre ele deve ser nula\u00a0<\/b><\/span><\/span><\/span>e, para que isso ocorra, a intensidade da for\u00e7a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

aplicada pelo guindaste deve ser igual ao peso do entulho\u00a0<\/b><\/span><\/span><\/span>— F=P=mg=m.10\u00a0<\/b><\/span><\/span><\/span>— P=F=10m.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo do\u00a0<\/b><\/span><\/span><\/span>trabalho\u00a0<\/b><\/span><\/span><\/span>realizado pela for\u00e7a F=P — W=F.d.cos\u03b1=10m.15.cos0<\/b><\/span><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/sup><\/span>—\u00a0<\/b><\/span><\/span><\/span>W=150.m<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pela\u00a0<\/b><\/span><\/span><\/span>tabela peri\u00f3dica,<\/b><\/span><\/span><\/span>\u00a0voc\u00ea tem que as massas dos is\u00f3topos do\u00a0<\/b><\/span><\/span><\/span>chumbo (Pb), da prata (Ag), do h\u00e9lio (He) e do hidrog\u00eanio (H<\/b><\/span><\/span><\/span>), s\u00e3o respectivamente:\u00a0<\/b><\/span><\/span><\/span>207,2, 107,8, 4,002 e 1,008<\/b><\/span><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/span>
\n<\/b><\/span><\/span><\/span>Assim,\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\u00a0> m<\/b><\/span><\/span><\/span>Ag<\/b><\/span><\/span><\/sub><\/span>\u00a0> m<\/b><\/span><\/span><\/span>He<\/b><\/span><\/span><\/sub><\/span>\u00a0> m<\/b><\/span><\/span><\/span>H<\/b><\/span><\/span><\/sub><\/span>
\n<\/b><\/span><\/span><\/sub><\/span>Pela express\u00e3o da energia cin\u00e9tica\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0<\/b><\/span><\/span><\/span>voc\u00ea verifica que, sendo a\u00a0<\/b><\/span><\/span><\/span>velocidade a mesma\u00a0<\/b><\/span><\/span><\/span>para os 4 feixes de \u00e1tomos, a<\/b><\/span><\/span><\/span>energia cin\u00e9tica \u00e9 diretamente proporcional \u00e0 massa<\/b><\/span><\/span><\/span>. Assim, a rela\u00e7\u00e3o pedida \u00e9:\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\u00a0> E<\/b><\/span><\/span><\/span>Ag<\/b><\/span><\/span><\/sub><\/span>\u00a0> E<\/b><\/span><\/span><\/span>He<\/b><\/span><\/span><\/sub><\/span>\u00a0> E<\/b><\/span><\/span><\/span>H<\/b><\/span><\/span><\/sub><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A energia armazenada por um corpo de massa\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>\u00a0devido ao fato de ele se encontrar a certa altura\u00a0<\/b><\/span><\/span><\/span>h\u00a0<\/b><\/span><\/span><\/span>recebe o nome de\u00a0energia potencial\u00a0gravitacional\u00a0(<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>),\u00a0de express\u00e3o matem\u00e1tica:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

onde:<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0 energia potencial gravitacional \u2013 no SI, medida em\u00a0<\/b><\/span><\/span><\/span>joules\u00a0(<\/b><\/span><\/span><\/span><\/span>J<\/b><\/span><\/span><\/span><\/span>)<\/b><\/span><\/span><\/span><\/span><\/p>\n

m\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0 massa \u2013 no SI, medida em quilograma\u00a0(<\/b><\/span><\/span><\/span>kg<\/b><\/span><\/span><\/span>)<\/b><\/span><\/span><\/span><\/p>\n

g\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0 acelera\u00e7\u00e3o da gravidade local \u2013 no SI, medida em\u00a0<\/b><\/span><\/span><\/span>m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0ou\u00a0N\/kg<\/b><\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span><\/span>segundo trecho do passeio<\/b><\/span><\/span><\/span>\u00a0corresponde \u00e0 ida do bondinho de massa\u00a0<\/b><\/span><\/span><\/span>m=5000kg<\/b><\/span><\/span><\/span>\u00a0do Morro da Urca (h<\/b><\/span><\/span><\/span>U<\/b><\/span><\/span><\/sub><\/span>=220m) ao P\u00e3o de A\u00e7\u00facar (<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=400m<\/b><\/span><\/span><\/span>) — energia potencial gravitacional do bondinho no Morro da<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Urca,em rela\u00e7\u00e3o ao solo —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>pU<\/b><\/span><\/span><\/sub><\/span>=m.g.h<\/b><\/span><\/span><\/span>U<\/b><\/span><\/span><\/sub><\/span>=5000x10x220=11000000<\/b><\/span><\/span><\/span>=11.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u00a0J<\/b><\/span><\/span><\/span>\u00a0— — energia potencial gravitacional do bondinho no P\u00e3o de A\u00e7\u00facar, em rela\u00e7\u00e3o ao solo —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>pP<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span>\u00a0m.g.h<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=5000x10x400=<\/b><\/span><\/span><\/span><\/p>\n

20000000<\/b><\/span><\/span><\/span>=20.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u00a0J<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>varia\u00e7\u00e3o de energia potencial gravitacional<\/b><\/span><\/span><\/span>\u00a0pedida —\u00a0<\/b><\/span><\/span><\/span>\u0394E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=20.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 11.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

9.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u00a0J<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo do\u00a0<\/b><\/span><\/span><\/span>trabalho da for\u00e7a de atrito (F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>),<\/b><\/span><\/span><\/span>\u00a0contr\u00e1ria ao deslocamento, e de intensidade fornecida F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=380N num deslocamento de\u00a0<\/b><\/span><\/span><\/span>\u0394S=d=72km=72.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span>\u00a0— W<\/b><\/span><\/span><\/span>Fat<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>.d.cos180<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=380.72.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.(-1) —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

W<\/b><\/span><\/span><\/span>Fat<\/b><\/span><\/span><\/sub><\/span>= – 27 360.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0J<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>= – 2,736.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>J — em m\u00f3dulo —\u00a0<\/b><\/span><\/span><\/span>|W<\/b><\/span><\/span><\/span>Fat<\/b><\/span><\/span><\/sub><\/span>|= 2,736.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>J (energia dissipada devido ao atrito com o ar)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>No instante do salto o atleta tem\u00a0<\/b><\/span><\/span><\/span>duas velocidades<\/b><\/span><\/span><\/span>, uma\u00a0<\/b><\/span><\/span><\/span>horizontal\u00a0<\/b><\/span><\/span><\/span>de intensidade\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>=10m\/s\u00a0<\/b><\/span><\/span><\/span>e outra,\u00a0<\/b><\/span><\/span><\/span>vertical\u00a0<\/b><\/span><\/span><\/span>de intensidade\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>que produz uma energia cin\u00e9tica de E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=70% de uma energia de E=500J, ou seja, produz uma E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0vertical de E<\/b><\/span><\/span><\/span>Cv<\/b><\/span><\/span><\/sub><\/span>=(70\/100)x500=0,7×500 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>Cv<\/b><\/span><\/span><\/sub><\/span>=350J\u00a0<\/b><\/span><\/span><\/span>— essa energia cin\u00e9tica origina uma\u00a0<\/b><\/span><\/span><\/span>velocidade vertical<\/b><\/span><\/span><\/span>\u00a0de E<\/b><\/span><\/span><\/span>Cv<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 — 350=70.V<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 — 70V<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2×350 — V<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=700\/70 —\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>=\u221a(10) m\/s\u00a0<\/b><\/span><\/span><\/span>—<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

ent\u00e3o, no instante do salto voc\u00ea tem duas velocidades — uma horizontal de intensidade fornecida\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>=10m\/s\u00a0<\/b><\/span><\/span><\/span>e outra\u00a0<\/b><\/span><\/span><\/span>vertical de intensidade V<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>=\u221a(10)m\/s\u00a0<\/b><\/span><\/span><\/span>— a velocidade pedida,\u00a0<\/b><\/span><\/span><\/span>velocidade resultante<\/b><\/span><\/span><\/span>, no momento do salto corresponde \u00e0\u00a0<\/b><\/span><\/span><\/span>soma vetorial\u00a0<\/b><\/span><\/span><\/span>dessas duas velocidades — aplicando Pit\u00e1goras — V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>= 10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ (\u221a10)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=100 + 10—\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span>\u221a(110)\u2248<\/b><\/span><\/span><\/span>10<\/b><\/span><\/span><\/span>,488m\/s\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

R- E<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>For\u00e7a resultante\u00a0<\/b><\/span><\/span><\/span>sobre o autom\u00f3vel pela segunda lei de Newton — F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a=1500x(1\/12) —\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=125N<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

trabalho da for\u00e7a resultante no deslocamento de d=1350m — W=F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>.d=125×1350 — W=168 750J\u00a0<\/b><\/span><\/span><\/span>W=1,678.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>J —\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Trata-se de um sistema dissipativo que surge\u00a0quando o trabalho \u00e9 realizado por for\u00e7as\u00a0<\/b><\/span><\/span><\/span>dissipativas<\/b><\/span><\/span><\/span>(for\u00e7a de atrito, for\u00e7a de resist\u00eancia do ar, etc.) no qual, parte da energia mec\u00e2nica do sistema \u00e9 dissipada nas formas de energia t\u00e9rmica, sonora, etc. Assim a energia mec\u00e2nica do sistema, diminui e, nesse caso, o\u00a0<\/b><\/span><\/span><\/span>trabalho das for\u00e7as n\u00e3o conservativas W <\/b><\/span><\/span><\/span>(for\u00e7a de atrito, for\u00e7a de resist\u00eancia do ar, etc.) \u00e9 igual \u00e0\u00a0\u00a0<\/b><\/span><\/span><\/span>energia total dissipada<\/b><\/span><\/span><\/span>, ou seja, \u00e9 igual \u00e0\u00a0<\/b><\/span><\/span><\/span>varia\u00e7\u00e3o da energia mec\u00e2nica\u00a0<\/b><\/span><\/span><\/span>—<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mecanicafinal<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>mecanicainicial<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>for\u00e7asdissipativas<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>Inicialmente voc\u00ea pode calcular a dist\u00e2ncia \u2206S percorrida pelo bloco entre os pontos A e B pela \u00e1rea do gr\u00e1fico Vxt fornecido (figura 2) — \u00e1rea do tri\u00e2ngulo=AB=\u2206S=b.h\/2=0,5×4\/2=1m — c\u00e1lculo da altura h entre os pontos A e B<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(figura 1) — sen37<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=cateto oposto\/hipotenusa — 0,6=h\/1 — h=0,6m — para a subida —\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>for\u00e7asdissipativas=<\/b><\/span><\/span><\/sub><\/span>E<\/b><\/span><\/span><\/span>mecanicafinal<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>mecanicainicial<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>— W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0— veja pelo gr\u00e1fico da figura 2 que a velocidade inicial em A \u00e9 4m\/s e a em B \u00e9 zero — E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=mgh<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ mV<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.10.0 + 2.4<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0 + 16 — E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=16J — E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=mgh<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0+ mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.10.0,6 + 2.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 — E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=12J — W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=16 \u2013 12=4J — esse trabalho \u00e9 o mesmo tanto para a subida quanto para a descida — para a descida —\u00a0—<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>for\u00e7asdissipativas=<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mecanicafinal<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>mecanicainicial<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>— W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>\u00a0— veja pelo gr\u00e1fico da figura 2 que a velocidade inicial em A \u00e9 4m\/s e a em B \u00e9 zero — E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=mgh<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ mV<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.10.0 + 2.V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 — E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=mgh<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0+ mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.10.0,6 + 2.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 — E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=12J — W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=16 \u2013 12=4J — esse trabalho \u00e9 o mesmo tanto para a subida quanto para a descida — para a descida — W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>\u00a0— 4=12 \u2013 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=\u221a8=2\u221a2m\/s.<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Energia mec\u00e2nica no ponto A\u00a0<\/b><\/span><\/span><\/span>de onde o menino parte do\u00a0<\/b><\/span><\/span><\/span>repouso<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=m.g.h\u00a0<\/b><\/span><\/span><\/span>— energia mec\u00e2nica no ponto B\u00a0<\/b><\/span><\/span><\/span>de onde chega com\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=2m\/s\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=m.4\/<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/span>— E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=2m\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>\u00a0= E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span><\/p>\n

mgh = 2m — h=2\/g=2\/10 —\u00a0<\/b><\/span><\/span><\/span>h=0,2m.<\/b><\/span><\/span><\/span><\/p>\n

Agora ele sai do ponto A com velocidade de V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=1,5m\/s\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh<\/b><\/span><\/span><\/span>=m.1,5<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.10.0,2 — E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>= 1,125m + 2m —<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=2,125m\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0<\/b><\/span><\/span><\/span>— E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>\u00a0=E<\/b><\/span><\/span><\/span>mB\u00a0<\/b><\/span><\/span><\/sub><\/span>— 2,125m = mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 — V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=\u221a4,25 —\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=2.06m\/s<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O\u00a0<\/b><\/span><\/span><\/span>trabalho da for\u00e7a peso\u00a0<\/b><\/span><\/span><\/span>\u00e9\u00a0<\/b><\/span><\/span><\/span>negativo na subida, positivo na descida\u00a0<\/b><\/span><\/span><\/span>e nulo num deslocamento horizontal e fornecido por\u00a0<\/b><\/span><\/span><\/span>W=P.h ou W=m.g.h.<\/b><\/span><\/span><\/span><\/p>\n

O trabalho da for\u00e7a peso\u00a0<\/b><\/span><\/span><\/span>entre\u00a0<\/b><\/span><\/span><\/span>dois pontos A e B, n\u00e3o depende da trajet\u00f3ria<\/b><\/span><\/span><\/span>, mas apenas da\u00a0<\/b><\/span><\/span><\/span>altura<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(h) vertical entre A e B e \u00e9 fornecido por\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>BA<\/b><\/span><\/span><\/sub><\/span>=+P.h na descida e por W<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>= -P.h na subida por qualquer trajet\u00f3ria\u00a0<\/b><\/span><\/span><\/span>(d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0ou d<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>).<\/b><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio a altura\u00a0<\/b><\/span><\/span><\/span>h vale\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00ad<\/b><\/span><\/span><\/sub><\/span>=h\/8\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>h=<\/b><\/span><\/span><\/span>8.0,5=<\/b><\/span><\/span><\/span>4m\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>m\u00f3dulo do trabalho do peso — W=P.h=20.4=80J.<\/b><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

Din\u00e2mica Impulsiva<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo da\u00a0<\/b><\/span><\/span><\/span>velocidade V<\/b><\/span><\/span><\/span>\u00a0com que o sistema\u00a0<\/b><\/span><\/span><\/span>bloco + proj\u00e9til<\/b><\/span><\/span><\/span>\u00a0inicia a compress\u00e3o da mola pela conserva\u00e7\u00e3o da quantidade de movimento.<\/b><\/span><\/span><\/span><\/p>\n

Choque entre o proj\u00e9til e o bloco:<\/b><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span><\/span>=180g<\/b><\/span><\/span><\/span><\/span>=0,18kg\u00a0<\/b><\/span><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/span>=20g<\/b><\/span><\/span><\/span><\/span>=0,02kg\u00a0<\/b><\/span><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/span>p<\/b><\/span><\/span><\/span><\/sub><\/span>=200m\/s — V=?<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Pelo princ\u00edpio da conserva\u00e7\u00e3o da quantidade de movimento —\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span>\u00a0= Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

4 = 0,2V — V=4\/0,2 —\u00a0<\/b><\/span><\/span><\/span>V=20m\/s\u00a0<\/b><\/span><\/span><\/span>(velocidade com que o sistema (proj\u00e9til + bloco) come\u00e7a a comprimir a mola.<\/b><\/span><\/span><\/span><\/p>\n

Agora vamos utilizar o teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica no sistema (bloco-proj\u00e9til) de massa M<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=(0,18 + 0,02)=0,2kg em duas situa\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/p>\n

* Energia mec\u00e2nica de M<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>imediatamente antes\u00a0<\/b><\/span><\/span><\/span>de come\u00e7ar a comprimir a mola onde a energia\u00a0<\/b><\/span><\/span><\/span>potencial el\u00e1stica \u00e9 nula\u00a0<\/b><\/span><\/span><\/span>(mola descomprimida) s\u00f3 possuindo\u00a0<\/b><\/span><\/span><\/span>energia cin\u00e9tica\u00a0<\/b><\/span><\/span><\/span>— E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=M<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2= 0,2.(20)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=40J<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

* Energia mec\u00e2nica de M<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>ap\u00f3s a mola estar comprimida de\u00a0<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>e M<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>\u00a0parado (V<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=0) onde a\u00a0<\/b><\/span><\/span><\/span>energia<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>cin\u00e9tica \u00e9 nula\u00a0<\/b><\/span><\/span><\/span>e o sistema s\u00f3 possui\u00a0<\/b><\/span><\/span><\/span>energia potencial el\u00e1stica\u00a0<\/b><\/span><\/span><\/span>— E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0<\/b><\/span><\/span><\/span>— E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Sendo o sistema conservativo —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>\u00a0= E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>— 40 = 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0— x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=40\/10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0— x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=0,04 —\u00a0<\/b><\/span><\/span><\/span>x=0,2m<\/b><\/span><\/span><\/span><\/p>\n

X=20cm.<\/b><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Depois de unidos, o carrinho com compras de massa 30kg e o vazio de massa 20kg se movem<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

engatados, com massa total m=50kg, e com sua velocidade passando de V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0para V=0 (para), percorrendo \u0394S=6m em t=4s.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da velocidade\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0do conjunto:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Pelo enunciado,\u00a0<\/b><\/span><\/span><\/span>antes\u00a0<\/b><\/span><\/span><\/span>do acoplamento o carrinho de\u00a0<\/b><\/span><\/span><\/span>massa 30kg e velocidade V\u2019\u00a0<\/b><\/span><\/span><\/span>se move em dire\u00e7\u00e3o ao carrinho de massa 20kg que estava parado e, imediatamente\u00a0<\/b><\/span><\/span><\/span>ap\u00f3s o acoplamento<\/b><\/span><\/span><\/span>\u00a0ambos, de\u00a0<\/b><\/span><\/span><\/span>massa 50kg\u00a0<\/b><\/span><\/span><\/span>se movem\u00a0<\/b><\/span><\/span><\/span>unidos com velocidade V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=3m\/s.\u00a0<\/b><\/span><\/span><\/span>Aplicando o princ\u00edpio da conserva\u00e7\u00e3o da quantidade de movimento:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Antes — Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span>=30V\u2019 + 20.0 — Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span>=30V\u2019<\/b><\/span><\/span><\/span><\/p>\n

Depois — Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>=(30 + 20).3 — Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>=150kgm\/s<\/b><\/span><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span>\u00a0= Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>\u00a0— 30V\u2019 = 150 —\u00a0<\/b><\/span><\/span><\/span>V\u2019=5m\/s<\/b><\/span><\/span><\/span><\/p>\n

R- A<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como a\u00a0<\/b><\/span><\/span><\/span>massa M<\/b><\/span><\/span><\/span>ext<\/b><\/span><\/span><\/sub><\/span>\u00a0do extintor n\u00e3o foi fornecida voc\u00ea deve avaliar uma massa que esteja dentro de um<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>intervalo razo\u00e1vel, por exemplo,\u00a0<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/span>ext<\/b><\/span><\/span><\/sub><\/span>=20kg<\/b><\/span><\/span><\/span>\u00a0— voc\u00ea pode utilizar a conserva\u00e7\u00e3o da quantidade de movimento para avaliar a velocidade pedida j\u00e1 que o extintor (sistema) est\u00e1 isolado —\u00a0<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/span>\u00a0de lan\u00e7ar o\u00a0<\/b><\/span><\/span><\/span>jato de CO<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, o sistema est\u00e1 em repouso (<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=0<\/b><\/span><\/span><\/span>) — depois que foi lan\u00e7ado a massa\u00a0<\/b><\/span><\/span><\/span>m=50g=0,05kg<\/b><\/span><\/span><\/span>\u00a0com velocidade de\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>CO2<\/b><\/span><\/span><\/sub><\/span>=20m\/s<\/b><\/span><\/span><\/span>, o extintor de massa\u00a0<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/span>ext<\/b><\/span><\/span><\/sub><\/span>=20kg<\/b><\/span><\/span><\/span>\u00a0recua com velocidade\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>ext<\/b><\/span><\/span><\/sub><\/span>\u00a0tal que:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0Em todo\u00a0<\/b><\/span><\/span><\/span>choque el\u00e1stico (perfeitamente el\u00e1stico)\u00a0<\/b><\/span><\/span><\/span>se os corpos possuirem a\u00a0<\/b><\/span><\/span><\/span>mesma massa\u00a0<\/b><\/span><\/span><\/span>eles\u00a0<\/b><\/span><\/span><\/span>trocam suas velocidades.<\/b><\/span><\/span><\/span><\/p>\n

Ap\u00f3s o choque a\u00a0<\/b><\/span><\/span><\/span>bola incidente para\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>a bola alvo segue com velocidade de 2m\/s.<\/b><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Estando o\u00a0<\/b><\/span><\/span><\/span>garoto em repouso dentro do barco<\/b><\/span><\/span><\/span>, o\u00a0<\/b><\/span><\/span><\/span>sistema<\/b><\/span><\/span><\/span>\u00a0est\u00e1 em\u00a0<\/b><\/span><\/span><\/span>equil\u00edbrio est\u00e1tico na vertical\u00a0<\/b><\/span><\/span><\/span>e a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a resultante\u00a0\"\"sobre ele \u00e9 nula.<\/b><\/span><\/span><\/span><\/p>\n

For\u00e7as verticais\u00a0<\/b><\/span><\/span><\/span>que agem sobre\u00a0<\/b><\/span><\/span><\/span>sistema<\/b><\/span><\/span><\/span>\u00a0(garoto mais barco) de massa\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=50 + 150=200kg:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Agora voc\u00ea vai trabalhar\u00a0<\/b><\/span><\/span><\/span>na horizontal<\/b><\/span><\/span><\/span>, utilizando o\u00a0<\/b><\/span><\/span><\/span>princ\u00edpio da conserva\u00e7\u00e3o da\u00a0<\/b><\/span><\/span><\/span>quantidade de movimento do sistema garoto-barco:<\/b><\/span><\/span><\/span><\/p>\n

Antes do salto:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Depois do salto:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Pelo\u00a0<\/b><\/span><\/span><\/span>princ\u00edpio da conserva\u00e7\u00e3o da quantidade de movimento:<\/b><\/span><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span>\u00a0= Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>— 0 = 45 \u2013 150V<\/b><\/span><\/span><\/span>rec<\/b><\/span><\/span><\/sub><\/span>\u00a0— V<\/b><\/span><\/span><\/span>rec<\/b><\/span><\/span><\/sub><\/span>=45\/150 —\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>rec<\/b><\/span><\/span><\/sub><\/span>=0,3m\/s<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Inicialmente voc\u00ea tem o n\u00facleo de pol\u00f4nio inicialmente em repouso (V<\/b><\/span><\/span><\/span>Po<\/b><\/span><\/span><\/sub><\/span>=0) e, assim a quantidade de<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

movimento inicial \u00e9 nula pois, Q<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>po<\/b><\/span><\/span><\/sub><\/span>=204u.0 — Q<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=0 — depois o n\u00facleo de pol\u00f4nio se transforma em<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

um n\u00facleo de chumbo com velocidade de m\u00f3dulo V<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\u00a0e em uma part\u00edcula alfa com velocidade de m\u00f3dulo V<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span>\u00a0— a quantidade de movimento do sistema depois da transmuta\u00e7\u00e3o ser\u00e1 —\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span>200u.V<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span><\/p>\n

+ 4u.V<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span>— pelo Princ\u00edpio da Conserva\u00e7\u00e3o da quantidade de movimento — Q<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0— 0 = 200u.V<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 4u.V<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span><\/p>\n

4u.V<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span>\u00a0= – 200uV<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span>\u00a0= – 50 V<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\u00ad<\/b><\/span><\/span><\/span>— energia cin\u00e9tica do n\u00facleo de chumbo — E<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\/2 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>=200u.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\/2 (I)\u00a0<\/b><\/span><\/span><\/span>— energia cin\u00e9tica da part\u00edcula alfa — E<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span>\/2 — E<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span>=4u.(-50V<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = 4ux 2500. V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\/2<\/b><\/span><\/span><\/span>— E<\/b><\/span><\/span><\/span>\u03b1\u00a0<\/b><\/span><\/span><\/sub><\/span>=10000u.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\/2 (II)\u00a0<\/b><\/span><\/span><\/span>— (I)\/(II) — E<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\/E<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span>=(200u.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\/2)\/<\/b><\/span><\/span><\/span>(<\/b><\/span><\/span><\/span>10000u.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\/2) — E<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\/E<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span>\u00a0=<\/b><\/span><\/span><\/span><\/p>\n

(200u.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\/2)x(2\/10000u.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>) — E<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>\/E<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span>\u00a0=200\/10000=1\/50 — E<\/b><\/span><\/span><\/span>Pb<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/sub><\/span>\/50 —\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

08-\u00a0<\/b><\/span><\/span><\/span>a) Teorema do Impulso: “O\u00a0<\/b><\/span><\/span><\/span>impulso da for\u00e7a resultante\u00a0<\/b><\/span><\/span><\/span>de um sistema de for\u00e7as que age sobre um corpo\u00a0<\/b><\/span><\/span><\/span>\u00e9 igual \u00e0 varia\u00e7\u00e3o da quantidade de movimento do corpo<\/b><\/span><\/span><\/span>“<\/b><\/span><\/span><\/span><\/p>\n

\"\"\"\"\"\"\"\"<\/p>\n

\u00a0<\/span>Pelo\u00a0<\/b><\/span><\/span><\/span>enunciado<\/b><\/span><\/span><\/span>, o m\u00f3dulo da velocidade da bola de massa\u00a0<\/b><\/span><\/span><\/span>m=300kg\u00a0<\/b><\/span><\/span><\/span>no ponto B, imediatamente antes de atingir a estrutura do pr\u00e9dio \u00e9 de\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=2m\/s<\/b><\/span><\/span><\/span>, o choque dura\u00a0<\/b><\/span><\/span><\/span>\u0394t=0,02s\u00a0<\/b><\/span><\/span><\/span>e ap\u00f3s esse intervalo de tempo a bola para instantaneamente\u00a0<\/b><\/span><\/span><\/span>V=0\u00a0<\/b><\/span><\/span><\/span>— aplicando o teorema do impulso:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Se o enunciado afirma que no ponto A a\u00a0<\/b><\/span><\/span><\/span>velocidade da bola \u00e9 nula<\/b><\/span><\/span><\/span>, ele est\u00e1 afirmando que a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a resultante centr\u00edpeta nesse ponto tamb\u00e9m \u00e9 nula<\/b><\/span><\/span><\/span>, pois F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=m.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=0 — para que isso ocorra, a intensidade da componente normal (radial) do peso\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=1440N\u00a0<\/b><\/span><\/span><\/span>deve anular a for\u00e7a de tra\u00e7\u00e3o no fio que \u00e9 pedida<\/b><\/span><\/span><\/span>, ou seja, F<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span>T=1440N\u00a0<\/b><\/span><\/span><\/span>— observa\u00e7\u00e3o: \u00c9 a componente tangencial do peso P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=Psen\u03b8 que faz a bola descer.<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Observe que, nesse deslocamento de\u00a0<\/b><\/span><\/span><\/span>d=20cm=0,2m\u00a0<\/b><\/span><\/span><\/span>fornecido o exerc\u00edcio pede o\u00a0<\/b><\/span><\/span><\/span>trabalho da for\u00e7a de atrito\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0de intensidade F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcN=\u03bcP=\u03bcmg=0,4.3.10 —\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=12N<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

O trabalho da for\u00e7a de atrito\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>Fat<\/b><\/span><\/span><\/sub><\/span>, onde\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=12N<\/b><\/span><\/span><\/span>, no deslocamento de d=0,2m e\u00a0<\/b><\/span><\/span><\/span>\u03b1=180<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00e9 fornecido por<\/b><\/span><\/span><\/span><\/p>\n

W<\/b><\/span><\/span><\/span>Fat<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>.d.cos\u03b1<\/b><\/span><\/span><\/span>=12.0,2.cos180<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=1,4x(-1) —\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>Fat<\/b><\/span><\/span><\/sub><\/span>= – 2,4J<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0a)<\/b><\/span><\/span><\/span>\u00a0C\u00e1lculo da velocidade\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>1i<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>com que a part\u00edcula de massa\u00a0<\/b><\/span><\/span><\/span>M que partiu de P\u00a0<\/b><\/span><\/span><\/span>com velocidade\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=0\u00a0<\/b><\/span><\/span><\/span>e chega em que\u00a0<\/b><\/span><\/span><\/span>Q no in\u00edcio da pista horizontal<\/b><\/span><\/span><\/span>, pela conserva\u00e7\u00e3o da energia mec\u00e2nica (figura 1):<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Energia mec\u00e2nica em\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>— E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span>\u00a0= MV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>+ Mgh=M.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>+ M.10.1,25 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span>\u00a0= 12,5M.<\/b><\/span><\/span><\/span><\/p>\n

Energia mec\u00e2nica em\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>— E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>\u00a0= MV<\/b><\/span><\/span><\/span>1i<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>+ Mgh= MV<\/b><\/span><\/span><\/span>1i<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>+ M.10.0 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>\u00a0= MV<\/b><\/span><\/span><\/span>1i<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span>= E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>— 12,5M = MV<\/b><\/span><\/span><\/span>1i<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>— V<\/b><\/span><\/span><\/span>1i<\/b><\/span><\/span><\/sub><\/span>=\u221a25 —\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>1i<\/b><\/span><\/span><\/sub><\/span>=5m\/s\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>como n\u00e3o existe atrito \u00e9 com essa velocidade com que ela chega at\u00e9 a part\u00edcula 2 imediatamente antes da colis\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

b)\u00a0<\/b><\/span><\/span><\/span>Observe na figura abaixo a situa\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>antes da colis\u00e3o e depois da colis\u00e3o\u00a0<\/b><\/span><\/span><\/span>onde ser\u00e1 calculada a quantidade de movimento do sistema antes e depois da colis\u00e3o<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Antes — Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span>=M.5 + m.0\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span>=5M\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>depois — Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>=M.4,5 + m.V<\/b><\/span><\/span><\/span>2f<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>=4,5M + V<\/b><\/span><\/span><\/span>2f<\/b><\/span><\/span><\/sub><\/span>m.<\/b><\/span><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span>\u00a0= Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>— 5M = 4,5M + V<\/b><\/span><\/span><\/span>2f<\/b><\/span><\/span><\/sub><\/span>m — 0,5M = V<\/b><\/span><\/span><\/span>2f<\/b><\/span><\/span><\/sub><\/span>m —\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>2f<\/b><\/span><\/span><\/sub><\/span>=0,5M\/m.<\/b><\/span><\/span><\/span><\/p>\n

Sendo a\u00a0<\/b><\/span><\/span><\/span>colis\u00e3o el\u00e1stica\u00a0<\/b><\/span><\/span><\/span>o coeficiente de restitui\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>e \u00e9 igual a 1\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>e=m\u00f3dulo da velocidade relativa depois\/m\u00f3dulo da velocidade relativa antes\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>1 = V<\/b><\/span><\/span><\/span>2f<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 4,5\/5 \u2013 0\u00a0<\/b><\/span><\/span><\/span>— 5 = V<\/b><\/span><\/span><\/span>2f<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 4,5 —\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>2f<\/b><\/span><\/span><\/sub><\/span>\u00a0= 9,5m\/s<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Com essa velocidade\u00a0<\/b><\/span><\/span><\/span>inicial de V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=9,5m\/s a\u00a0<\/b><\/span><\/span><\/span>part\u00edcula\u00a0<\/b><\/span><\/span><\/span>m\u00a0<\/b><\/span><\/span><\/span>atinge a\u00a0<\/b><\/span><\/span><\/span>altura m\u00e1xima em S onde V<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>=0\u00a0<\/b><\/span><\/span><\/span>—<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Energia mec\u00e2nica em R\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mR<\/b><\/span><\/span><\/sub><\/span>=m.(9,5)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.10.0<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mR<\/b><\/span><\/span><\/sub><\/span>=45,125m (I).\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>energia mec\u00e2nica em S\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mS<\/b><\/span><\/span><\/sub><\/span>=m.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>+ m.10.h<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mS<\/b><\/span><\/span><\/sub><\/span>=10mh<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>. (II).<\/b><\/span><\/span><\/span><\/p>\n

Igualando\u00a0<\/b><\/span><\/span><\/span>(I) com (II)\u00a0<\/b><\/span><\/span><\/span>— 45,125m = 10mh<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=4,5125m<\/b><\/span><\/span><\/span><\/p>\n

Primeira lei de Newton<\/b><\/span><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span> \u00a0<\/b><\/span><\/span><\/span>Princ\u00edpio\u00a0da\u00a0In\u00e9rcia\u00a0(Primeira Lei de Newton)<\/b><\/span><\/span><\/span><\/p>\n

\u201cTodo corpo que esteja em repouso ou em movimento retil\u00edneo e uniforme (movendo-se em trajet\u00f3ria reta com velocidade\u00a0 vetorial constante), tende a continuar nestes estados se a for\u00e7a resultante que age sobre ele for nula\u201d.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\"\"<\/span><\/p>\n

Quando a for\u00e7a resultante sobre ele for nula, se ele estiver em repouso, tende a continuar em repouso e se estiver em movimento com velocidade\u00a0\"\"\u00a0tender\u00e1 a se mover em linha reta com velocidade\u00a0\"\".\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- C.<\/b><\/span><\/span><\/span><\/p>\n

30\u00a0–<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O momento de in\u00e9rcia refere-se a uma resist\u00eancia \u00e0 mudan\u00e7a no movimento rotacional de um corpo\u00a0 —\u00a0 \u00e9 evidente que quanto mais pr\u00f3ximo o centro massa do corpo estiver do eixo de rota\u00e7\u00e3o, menor ser\u00e1 o momento de in\u00e9rcia, e quanto mais afastada a massa estiver do eixo de rota\u00e7\u00e3o, maior ser\u00e1 seu momento de in\u00e9rcia\u00a0 —\u00a0\u00a0R- B.<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>Isso ocorre porque voc\u00ea est\u00e1 na atmosfera que, por sua vez, por in\u00e9rcia, est\u00e1 acompanhando a Terra\u00a0 —\u00a0 isso se deve ao fato de existir uma for\u00e7a que atrai todos os corpos em dire\u00e7\u00e3o ao centro da Terra chamada for\u00e7a gravitacional ou for\u00e7a peso\u00a0 —\u00a0\u00a0R- D.<\/b><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Segunda lei de Newton<\/b><\/span><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>–\u00a0<\/b><\/span><\/span><\/span>a) Quando a caixa \u00e9 arrastada com velocidade constante, pelo princ\u00edpio da in\u00e9rcia a for\u00e7a resultante sobre ela \u00e9 nula (equil\u00edbrio din\u00e2mico)\u00a0 —\u00a0 nesse caso, a intensidade da for\u00e7a aplicada pelo fio deve anular a for\u00e7a de atrito\u00a0 —\u00a0 F = F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F=1N.<\/b><\/span><\/span><\/span><\/p>\n

b) Quando a intensidade da for\u00e7a aplicada pelo fio \u00e9 maior que F=1N, a caixa entra em movimento acelerado, mas a for\u00e7a de atrito continua sendo de F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=1N, pois, dentro de certos limites, independe da velocidade que, no caso, vai aumentando\u00a0 —\u00a0 pelo gr\u00e1fico, a for\u00e7a (F) de tra\u00e7\u00e3o no fio\u00a0 \u00e9 proporcional \u00e0 acelera\u00e7\u00e3o (a)\u00a0 —\u00a0\u00a0 quando, por exemplo, a acelera\u00e7\u00e3o for de 15m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>, a for\u00e7a ser\u00e1 de 4N\u00a0 —\u00a0 a for\u00e7a resultante sobre a caixa vale\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/sub><\/span>=F \u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=4 \u2013 1=3N\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Princ\u00edpio fundamental da Din\u00e2mica\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a\u00a0 —\u00a0 3=m.15\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0m=0,2 kg.<\/b><\/span><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>Na figura abaixo as for\u00e7as\u00a0\"\"\u00a0foram projetadas na vertical e na horizontal\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1x<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>cos\u03b1\u00a0 —\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span><\/span>1y<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>sen\u03b1\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2x<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>sen\u03b2\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2y<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>cos\u03b2.<\/b><\/span><\/span><\/span><\/p>\n

01. Correta\u00a0 —\u00a0 supondo que o corpo se mova para a direita\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>1x<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 F<\/b><\/span><\/span><\/span>2x<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>cos\u03b1 \u2013 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>sen\u03b2 \u2013 \u03bcN.<\/b><\/span><\/span><\/span><\/p>\n

02. Correta\u00a0 —\u00a0\u00a0\"\".<\/b><\/span><\/span><\/span><\/p>\n

04. Falsa\u00a0 —\u00a0 se a resultante do sistema for nula, o poder\u00e1 estar em repouso(equil\u00edbrio est\u00e1tico ou em movimento retil\u00edneo e uniforme (princ\u00edpio da in\u00e9rcia).<\/b><\/span><\/span><\/span><\/p>\n

08. Falsa\u00a0 —\u00a0 se ele estiver subindo\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>1y<\/b><\/span><\/span><\/sub><\/span>\u00a0+ F<\/b><\/span><\/span><\/span>2y<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 P\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>sen\u03b1 + F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>cos\u03b2 \u2013 P.<\/b><\/span><\/span><\/span><\/p>\n

16. Falsa\u00a0 —\u00a0 como existe for\u00e7a de atrito, ele deve atuar no sistema.<\/b><\/span><\/span><\/span><\/p>\n

R- (01 + 02)=03.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Terceira lei de Newton<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0Entre os animais que desenvolveram o tipo de v\u00f4o<\/b><\/span><\/span><\/span>\u00a0p\u00e1ra-quedismo<\/b><\/span><\/span><\/span>\u00a0est\u00e1 o sapo voador de Born\u00e9u,\u00a0Rhacophorus dulitensis\u00a0 —\u00a0 o p\u00e1ra-quedas \u00e9 formado atrav\u00e9s de membranas presentes nas patas\u00a0 —\u00a0 quando os animais descem das \u00e1rvores, age uma for\u00e7a aerodin\u00e2mica paralela a dire\u00e7\u00e3o do ar que passa por eles\u00a0 —\u00a0 essa for\u00e7a pode ser chamada de<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0for\u00e7a de resist\u00eancia\u00a0 — \u00a0o animal atinge uma velocidade constante quando esta for\u00e7a total contrabalancear o seu peso.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0O animal\u00a0<\/b><\/span><\/span><\/span>planador\u00a0<\/b><\/span><\/span><\/span>se desloca no ar em movimento descendente, sem realizar trabalho\u00a0 —\u00a0 sua trajet\u00f3ria \u00e9 retil\u00ednea\u00a0 —\u00a0 como animal planador podemos citar o esquilo voador (Glaucomys volans)\u00a0 —\u00a0 possui geralmente uma pele que se<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 possui geralmente uma pele que se estende das patas dianteiras at\u00e9 as traseiras, denominada pat\u00e1gio.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0V\u00f4os propulsionados\u00a0 —\u00a0 \u00e9 o tipo de v\u00f4o realizado por grande parte dos animais que voam e \u00e9 resultado da movimenta\u00e7\u00e3o de m\u00fasculos\u00a0 —\u00a0 devido a exist\u00eancia da for\u00e7a de arrastamento, um animal n\u00e3o consegue planar horizontalmente por muito tempo, devido a perda de velocidade\u00a0 —\u00a0 para voar horizontalmente com velocidade constante,uma for\u00e7a de impulso deve ser fornecida com dire\u00e7\u00e3o igual, sentido contr\u00e1rio e mesma intensidade que a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

for\u00e7a de arrastamento\u00a0 —\u00a0 essa pot\u00eancia \u00e9 fornecida pelos m\u00fasculos que impulsionam e movimentam as asas.<\/b><\/span><\/span><\/span><\/p>\n

R- B.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Exerc\u00edcios sobre conceitos das tr\u00eas leis de Newton<\/b><\/span><\/span><\/span><\/p>\n

57-<\/b><\/span><\/span><\/span> <\/b>–\u00a0Dados\u00a0 —\u00a0 m=1200kg\u00a0 —\u00a0 Vo=108\/3,6=30m\/s\u00a0 —\u00a0 V=0 (p\u00e1ra)\u00a0 —\u00a0 \u2206S=200m\u00a0 —\u00a0 c\u00e1lculo da acelera\u00e7\u00e3o de retardamento pela equa\u00e7\u00e3o de Torricelli\u00a0 —\u00a0 V2\u00a0= Vo2\u00a0\u2013 2.a.\u2206S\u00a0 —\u00a0 02\u00a0= 302\u00a0\u2013 2.a.200\u00a0 —\u00a0 a=900\/400=2.25m\/s2\u00a0 —\u00a0 for\u00e7a resultante durante a frenagem\u00a0 —\u00a0 FR=m.a=1200.2,25\u00a0 —\u00a0 FR=2700N\u00a0 —\u00a0 essa for\u00e7a \u00e9 a for\u00e7a de atrito que est\u00e1 segurando o carro\u00a0 —\u00a0 Fat=2700N\u00a0 —\u00a0 trabalho da for\u00e7a de atrito (realizado pelos freios)\u00a0 —\u00a0 WFat=Fat.d.cos180o=2700.200.(-1)\u00a0 —\u00a0 WFat= – 540000J=5,4.105\u00a0J\u00a0 —\u00a0\u00a0R- B.\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

\u00a058-<\/b><\/span><\/span><\/span> \u00a0<\/b><\/span><\/span><\/span>A intensidade da for\u00e7a resultante sobre os dois lutadores juntos, de massa M=200 + 200=400kg \u00e9 de F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=1000 \u2013 950=950N\u00a0 —\u00a0 c\u00e1lculo da acelera\u00e7\u00e3o\u00a0\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=M.a\u00a0 —\u00a0 50=400.a\u00a0 —\u00a0 a=0,125m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

como eles se encontram no centro do c\u00edrculo, para sair do mesmo eles devem percorrer R=\u2206S=2m\u00a0 —\u00a0 \u2206S=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + a.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 2=0.t + 1,25.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=4\/0,125\u00a0 —\u00a0 t=\u221a32\u00a0 —\u00a0 t\u22485,6s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Tipos de for\u00e7a<\/b><\/span><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Colocando as for\u00e7as que agem sobre cada cinta, que correspondem ao acr\u00e9scimo da for\u00e7a de contato entre a plataforma e o pneu, que no caso \u00e9 a for\u00e7a (T) que traciona ou tensiona o fio em cada <\/b><\/span><\/span><\/span><\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\"\"<\/span><\/p>\n

lado do pneu\u00a0\u00a0 —\u00a0\u00a0\u00a0 decompondo essas for\u00e7as na horizontal e na vertical (observe a seq\u00fc\u00eancia da figura)\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=T.sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=0,5.T\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=T.cos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=(\u221a3\/2).T\u00a0 —\u00a0observe que, na horizontal as componentes (T<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>) se anulam\u00a0 —\u00a0 na vertical sobram, para baixo as duas for\u00e7as T<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0que pressionam a plataforma para baixo e que, somadas, fornecem\u00a0 —\u00a0 F = 2T<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=2x(\u221a3\/2)T\u00a0\u00a0 —\u00a0 F =\u221a3.T\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D.\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O exerc\u00edcio quer o diagrama das for\u00e7as que as dobradi\u00e7as aplicam na porta:<\/b><\/span><\/span><\/span><\/p>\n

Na horizontal<\/b><\/span><\/span><\/span>: ao girar, a porta traciona\u00a0 (puxa) para a direita a dobradi\u00e7a superior e esta, pelo princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o reage na porta com uma for\u00e7a de mesma intensidade, mesma dire\u00e7\u00e3o, mas sentido contr\u00e1rio,<\/b><\/span><\/span><\/span>\u00a0tracionando-a (puxando-a) para a esquerda com uma for\u00e7a\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 analogamente ao girar, a porta comprime\u00a0 (empurra) para a esquerda a dobradi\u00e7a superior e esta, pelo princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o reage na porta com uma for\u00e7a de mesma intensidade, mesma dire\u00e7\u00e3o mas sentido contr\u00e1rio,<\/b><\/span><\/span><\/span>\u00a0comprimindo-a (empurrando-a) para a direita com uma for\u00e7a\u00a0\"\"<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Na vertical,<\/b><\/span><\/span><\/span>\u00a0devido ao seu peso, a porta aplica em cada dobradi\u00e7a uma for\u00e7a puxando-as para baixo e, elas reagem\u00a0<\/b><\/span><\/span><\/span>sobre a porta com for\u00e7as\u00a0\"\"\u00a0e\u00a0\"\", verticais e para cima.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A soma vetorial dessas for\u00e7as, fornece as for\u00e7as resultantes\u00a0\"\"\u00a0e\u00a0\"\"que cada dobradi\u00e7a exerce sobre a porta (figura)\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Peso e massa<\/b><\/span><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O peso do corpo em Marte \u00e9 o produto de sua massa que \u00e9 a mesma em qualquer lugar ou planeta pela acelera\u00e7\u00e3o da gravidade na superf\u00edcie do planeta Marte\u00a0 —\u00a0 P=m.g=3,10×3,69=11,439N\u00a0 —\u00a0 com tr\u00eas algarismos significativos\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

P=11,4N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Com duas pessoas no carro a deforma\u00e7\u00e3o do sistema de molas \u00e9 x=2cm=2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m e\u00a0 a for\u00e7a que deformadora \u00e9 o peso das duas pessoas\u00a0 —\u00a0 P=2mg=2.60.10=12.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0 esse peso corresponde \u00e0 for\u00e7a el\u00e1stica\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>P=kx\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

12.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=k.2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 k=6.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>N\/m (constante el\u00e1stica do sistema de molas que \u00e9 constante independente da for\u00e7a deformadora)\u00a0 —\u00a0 com\u00a0 4 pessoas a deforma\u00e7\u00e3o \u00e9 x\u2019 e o peso \u00e9 P\u2019=4mg=4.70.10=28.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0\u00a0 P\u2019=kx\u2019\u00a0 —\u00a0 28.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=6.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>x\u2019\u00a0 —\u00a0 x\u2019=4,66.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m=4,66cm\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Leis de Newton do MUV<\/b><\/span><\/span><\/span><\/p>\n

48-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) C\u00e1lculo da acelera\u00e7\u00e3o do m\u00f3vel no trecho AB\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a.t\u00a0 —\u00a0 30 = 0 + a.10\u00a0 —\u00a0 a=3m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 como n\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

tem atrito a for\u00e7a resultante vale F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a=1200.3\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=3600N ou\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=3,6.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>N.<\/b><\/span><\/span><\/span><\/p>\n

b)\u00a0 Parcela do peso na dire\u00e7\u00e3o do movimento que diminui a velocidade do m\u00f3vel durante a subida e que \u00e9 a for\u00e7a resultante sobre ele\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=Psen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=m,g.sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=1200.10.1\/2\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=6000N\u00a0 —\u00a0 acelera\u00e7\u00e3o de subida\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

m.a\u00a0 —\u00a0 6000=1200.a\u00a0 —\u00a0 a=5m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(retardamento)\u00a0 —\u00a0 \u2206S=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t + a.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 o tempo de subida \u00e9 t=(15 \u2013 10)=5s\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\u00a0\u2206S=30.5 + (-5).5<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0 \u2206S=87,5m.<\/b><\/span><\/span><\/span><\/p>\n

No trecho BC a \u00fanica for\u00e7a que age sobre ele, na dire\u00e7\u00e3o do movimento \u00e9 a parcela do peso (P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=P.sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>) e, como n\u00e3o existe atrito, o sistema \u00e9 conservativo\u00a0 —\u00a0 assim, a energia mec\u00e2nica total do m\u00f3vel em B (somente a cin\u00e9tica) \u00e9<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

igual \u00e0 energia mec\u00e2nica total do m\u00f3vel em C\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>= mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=1200.30<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=540000J ou\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Plano inclinado sem atrito<\/b><\/span><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo o atrito com a plataforma da rampa desprezado a \u00fanica for\u00e7a respons\u00e1vel pela descida das pessoas \u00e9 a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

parcela do peso\u00a0\"\"\u00a0\u00a0paralela \u00e0 superf\u00edcie da plataforma, que \u00e9 igual \u00e0 for\u00e7a resultante\u00a0\"\"\u00a0— P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=Psen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=m.g.1\/2=m.10\/2\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

5m (I)\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a (II)\u00a0 —\u00a0 (I)=(II)\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 5m = ma\u00a0 —\u00a0 a=5m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(acelera\u00e7\u00e3o de descida das pessoas)\u00a0 —\u00a0 c\u00e1lculo da dist\u00e2ncia percorrida \u2206S pela rampa\u00a0 —\u00a0 sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=31,25\/\u2206S\u00a0 —\u00a0 0,5=31,25\/\u2206S\u00a0 —\u00a0 \u2206S=62,5m\u00a0 —\u00a0 \u2206S=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t +<\/b><\/span><\/span><\/span><\/p>\n

At<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>—\u00a0 pelo enunciado, partiu do repouso V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 62,5=0.t + 5t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=125\/5\u00a0 —\u00a0 t=\u221a25\u00a0 —\u00a0 t=5s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A.<\/b><\/span><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo o atrito desprezado, o sistema \u00e9 conservativo\u00a0 —\u00a0 usando o teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0 energia mec\u00e2nica em P\u00a0 — EmP\u00a0= mVP2\/2 + m.g.h = m.32\/2 + m.10.0,8\u00a0 —\u00a0 EmP\u00a0= 12,5m\u00a0 —\u00a0 Energia mec\u00e2nica em Q\u00a0 —\u00a0 EmQ\u00a0= mVQ2\/2 + m.g.h = m.VQ2\/2 + m.10.0\u00a0 —\u00a0 EmQ=mVQ2\/2\u00a0 —\u00a0 EmP\u00a0= EmQ\u00a0 —\u00a0 12,5m = mVQ2\/2\u00a0 —\u00a0 VQ=\u221a(25)\u00a0 —\u00a0\u00a0VQ=5,0 m\/s.<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I. Verdadeira\u00a0 —\u00a0 em (I) ela est\u00e1 em repouso, equil\u00edbrio est\u00e1tico e F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=0 o que implica que T=P=mg\u00a0 —\u00a0 em (II) ela<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

est\u00e1 em movimento circular surgindo sobre ela uma for\u00e7a resultante centr\u00edpeta (dirigida para o centro da circunfer\u00eancia) e de intensidade F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R, com F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=T \u2013 P\u00a0 —\u00a0 T=P + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 observe, nesse caso, que T>P.<\/b><\/span><\/span><\/span><\/p>\n

II. Verdadeira\u00a0 —\u00a0 As for\u00e7as que agem sobre um corpo apoiado sobre um plano inclinado s\u00e3o seu peso\"\", vertical e para baixo e a for\u00e7a normal\u00a0\"\", perpendicular \u00e0 superf\u00edcie de contato entre o bloco e o plano.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0391 \u00e9 o \u00e2ngulo de inclina\u00e7\u00e3o do plano. Como\u00a0\"\"\u00a0e\u00a0\"\"\u00a0n\u00e3o tem a mesma dire\u00e7\u00e3o, vamos decompor o peso\u00a0\"\"\u00a0em duas parcelas:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0parcela do peso paralela \u00e0 superf\u00edcie do plano inclinado (respons\u00e1vel pela tentativa de descida do bloco).<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0parcela do peso perpendicular \u00e0 superf\u00edcie do plano inclinado (for\u00e7a que comprime o bloco contra o plano)<\/b><\/span><\/span><\/span><\/p>\n

Observe que P<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 menor que o peso, pois \u00e9 parcela do mesmo.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

III. Verdadeira\u00a0 —\u00a0 se a resist\u00eancia do ar \u00e9 nula, a acelera\u00e7\u00e3o de queda \u00e9 a da gravidade g, que \u00e9 constante, pois elas chegam juntas ao solo, quando largadas de mesma altura\u00a0 —\u00a0 P=mg\u00a0 —\u00a0 g=P\/m=constante.<\/b><\/span><\/span><\/span><\/p>\n

R- C.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Lei de Hooke<\/b><\/span><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Com duas pessoas no carro a deforma\u00e7\u00e3o do sistema de molas \u00e9 x=2cm=2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m e\u00a0 a for\u00e7a que deformadora \u00e9 o peso das duas pessoas\u00a0 —\u00a0 P=2mg=2.60.10=12.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0 esse peso corresponde \u00e0 for\u00e7a el\u00e1stica\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>P=kx\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

12.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=k.2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 k=6.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>N\/m (constante el\u00e1stica do sistema de molas que \u00e9 constante independente da for\u00e7a deformadora)\u00a0 —\u00a0 com\u00a0 4 pessoas a deforma\u00e7\u00e3o \u00e9 x\u2019 e o peso \u00e9 P\u2019=4mg=4.70.10=28.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0\u00a0 P\u2019=kx\u2019\u00a0 —\u00a0 28.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=6.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>x\u2019\u00a0 —\u00a0 x\u2019=4,66.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m=4,66cm\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D.<\/b><\/span><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Determinando\u00a0 o valor da constante el\u00e1stica k da mola pela figura 1\u00a0 —\u00a0 quando x=4,0cm=4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m, F=20N\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=kx\u00a0 —\u00a0 20=k.4.10<\/b><\/span><\/span><\/span>-2\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>—\u00a0 k=5.10<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>N\/m\u00a0 —\u00a0 for\u00e7a aplicada pela mola quando o sistema est\u00e1 em equil\u00edbrio e a mola deformada de x=6,4cm=6,4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 F=kx=5.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.6,4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=32N\u00a0 —\u00a0 observe a seq\u00fc\u00eancia das figuras abaixo onde<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

foram colocadas todas as for\u00e7as no ponto de interse\u00e7\u00e3o dos fios (I)\u00a0 —\u00a0 em (II) foram decompostas as for\u00e7as inclinadas\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=Tcos\u03b1=0,8T\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=Tsen\u03b1=0,6T\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=Fcos\u03b2=32.0,6=19,2N\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=Fsen\u03b2=32.0,8=25,6N\u00a0 —\u00a0 em (III) est\u00e3o colocadas apenas as for\u00e7as que agem nas dire\u00e7\u00f5es horizontal e vertical\u00a0 —\u00a0 estando o sistema em equil\u00edbrio essas for\u00e7as se anulam em cada dire\u00e7\u00e3o\u00a0 —\u00a0 na horizontal\u00a0 —\u00a0 0,8T=19,2\u00a0 —\u00a0 T=24N\u00a0 —\u00a0 na vertical\u00a0 —\u00a0 0,6T + 25,6 = 10m\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

0,6.24 + 25,6 = 10m\u00a0 —\u00a0 14,4 + 25,6 = 10m\u00a0 —\u00a0 m=40\/10=4kg\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Movimento Harm\u00f4nico Simples (MHS)<\/b><\/span><\/span><\/span><\/p>\n

Din\u00e2mica do MHS \u2013 Sistema massa-mola<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A. Falsa\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=kx\u00a0 —\u00a0 10=k.0,1\u00a0 —\u00a0 k=100n\/m\u00a0 —\u00a0 k=10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>N\/m.<\/b><\/span><\/span><\/span><\/p>\n

B. Falsa\u00a0 —\u00a0 o per\u00edodo de oscila\u00e7\u00e3o do sistema massa-mola vale\u00a0 —\u00a0 T=2\u03c0\u221a(m\/k)=2\u03c0\u221a(0,25\/10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 T=2\u03c0\u221a(25.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 T=2\u03c0.5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 T\u224830.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>s.<\/b><\/span><\/span><\/span><\/p>\n

C. Correta\u00a0 —\u00a0 a amplitude A vale A=20cm=2.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 energia mec\u00e2nica nos extremos\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>me<\/b><\/span><\/span><\/sub><\/span>=k.A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.(2.10<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/sup><\/span><\/p>\n

\"\"<\/p>\n

1<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>me<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.4\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>me<\/b><\/span><\/span><\/sub><\/span>=2J\u00a0 —\u00a0 a velocidade \u00e9 m\u00e1xima no ponto m\u00e9dio onde a energia cin\u00e9tica \u00e9 m\u00e1xima\u00a0 —\u00a0 teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>me<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 2=0,25.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V=16\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>=4m\/s.<\/b><\/span><\/span><\/span><\/p>\n

D. Falsa\u00a0 —\u00a0 F=ma\u00a0 —\u00a0 F=kx\u00a0 —\u00a0 m.a=kx\u00a0 —\u00a0 0,25.a=10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.2.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 a=20\/0,25\u00a0 —\u00a0 a=80m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

R- C.<\/b><\/span><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Na situa\u00e7\u00e3o inicial (ponto I), de altura m\u00e1xima, a for\u00e7a el\u00e1stica vale F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=3N\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=kx\u00a0 —\u00a0 3=100.x\u00a0 —\u00a0 x=0,03m\u00a0\u00a0 —\u00a0 nesse ponto, a mola est\u00e1 comprimida de x=0,03m ao em rela\u00e7\u00e3o \u00e0 sua posi\u00e7\u00e3o natural\u00a0 —\u00a0 na posi\u00e7\u00e3o de equil\u00edbrio (ponto O) a for\u00e7a resultante \u00e9 nula e P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=mgsen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=F<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/sub><\/span>=kd\u00a0 —\u00a0 1.10.1\/2 = 100.d\u00a0 —\u00a0 d=0,05m\u00a0 —\u00a0 nesse ponto a mola est\u00e1 esticada de d=0,05m em rela\u00e7\u00e3o \u00e0 sua posi\u00e7\u00e3o natural\u00a0\u00a0 —\u00a0 sendo a amplitude A a dist\u00e2ncia entre a posi\u00e7\u00e3o natural (ponto O) e a compress\u00e3o m\u00e1xima (pontoI), ent\u00e3o A=0,03 + 0,05=0,08m\u00a0 —\u00a0 desprezados os atritos, a for\u00e7a el\u00e1stica e a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

for\u00e7a peso s\u00e3o for\u00e7as conservativas, ent\u00e3o o sistema \u00e9 conservativo e a energia mec\u00e2nica do sistema \u00e9 constante\u00a0 —\u00a0 colocando o n\u00edvel de altura zero na posi\u00e7\u00e3o de equil\u00edbrio (ponto O) onde a mola est\u00e1 na posi\u00e7\u00e3o natural (n\u00e3o deformada), a energia mec\u00e2nica a\u00ed \u00e9 s\u00f3 a cin\u00e9tica, pois a potencial gravitacional \u00e9 nula (h\u2019=0) e a potencial el\u00e1stica tamb\u00e9m \u00e9 nula (mola na posi\u00e7\u00e3o natural, sem deforma\u00e7\u00e3o)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mO<\/b><\/span><\/span><\/sub><\/span>=mv<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 c\u00e1lculo da altura no ponto I (veja figura)\u00a0 —\u00a0 sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=h\/0,08\u00a0 —\u00a0 0,5=h\/0.08\u00a0 —\u00a0 h=0,04m\u00a0 —\u00a0 no ponto I a energia mec\u00e2nica \u00e9 a potencial el\u00e1stica armazenada na mola (E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>=kA<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2) somada com a potencial gravitacional (E<\/b><\/span><\/span><\/span>pg<\/b><\/span><\/span><\/sub><\/span>=mgh)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mI<\/b><\/span><\/span><\/sub><\/span>=kA<\/b><\/span><\/span><\/span>2\/<\/b><\/span><\/span><\/sup><\/span>2 + mgh=100.(0,08)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span><\/p>\n

+ 1.10.0,04\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>pI<\/b><\/span><\/span><\/sub><\/span>= 0,32 + 0,4=0,36J\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mI<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mO<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0,36=mv<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>—\u00a0 0,36=1.v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V=\u221a(0,64)\u00a0 —\u00a0 v=0,8m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Fun\u00e7\u00e3o hor\u00e1ria da elonga\u00e7\u00e3o<\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O per\u00edodo de um sistema massa-mola efetuando um MHS \u00e9 fornecido por T=2\u03c0\u221am\/k\u00a0 —\u00a0 observe que se o sistema massa-mola estiver oscilando num plano inclinado, na horizontal ou na vertical, se a massa (m) do corpo e a constante el\u00e1stica da mola (k) forem as mesmas, o que \u00e9 o caso, o per\u00edodo de oscila\u00e7\u00e3o T ser\u00e1 tamb\u00e9m o mesmo\u00a0 — \u00a0R- D.<\/b><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O per\u00edodo\u00a0T\u00a0desse MHS \u00e9 fornecido pela express\u00e3o abaixo\u00a0 —\u00a0 substituindo m=0,08kg, k=0,5 N\/m e\u00a0 \u03c0 = 3,14 na<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0express\u00e3o T=2\u03c0\u221a(m\/k)\u00a0 —\u00a0 T=2.3,14.\u221a0,08\/0,5)=6,28.\u221a(0,16)\u00a0 —\u00a0 T=6,28.0,4=2,512s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

For\u00e7a de resist\u00eancia no ar<\/b><\/span><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) S\u00e3o dados\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=6000m\/s\u00a0 —\u00a0 V=0\u00a0\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 t=7min=7.60=420s\u00a0 —\u00a0 c\u00e1lculo da acelera\u00e7\u00e3o de<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

retardamento\u00a0 —\u00a0 a=(V \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\/(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)=(0 \u2013 6000)\/(420 \u2013 0)\u00a0 —\u00a0 a=-14,3s\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a=1000.(- 14,3)= – 14300N\u00a0 —\u00a0 em m\u00f3dulo\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=1,43.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>N.<\/b><\/span><\/span><\/span><\/p>\n

b) Pelos dados ela entrou na atmosfera marciana com V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=6000m\/s em uma altitude de ho=125km=125.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>m e na altura<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

h=100km=100.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>m ela teve velocidade de V=4000m\/s\u00a0 —\u00a0 Teorema da Energia Mec\u00e2nica\u00a0 —\u00a0 Em todo sistema dissipativo, o trabalho das for\u00e7as n\u00e3o conservativas (for\u00e7a de atrito, for\u00e7a de resist\u00eancia do ar, etc.) \u00e9 igual \u00e0\u00a0 energia total dissipada, ou seja, \u00e9 igual \u00e0 varia\u00e7\u00e3o da energia mec\u00e2nica \u2013\u00a0E<\/b><\/span><\/span><\/span>mfinal<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>minicial<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>for\u00e7as dissipativas<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 a for\u00e7a dissipativa<\/b><\/span><\/span><\/span><\/p>\n

nesse caso \u00e9 a for\u00e7a de atrito F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>mo<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh=10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.(4.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10.100.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>mo<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.(6.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10.125.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=19,25.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>= – 10,25.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>\u2248 – 1,0.10<\/b><\/span><\/span><\/span>10<\/b><\/span><\/span><\/sup><\/span>J.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Pelo enunciado a intensidade da for\u00e7a vertical e para cima que o vento exerce sobre a gota vale F<\/b><\/span><\/span><\/span>vento<\/b><\/span><\/span><\/sub><\/span>=b.r com b=1,6.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>N\/m\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=m\/V\u00a0 —\u00a0 volume da gota considerada esf\u00e9rica de raio r\u00a0 —\u00a0 V=(4\/3).\u03c0.r<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=m\/(4\/3).\u03c0.r<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 m=\u03c1<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.\u03c0.r<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.3.(4\/3)r<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 m=4.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.r<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 na imin\u00eancia de come\u00e7ar a cair, o peso da gota<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(vertical e para baixo) deve ser igual \u00e0 for\u00e7a aplicada pelo vento (vertical e para cima)\u00a0 —\u00a0 P=F<\/b><\/span><\/span><\/span>vento<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 m.g = b.r\u00a0 —\u00a0 4.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.r<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10 = 1,6.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.r\u00a0 —\u00a0 r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>= 1,6.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\/4.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>=0,4.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 r=\u221a(4.10<\/b><\/span><\/span><\/span>-8<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 r=2.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>m=0,2mm\u00a0 —\u00a0 esse deve ser o raio da gota quando ela est\u00e1 na imin\u00eancia de cair\u00a0 —\u00a0 para que ela caia voc\u00ea deve ter\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>r > 0,2mm.<\/b><\/span><\/span><\/span><\/p>\n

b) C\u00e1lculo da massa dessa \u00e1gua que ocupa um volume V de \u00e1gua contida numa \u00e1rea de S=1m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0e altura h=20mm=<\/b><\/span><\/span><\/span><\/p>\n

20.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>m=2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 V=S.h=1.2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=m\/V\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=m\/2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 m=20kg (\u00e9 essa massa de \u00e1gua que ao se chocar com o solo provoca nele um impulso vertical e para baixo)\u00a0 —\u00a0 I=m.\u2206V\u00a0 —\u00a0 como o choque \u00e9 inel\u00e1stico, a varia\u00e7\u00e3o de velocidade da \u00e1gua com o solo foi de \u2206V=2,5m\/s\u00a0 —\u00a0 I=20.2,5\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0I=50N.s,\u00a0<\/b><\/span><\/span><\/span>\u00a0impulso sobre o solo, vertical e para baixo.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

For\u00e7a de atrito<\/b><\/span><\/span><\/span><\/p>\n

66-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>For\u00e7a que ele aplica no pau de sebo comprimindo-o e trocando com ele uma for\u00e7a perpendicular ao poste de F=800N\u00a0 —\u00a0 pelo princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o o poste aplica nele uma for\u00e7a de mesma intensidade (N=800N), mesma dire\u00e7\u00e3o mas sentido contr\u00e1rio\u00a0 —\u00a0 na vertical, existe sobre ele duas for\u00e7as\u00a0 —\u00a0 seu peso vertical e para baixo e uma<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

for\u00e7a de atrito, vertical e para cima que o impede de cair, anulando o peso\u00a0 —\u00a0 P=m.g=60.10\u00a0 —\u00a0 P=600N\u00a0 —\u00a0 N=800N\u00a0 —F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcN=<\/b><\/span><\/span><\/span><\/p>\n

\u03bc<\/b><\/span><\/span><\/span>.800\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub><\/span>=800<\/b><\/span><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub><\/span>=P\u00a0 —\u00a0 800<\/b><\/span><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>=600\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>=600\/800\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>=0,75\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>R- C.<\/b><\/span><\/span><\/span><\/span><\/p>\n

67-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Considerando os dois blocos como um \u00fanico de massa M=(4 + 1)=5kg e sob a\u00e7\u00e3o da for\u00e7a de F=10N eles se movem com a mesma acelera\u00e7\u00e3o de F=Ma\u00a0 —\u00a0 10=5.a\u00a0 —\u00a0 a=2m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(acelera\u00e7\u00e3o de cada bloco)\u00a0 —\u00a0 sobre o bloco 2 agem as for\u00e7as de intensidades F=10N (horizontal e para a direita) e a for\u00e7a de atrito F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0(horizontal e para a esquerda)\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

— for\u00e7a resultante sobre o bloco 2\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=F \u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=10 – F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.a\u00a0 —\u00a0 10 \u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=1.2\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=8N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E.<\/b><\/span><\/span><\/span><\/p>\n

68-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 m=0,4kg\u00a0 —\u00a0 t=0,8s\u00a0 —\u00a0 \u2206S=1,6m\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 calculo da acelera\u00e7\u00e3o do gar\u00e7om, copo e bandeja\u00a0 —\u00a0\u00a0 \u2206S=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 1,6=0.0,8 + a.(0,8)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 a=1,6\/0,32\u00a0 —\u00a0 a=5m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 o copo de massa m=0,4kg que se move com a=5m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0fica sob a\u00e7\u00e3o de uma for\u00e7a resultante\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=ma=0,4.5=2N\u00a0 —\u00a0 observe na figura abaixo onde foram colocadas<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

todas as for\u00e7as que agem sobre o copo que essa for\u00e7a de 2N \u00e9 a for\u00e7a de atrito que obriga o copo a<\/b><\/span><\/span> acompanhar a bandeja e o gar\u00e7om\u00a0 —\u00a0 FR=Fat=2N\u00a0 —\u00a0\u00a0R- A.<\/b><\/span><\/span><\/p>\n

69-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 Vo=5m\/s\u00a0 —\u00a0 V=0\u00a0 —\u00a0 \u2206S=5m\u00a0 —\u00a0 c\u00e1lculo da acelera\u00e7\u00e3o a do bloco utilizando a equa\u00e7\u00e3o de Torricelli\u00a0 —\u00a0 V2\u00a0= Vo2\u00a0+ 2.a. \u2206S\u00a0 —\u00a0 02\u00a0= 52\u00a0+ 2.a.5\u00a0 —\u00a0 a = – 2,5m\/s2\u00a0(negativo, est\u00e1 retardando)\u00a0 —\u00a0 m\u00f3dulo da for\u00e7a resultante sobre o bloco de massa m\u00a0 —\u00a0 FR\u00a0= m.a\u00a0 —\u00a0 FR=2,5m\u00a0 —\u00a0 mas, a \u00fanica for\u00e7a sobre o bloco na dire\u00e7\u00e3o do movimento \u00e9 a for\u00e7a de atrito que o ret\u00e9m at\u00e9 parar\u00a0 —\u00a0 Fat=\u03bc.N= \u03bc.P= \u03bc.mg=10\u03bcm\u00a0 —\u00a0 portanto, a for\u00e7a de atrito \u00e9 a pr\u00f3pria for\u00e7a resultante\u00a0 —\u00a0 Fat\u00a0= FR\u00a0 —\u00a0 10\u03bcm = 2,5m\u00a0 —\u00a0 \u03bc=0,25\u00a0 —\u00a0\u00a0R- E.<\/b><\/span><\/span><\/p>\n

70-<\/b><\/span><\/span><\/span> <\/b>I. Comportamento das for\u00e7as de atrito que agem em um carro<\/b><\/span><\/span><\/span>\u00a0sem o mecanismo ABS:<\/b><\/span><\/span><\/span><\/p>\n

Considere um bloco de massa m sujeito a uma for\u00e7a externa\u00a0\"\"de intensidade vari\u00e1vel<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Quando o corpo estiver em repouso e n\u00e3o houver for\u00e7a externa, F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 com o corpo permanecendo em repouso, aumentando a intensidade de\"\", a intensidade da for\u00e7a de atrito est\u00e1tico (\"\") tamb\u00e9m aumenta (figuras acima), at\u00e9 que o corpo fique na imin\u00eancia de movimento (figura abaixo).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Quando o corpo est\u00e1 na imin\u00eancia de movimento a intensidade da for\u00e7a de atrito est\u00e1tico \u00e9 m\u00e1xima\u00a0 —\u00a0 uma for\u00e7a de intensidade maior que\u00a0\"\"\u00a0faz com que o corpo entre em movimento e, a partir da\u00ed a for\u00e7a de atrito \u00e9 denominada din\u00e2mica ou cin\u00e9tica.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Depois que o bloco entra em movimento a for\u00e7a de atrito \u00e9 denominada for\u00e7a de atrito din\u00e2mica, e \u00e9 sempre a mesma, \u00a0independente da velocidade. A intensidade da for\u00e7a de atrito din\u00e2mica \u00e9 ligeiramente menor que a intensidade da for\u00e7a de atrito est\u00e1tica m\u00e1xima.<\/b><\/span><\/span><\/span><\/p>\n

O gr\u00e1fico abaixo representa todo o processo explicado acima.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

II. Comportamento das for\u00e7as de atrito que agem em um carro<\/b><\/span><\/span><\/span>\u00a0com o mecanismo ABS:<\/b><\/span><\/span><\/span><\/p>\n

Os freios antitravamento (ABS,\u00a0anti-lock braking system) ajudam a parar melhor\u00a0 —\u00a0 eles previnem o <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

travamento das rodas e proporcionam uma dist\u00e2ncia de frenagem mais curta em superf\u00edcies escorregadias, evitando o descontrole do ve\u00edculo\u00a0 —\u00a0 ele mant\u00e9m as rodas sempre na imin\u00eancia de deslizar, aproveitando melhor o atrito est\u00e1tico m\u00e1ximo, que \u00e9 maior que o atrito cin\u00e9tico (de deslizamento)\u00a0 —\u00a0 quando a for\u00e7a aplicada pelos freios atrav\u00e9s da press\u00e3o aplicada no pedal chega aumentada at\u00e9 as rodas, estando elas na imin\u00eancia de movimento (for\u00e7a de atrito de destaque), o sistema ABS libera instantaneamente a roda impedindo seu travamento e mantendo assim a for\u00e7a de atrito m\u00e1xima (for\u00e7a de atrito de destaque) que \u00e9 superior \u00e0 for\u00e7a de atrito cin\u00e9tica ou din\u00e2mica que surgiria, caso ele deslizasse\u00a0 —\u00a0 o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

processo \u00e9 repetido instant\u00e2nea e sucessivamente conforme o gr\u00e1fico acima\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Plano inclinado com atrito<\/b><\/span><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo da intensidade da for\u00e7a de atrito est\u00e1tico (imin\u00eancia de movimento)\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>=\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>.N.cos\u03b8= \u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>.P.cos\u03b8=0,75.100.0,8\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>=60N\u00a0 —\u00a0 c\u00e1lculo da intensidade da parcela do peso paralela ao plano inclinado que tende a puxar o bloco sobre o plano inclinado\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=Psen\u03b8=100.0,6=60N\u00a0 —\u00a0 observe que a for\u00e7a resultante sobre o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

bloco \u00e9 nula e, como ele estava em repouso, por in\u00e9rcia ele tende a continuar em repouso, mas est\u00e1 na imin\u00eancia de movimento)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>-As for\u00e7as que agem sobre um corpo apoiado sobre um plano inclinado s\u00e3o seu peso\"\", vertical e para baixo e a for\u00e7a normal\u00a0\"\", perpendicular \u00e0 superf\u00edcie de contato entre o bloco e o plano.<\/b><\/span><\/span><\/span><\/p>\n

\u03b1\u00a0\u00e9 o \u00e2ngulo de inclina\u00e7\u00e3o do plano. Como\u00a0\"\"\u00a0e\u00a0\"\"\u00a0n\u00e3o tem a mesma dire\u00e7\u00e3o, vamos decompor o peso\u00a0\"\"\u00a0em duas parcelas:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0parcela do peso paralela \u00e0 superf\u00edcie do plano inclinado (respons\u00e1vel pela tentativa de descida do bloco).<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0parcela do peso perpendicular \u00e0 superf\u00edcie do plano inclinado (for\u00e7a que comprime o bloco contra o plano)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

sena=cateto oposto\/hipotenusa\u00a0 —\u00a0 sena=PP\/P\u00a0 —\u00a0 PP=P.sena<\/b><\/span><\/span><\/span><\/p>\n

cosa=cateto adjacente\/hipotenusa\u00a0 —\u00a0 cosa=PN\/P\u00a0 —\u00a0 PN=P.cosa (segunda figura acima)<\/b><\/span><\/span><\/span><\/p>\n

\"\"\"\"<\/p>\n

A componente normal do peso\u00a0\"\"\u00a0anula a rea\u00e7\u00e3o normal do apoio\u00a0\"\", assim temos (terceira figura acima).<\/b><\/span><\/span><\/span><\/p>\n

Havendo atrito:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A for\u00e7a de atrito \u00e9 sempre contr\u00e1ria ao movimento ou \u00e0 sua tend\u00eancia e \u00e9 fornecida por F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=m\u03bcN\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>=N\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=mP<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcPcos\u03b1\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcmgcos\u03b1<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Resolvendo o exerc\u00edcio\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0=m.a\u00a0 —\u00a0 mgsen\u03b8 \u2013 \u03bcmgcos\u03b8 = ma\u00a0 —\u00a0 10.0,6 \u2013 <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

0,25.10.0,8 = a\u00a0 — a=6 \u2013 2\u00a0 —\u00a0 a=4,0m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 deslocamento \u2206S entre A e B (veja figura abaixo)\u00a0 —\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

cos\u03b8=cateto adjacente\/hipotenusa\u00a0 — 0,8 = BC\/AB\u00a0 —\u00a0 0,8=2,4\/\u2206S\u00a0 —\u00a0 \u2206S=3m\u00a0 —\u00a0 para calcular a velocidade em B (V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>) voc\u00ea pode usar a equa\u00e7\u00e3o de Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.a.\u2206S\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.4.3\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=5m\/s\u00a0 —\u00a0 varia\u00e7\u00e3o da quantidade de movimento entre A e B\u00a0 —\u00a0 \u2206Q = Q<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 Q<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 mV<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=50.5 \u2013 50.1\u00a0 —\u00a0 \u2206Q=200N.s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

For\u00e7as no movimento circular<\/b><\/span><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Express\u00e3o da intensidade da for\u00e7a resultante centr\u00edpeta\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 kart\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>K<\/b><\/span><\/span><\/sub><\/span>=M.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2R=0,5MV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 f\u00f3rmula 1\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>=3M.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=3MV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 stock-car\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>=6M.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/3R=2MV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 assim, F<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>\u00a0> F<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>\u00a0> F<\/b><\/span><\/span><\/span>K<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>R- B.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>50-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Em todo movimento circular existe sempre uma for\u00e7a resultante dirigida para o centro da circunfer\u00eancia denominada for\u00e7a resultante centr\u00edpeta (\"\"<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(\"\") de intensidade F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 no ponto B essa for\u00e7a tem dire\u00e7\u00e3o vertical e sentido para cima e de intensidade T \u2013 P =mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Trabalho<\/b><\/span><\/span><\/span><\/p>\n

48-\u00a0<\/b><\/span><\/span><\/span>Observe que durante todo os deslocamentos da mo\u00e7a num lugar plano, o \u00e2ngulo que a for\u00e7a <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

peso dos brincos forma com o deslocamento (na nossa figura, horizontal) \u00e9 sempre de 90<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>peso<\/b><\/span><\/span><\/sub><\/span>=P.d.cos90<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=p.d.0=0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E.<\/b><\/span><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como n\u00e3o existe atrito com o ar a intensidade da for\u00e7a resultante (F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>) sobre o conjunto (sat\u00e9lite-foguete) \u00e9 igual \u00e0 intensidade da for\u00e7a propulsora F<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>) menos a intensidade da for\u00e7a peso (P)\u00a0 —\u00a0 intensidade da for\u00e7a peso\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

P=m.g=1.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>.10=10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 P\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0= 5.10<\/b><\/span><\/span><\/span>7\u00a0<\/b><\/span><\/span><\/sup><\/span>\u2013 1.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>=4.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0 trabalho da for\u00e7a resultante\u00a0 —\u00a0\u00a0 W<\/b><\/span><\/span><\/span>FR\u00a0<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

F\u00ad<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>.d.cos0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=4.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>.2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.1\u00a0 —\u00a0\u00a0 W<\/b><\/span><\/span><\/span>FR<\/b><\/span><\/span><\/sub><\/span>=8.10<\/b><\/span><\/span><\/span>10<\/b><\/span><\/span><\/sup><\/span>\u00a0J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Pot\u00eancia<\/b><\/span><\/span><\/span><\/p>\n

51-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 m=70kg\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 V=72km\/h\/3,6=20m\/s\u00a0 —\u00a0 \u2206t=2min=2.60=120s\u00a0 —\u00a0 c\u00e1lculo do trabalho pela varia\u00e7\u00e3o de energia cin\u00e9tica\u00a0 —\u00a0 W=E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>co<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 mV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=70.20<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 70.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 W=14000J\u00a0 —\u00a0 P=W\/\u2206t\u00a0 — P=14000\/120\u00a0 —\u00a0 P=116,66W\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

52-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Correta\u00a0 —\u00a0 pelo enunciado a massa total que sobe \u00e9 a massa da g\u00f4ndola (500kg) somada com sua carga (240kg)\u00a0 —\u00a0 peso\u00a0\u00a0 —\u00a0 P=M.g=(500 + 240),g=740.10\u00a0 —\u00a0 P=7400N (peso de uma g\u00f4ndola carregada)\u00a0 —\u00a0 a g\u00f4ndola sobe com velocidade constante uma altura \u2206S=100m em\u00a0 \u2206t=40s e sua velocidade m\u00e9dia V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0ser\u00e1\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>= \u2206S\/\u2206t=100\/40\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=2,5m\/s\u00a0 —\u00a0 a subida \u00e9 um MRU, ent\u00e3o a for\u00e7a resultante sobre a g\u00f4ndola durante a subida \u00e9 nula o que implica que<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

a intensidade da for\u00e7a m\u00e9dia aplicada pelo motor equilibra a for\u00e7a peso\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=P=7400N\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=7400N\u00a0 —\u00a0 o trabalho da for\u00e7a m\u00e9dia aplicada pelo motor na subida vale W=F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.d,cos0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.d.1\u00a0 —\u00a0 W=F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.d\u00a0 —\u00a0 potencia m\u00e9dia\u00a0 — P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u2206t\u00a0\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.d\/\/\u2206t\u00a0 —\u00a0 d\/\/\u2206t=V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>,V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=7400.2,5\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=18500W.<\/b><\/span><\/span><\/span><\/p>\n

02. Correta\u00a0 —\u00a0 tempo que demora, em queda livre (a=g) para , a partir do repouso (V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0), atingir velocidade de V=122,4\/3,6\u00a0 —\u00a0 V=122,4\/3,6\u00a0 =34m\/s\u00a0 —\u00a0 V = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ g.t\u00a0 —\u00a0 34=0 + 10t\u00a0 —\u00a0 t=3,4s\u00a0 —\u00a0 c\u00e1lculo da acelera\u00e7\u00e3o de retardamento a quando, descendo na vertical, com a velocidade passando de V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=122,4\/3,6=34m\/s\u00a0 at\u00e9 parar, quando V=0 em \u2206t=8,4 \u2013 3,4=5s\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0– a.t\u00a0 —\u00a0 0 = 34 \u2013 a.5\u00a0 —\u00a0 a=34\/5\u00a0 —\u00a0 a=6,8m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 for\u00e7a desaceleradora m\u00e9dia\u00a0 —\u00a0 F\u00ad<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>= m.a=740.6,8\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=5032N.<\/b><\/span><\/span><\/span><\/p>\n

04. Falsa\u00a0 —\u00a0 \u00e9 de 3,4s (veja 02).<\/b><\/span><\/span><\/span><\/p>\n

08. Correta\u00a0 —\u00a0 a queda livre demorou t=3,4s, com a=g=10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0e V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 \u2206S=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0.3,4 + 10.(3,4)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2= 0 + 10.11,56\/2=10.5,78\u00a0 —\u00a0 \u2206S=57,8m.<\/b><\/span><\/span><\/span><\/p>\n

16. Falsa\u00a0 —\u00a0 \u00e9 g apenas na queda livre e n\u00e3o durante todo o percurso.<\/b><\/span><\/span><\/span><\/p>\n

R- (01 + 02 + 08)=11.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Energia<\/b><\/span><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Autocontra\u00e7\u00e3o gravitacional (potencial gravitacional)\u00a0 —\u00a0 nesse processo, os \u00edons de hidrog\u00eanio (pr\u00f3tons) contidos no seu interior adquirem\u00a0 velocidades muito altas (energia cin\u00e9tica)\u00a0 —\u00a0 o que os leva a atingirem temperaturas da ordem de milh\u00f5es de graus (energia t\u00e9rmica)\u00a0 —\u00a0 com isso, t\u00eam in\u00edcio rea\u00e7\u00f5es exot\u00e9rmicas de fus\u00e3o nuclear, nas quais n\u00facleos de hidrog\u00eanio s\u00e3o fundidos, gerando n\u00facleos de He\u00a0(H\u00e9lio) e propiciando a produ\u00e7\u00e3o da radia\u00e7\u00e3o (energia de massa)\u00a0 —\u00a0 parte dessa radia\u00e7\u00e3o atinge a Terra (sob forma de radia\u00e7\u00e3o eletromagn\u00e9tica) e \u00e9 a principal fonte de toda a energia que utilizamos\u00a0 —\u00a0\u00a0R- D.<\/b><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Massa do atleta\u00a0 —\u00a0 m=70kg\u00a0 —\u00a0 como pedido no exerc\u00edcio, pelo gr\u00e1fico 1, a terceira corrida demorou \u2206t=t=0,5h num percurso de \u2206S=7,5km\u00a0 —\u00a0 V=\u2206S\/\u2206t=7,5\/0,5\u00a0 —\u00a0 V=15km\/h\u00a0 —\u00a0 pra essa velocidade V=15km\/h, voc\u00ea verifica no gr\u00e1fico 2 que C<\/b><\/span><\/span><\/span>MET<\/b><\/span><\/span><\/sub><\/span>=60kJ\/kg.h\u00a0 —\u00a0 usando a express\u00e3o fornecida pelo enunciado do exerc\u00edcio\u00a0 —\u00a0 E=C<\/b><\/span><\/span><\/span>MET<\/b><\/span><\/span><\/sub><\/span>.m.t=60.70.<\/b><\/span><\/span><\/span><\/p>\n

0,5\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0E=2100kJ.<\/b><\/span><\/span><\/span><\/p>\n

b) Avaliando a \u00e1rea da sola de um t\u00eanis como um ret\u00e2ngulo de 35cm=0,35m de comprimento e 9cm=0,09m de largura<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0—\u00a0 S=0,35×0,09=30,5.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u22483.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 press\u00e3o (P<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span>)=for\u00e7a(no caso, peso P)\/\u00e1rea (S)\u00a0 —\u00a0 P=mg=70.10=700N\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

P\/S=700\/3.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span>\u22482,3.10<\/b><\/span><\/span><\/span>4\u00a0<\/b><\/span><\/span><\/sup><\/span>N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(Pa).<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Tipos de energia \u2013 Trabalho<\/b><\/span><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>Da mesma maneira que uma atiradeira (estilingue) transforma energia potencial el\u00e1stica em energia cin\u00e9tica\u00a0 devido \u00e0 uma distens\u00e3o da borracha (el\u00e1stico) o carrinho mencionado tamb\u00e9m transforma energia potencial el\u00e1stica em energia cin\u00e9tica devido \u00e0 compress\u00e3o da mola\u00a0 —\u00a0\u00a0R- E.\u00a0<\/b><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Energia mec\u00e2nica<\/b><\/span><\/span><\/span><\/p>\n

7<\/b><\/span><\/span><\/span>2-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Colocando o referencial na altura em que se encontra seu centro de massa, no in\u00edcio, quando ele finca a vara no ch\u00e3o para subir ele s\u00f3 possui energia cin\u00e9tica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=m.11<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=60,5m\u00a0 —\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

pelo enunciado, na altura M\u00e1xima, 80% de sua energia cin\u00e9tica \u00e9 convertida em energia potencial gravitacional\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>pg<\/b><\/span><\/span><\/sub><\/span>=0,8xE<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=0,8×60,5m=48,4m\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>pg<\/b><\/span><\/span><\/sub><\/span>=m.g.h=48,4m\u00a0 —\u00a0 m.10.h=48,4m\u00a0 —\u00a0 h=4,84m\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 R- E.<\/b><\/span><\/span><\/span><\/p>\n

73-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span><\/span>\u00a0As for\u00e7as que agem sobre um corpo apoiado sobre um plano inclinado s\u00e3o seu peso\"\", vertical e para baixo e a for\u00e7a normal\u00a0\"\", perpendicular \u00e0 superf\u00edcie de contato entre o bloco e o plano (figura abaixo).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u03b1\u00a0\u00e9 o \u00e2ngulo de inclina\u00e7\u00e3o do plano. Como\u00a0\"\"\u00a0e\u00a0\"\"\u00a0n\u00e3o tem a mesma dire\u00e7\u00e3o, vamos decompor o peso\u00a0\"\"\u00a0em duas parcelas:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0parcela do peso paralela \u00e0 superf\u00edcie do plano inclinado (respons\u00e1vel pela tentativa de descida do bloco).<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0parcela do peso perpendicular \u00e0 superf\u00edcie do plano inclinado (for\u00e7a que comprime o bloco contra o plano)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

sen\u03b1=cateto oposto\/hipotenusa\u00a0 —\u00a0 sen\u03b1=P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\/P\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=P.sena<\/b><\/span><\/span><\/span><\/p>\n

cos\u03b1=cateto adjacente\/hipotenusa\u00a0 —\u00a0 cos\u03b1=P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\/P\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>=P.cosa<\/b><\/span><\/span><\/span><\/p>\n

\"\"\"\"<\/p>\n

\u00a0<\/span><\/p>\n

As duas for\u00e7as acima substituem o peso e podemos tir\u00e1-lo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A componente normal do peso\u00a0\"\"\u00a0anula a rea\u00e7\u00e3o normal do apoio\u00a0\"\", assim temos:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como n\u00e3o existe atrito, o bloco desce com acelera\u00e7\u00e3o de intensidade a, tal que F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a\u00a0 — \u00a0\u00a0P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=m.a\u00a0 —\u00a0 mgsen\u03b1=ma\u00a0 —\u00a0\u00a0a=gsen\u03b1.<\/b><\/span><\/span><\/span><\/p>\n

Como o bloco est\u00e1 em equil\u00edbrio na dire\u00e7\u00e3o perpendicular \u00e0 superf\u00edcie do plano inclinado\u00a0 —\u00a0 N=P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>=Pcos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=10.\u221a3\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0N\u22485.1,7\u22488,5N.<\/b><\/span><\/span><\/span><\/p>\n

b) Veja em (a) que ele desce com acelera\u00e7\u00e3o a=gsen30<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sup><\/span>=10.1\/2\u00a0 — \u00a0a=5m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 aplicando Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+<\/b><\/span><\/span><\/span><\/p>\n

2.a.\u2206S\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.5.d\u00a0 —\u00a0 d=100\/10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>d=10m.<\/b><\/span><\/span><\/span><\/p>\n

c) Como n\u00e3o existe atrito o bloco 1 atinge o bloco 2 com velocidade de V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=10m\/s\u00a0 —\u00a0 antes do choque\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=10m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 quantidade de movimento do sistema antes do choque \u00a0—\u00a0 Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1.10 + 3.0\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span>=10kg.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

m\/s\u00a0 —\u00a0\u00a0 depois do choque\u00a0 —\u00a0 V\u2019<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=?\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=4m\/s\u00a0 —\u00a0 quantidade de movimento do sistema depois do choque\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>V\u2019<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1.V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u2019 + 3.4\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>= V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u2019 + 12\u00a0 —\u00a0 pelo princ\u00edpio da conserva\u00e7\u00e3o da quantidade de movimento\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 10 = V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u2019 + 12\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u2019= – 2m\/s\u00a0 (o sinal negativo significa que o bloco 1 retornou ap\u00f3s o choque).<\/b><\/span><\/span><\/span><\/p>\n

d) C\u00e1lculo da energia mec\u00e2nica do bloco 2 no ponto A de altura zero e onde V=4m\/s\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh=3.4<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 +<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

0\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=24J\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>2 + m.g.R=3.V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 3.10.0,6\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=\u221a4=2m\/s\u00a0 —\u00a0 no ponto mais alto B, a for\u00e7a resultante sobre o bloco \u00e9 a centr\u00edpeta, vertical e dirigida para o centro da circunfer\u00eancia (baixo), de intensidade\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 P \u2013 N = 3.2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/0,6\u00a0 —\u00a0 30 \u2013 N = 20\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0N=10N.<\/b><\/span><\/span><\/span><\/p>\n

74-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Utilizando o teorema da conserva\u00e7\u00e3o da quantidade de movimento do sistema\u00a0 —\u00a0 antes do disparo —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=(M + m)V=(M + m).0\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 depois do disparo\u00a0 —\u00a0 o canh\u00e3o de massa M se move para a direita com velocidade<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

V e o proj\u00e9til de massa m para a\u00a0 com velocidade ( – v)\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>depois\u00a0<\/b><\/span><\/span><\/sub><\/span>=MV \u2013 mv\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>\u00a0= Q<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0 = MV \u2013mv\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

V=mv\/M\u00a0 —\u00a0\u00a0 ap\u00f3s o disparo o canh\u00e3o de massa M se desloca para a direita com velocidade inicial V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=mv\/M at\u00e9 parar V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=0, percorrendo uma dist\u00e2ncia d\u00a0 —\u00a0 a energia dissipada pela varia\u00e7\u00e3o de energia cin\u00e9tica (\u2206E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>) do canh\u00e3o \u00e9 igual \u00e0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

energia dissipada pela for\u00e7a de atrito (W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0\u00a0 \u2206E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0= W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 MV<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 MV<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>2 = F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>.d.cos180<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0 M.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 M(mv\/M)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>.d.(-1)\u00a0 —\u00a0\u00a0 0 \u2013 M.(m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)\/2 = – \u03bcNd\u00a0 —\u00a0 N=P=Mg\u00a0 —\u00a0 M.m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= \u03bcMgd\u00a0 —\u00a0 \u03bcMgd = m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>v<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>\/2M\u00a0 —\u00a0 \u03bc = m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>gd\u00a0 —\u00a0 \u03bc = (mv\/M)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2gd\u00a0 —<\/b><\/span><\/span><\/span>\u00a0R- A.<\/b><\/span><\/span><\/span><\/p>\n

75-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A energia mec\u00e2nica antes do disparo que vale Ema=3,75J e \u00e9 a soma das energia cin\u00e9tica Ec=mV2\/2 com a potencial el\u00e1stica armazenada na mola (Epel)\u00a0 —\u00a0 Ema\u00a0= Ec\u00a0+ Epel\u00a0 —\u00a0 3,75=mV2\/2 + Epel\u00a0 —\u00a0 3,75 = (3 + 3).12\/2 + Epel\u00a0 —\u00a0\u00a0Epel=0,75J\u00a0 —\u00a0 agora voc\u00ea deve utilizar o teorema da conserva\u00e7\u00e3o da quantidade de movimento\u00a0 —\u00a0 antes do disparo do gatilho\u00a0 —\u00a0 Qsa=mA.VA\u00a0+ mB.VB=3.1 + 3.1=3kgm\/s\u00a0 —\u00a0 depois do disparo do gatilho\u00a0 —\u00a0 Qsd=mAVA\u00a0+ mBVB=3VA\u00a0+ 3.1,5=3VA\u00a0+ 4,5\u00a0 —\u00a0 Qsa\u00a0= Qsd\u00a0 —\u00a0 3=3VA\u00a0+ 4,5\u00a0 —\u00a0\u00a0VA=1,5\/3=0,5m\/s.<\/b><\/span><\/span><\/p>\n

76-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Dados\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>16<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=30km\/s=30.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=3.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>m\/s\u00a0 —\u00a0 quantidade de movimento inicial do meteoro\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>16<\/b><\/span><\/span><\/sup><\/span>.3.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>=3.10<\/b><\/span><\/span><\/span>20\u00a0<\/b><\/span><\/span><\/sup><\/span>kg.m\/s<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

b) Defini\u00e7\u00e3o de energia cin\u00e9tica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>\u00a0= m<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=10<\/b><\/span><\/span><\/span>16<\/b><\/span><\/span><\/sup><\/span>.(3.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>16<\/b><\/span><\/span><\/sup><\/span>.9.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>=4,5.10<\/b><\/span><\/span><\/span>24<\/b><\/span><\/span><\/sup><\/span>\u00a0J.<\/b><\/span><\/span><\/span><\/p>\n

c)\u00a0<\/b><\/span><\/span><\/span>Utilizando o princ\u00edpio da conserva\u00e7\u00e3o da quantidade de movimento do sistema (Terra-meteoro) antes e depois do fen\u00f4meno (colis\u00e3o)\u00a0 — antes da colis\u00e3o\u00a0 —\u00a0\u00a0 P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\u00a0= P<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>\u00a0+ P<\/b><\/span><\/span><\/span>Ti<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mas, V<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>=0 (antes da colis\u00e3o, a Terra n\u00e3o se move na dire\u00e7\u00e3o perpendicular \u00e0 sua trajet\u00f3ria, veja figura)\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>16<\/b><\/span><\/span><\/sup><\/span>.3.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 6.10<\/b><\/span><\/span><\/span>24<\/b><\/span><\/span><\/sup><\/span>.0\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=3.10<\/b><\/span><\/span><\/span>20<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 0\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=3.10<\/b><\/span><\/span><\/span>20<\/b><\/span><\/span><\/sup><\/span>kg.m\/s\u00a0 —\u00a0 depois da colis\u00e3o\u00a0 —\u00a0 trata-se de uma colis\u00e3o inel\u00e1stica (Terra e meteoro se movem juntos, com a mesma velocidade V)\u00a0 — P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0= P<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>\u00a0+ P<\/b><\/span><\/span><\/span>Tf<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.V + m<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>.V=(m<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>),V\u00a0 —\u00a0 como a massa do meteoro \u00e9 insignificante em rela\u00e7\u00e3o \u00e0 massa da Terra, voc\u00ea pode considerar (m<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>)\u2248m<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>.V=6.10<\/b><\/span><\/span><\/span>24<\/b><\/span><\/span><\/sup><\/span>V\u00a0 —\u00a0 pelo princ\u00edpio da conserva\u00e7\u00e3o da quantidade de movimento\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\u00a0= P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 3.10<\/b><\/span><\/span><\/span>20<\/b><\/span><\/span><\/sup><\/span>=6.10<\/b><\/span><\/span><\/span>24<\/b><\/span><\/span><\/sup><\/span>V\u00a0 —\u00a0 V=3.10<\/b><\/span><\/span><\/span>20<\/b><\/span><\/span><\/sup><\/span>\/6.10<\/b><\/span><\/span><\/span>24<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V=5,0.10<\/b><\/span><\/span><\/span>-5\u00a0<\/b><\/span><\/span><\/sup><\/span>m\/s<\/b><\/span><\/span><\/span>\u00a0(velocidade \u2018Terra + meteoro\u201d , no nosso exemplo para a direita, imediatamente ap\u00f3s a colis\u00e3o).<\/b><\/span><\/span><\/span><\/p>\n

c) C\u00e1lculo da energia cin\u00e9tica final do sistema (Terra + meteoro)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>(m + T)<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=m<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=6.10<\/b><\/span><\/span><\/span>24<\/b><\/span><\/span><\/sup><\/span>.(5.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span>=7,5.10<\/b><\/span><\/span><\/span>15<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 essa energia cin\u00e9tica final do sistema (7,5.10<\/b><\/span><\/span><\/span>15<\/b><\/span><\/span><\/sup><\/span>J) \u00e9 muito menor do que a energia cin\u00e9tica inicial do meteoro (4,5.10<\/b><\/span><\/span><\/span>24<\/b><\/span><\/span><\/sup><\/span>J)\u00a0 —\u00a0 assim, a energia dissipada no choque \u00e9 praticamente igual \u00e0 energia cin\u00e9tica inicial do meteoro\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=4,5.10<\/b><\/span><\/span><\/span>24<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 megaton \u2013 4.10<\/b><\/span><\/span><\/span>15<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 4,4.10<\/b><\/span><\/span><\/span>24<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=4,4.10<\/b><\/span><\/span><\/span>24<\/b><\/span><\/span><\/sup><\/span>\/4.10<\/b><\/span><\/span><\/span>15<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=1,125.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0megatons<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Impulso e quantidade de movimento<\/b><\/span><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Utilizando o teorema da conserva\u00e7\u00e3o da quantidade de movimento do sistema\u00a0 —\u00a0 antes do disparo —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=(M + m)V=(M + m).0\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 depois do disparo\u00a0 —\u00a0 o canh\u00e3o de massa M se move para a direita com velocidade<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

V e o proj\u00e9til de massa m para a\u00a0 com velocidade ( – v)\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>depois\u00a0<\/b><\/span><\/span><\/sub><\/span>=MV \u2013 mv\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>\u00a0= Q<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0 = MV \u2013mv\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

V=mv\/M\u00a0 —\u00a0\u00a0 ap\u00f3s o disparo o canh\u00e3o de massa M se desloca para a direita com velocidade inicial V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=mv\/M at\u00e9 parar V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=0, percorrendo uma dist\u00e2ncia d\u00a0 —\u00a0 a energia dissipada pela varia\u00e7\u00e3o de energia cin\u00e9tica (\u2206E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>) do canh\u00e3o \u00e9 igual \u00e0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

energia dissipada pela for\u00e7a de atrito (W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0\u00a0 \u2206E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0= W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 MV<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 MV<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>2 = F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>.d.cos180<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0 M.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 M(mv\/M)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>.d.(-1)\u00a0 —\u00a0\u00a0 0 \u2013 M.(m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)\/2 = – \u03bcNd\u00a0 —\u00a0 N=P=Mg\u00a0 —\u00a0 M.m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= \u03bcMgd\u00a0 —\u00a0 \u03bcMgd = m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>v<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>\/2M\u00a0 —\u00a0 \u03bc = m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>gd\u00a0 —\u00a0 \u03bc = (mv\/M)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2gd\u00a0 —<\/b><\/span><\/span><\/span>\u00a0R- A.<\/b><\/span><\/span><\/span><\/p>\n

40- Teorema do Impulso<\/b><\/span><\/span><\/span>: \u201c O impulso da for\u00e7a resultante de um sistema de for\u00e7as que age sobre um corpo \u00e9 igual \u00e0 varia\u00e7\u00e3o da quantidade de movimento do corpo\u201d<\/b><\/span><\/span><\/span><\/p>\n

\"\"\"\"\"\"\"\"<\/p>\n

\u00a0Se uma bola de massa m incide perpendicularmente numa parede com velocidade de m\u00f3dulo V e retorna na mesma dire\u00e7\u00e3o, mas em sentido contr\u00e1rio com velocidade de mesmo m\u00f3dulo, teremos:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

F.\u2206t = m.V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 m.V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F.\u2206t = m.(+V) \u2013 m.(-V)\u00a0 —\u00a0 F \u2206t = m.V \u2013 (- m.V)\u00a0 —\u00a0 F.\u2206t = mV + mV\u00a0 —\u00a0\u00a0F \u2206t = 2mV\u00a0 \u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=-40m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=+30m\/s\u00a0 —\u00a0 F.\u2206t = m.V<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 m.V<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F.\u2206t = m.(+30) \u2013 m.(-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

40)<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span>—\u00a0 F.\u2206t = 30m + 40m\u00a0 —\u00a0 F.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=70×6.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F=420,10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\/10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F=420N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R-D.<\/b><\/span><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Teorema do Impulso:\u201c O impulso da for\u00e7a resultante de um sistema de for\u00e7as que age sobre um corpo \u00e9 igual \u00e0 varia\u00e7\u00e3o da quantidade de movimento do corpo\u201d<\/b><\/span><\/span><\/span><\/p>\n

\"\"\"\"\"\"\"\"<\/p>\n

Muitos autom\u00f3veis disp\u00f5e de um eficiente sistema\u00a0 de prote\u00e7\u00e3o para o motorista, que consiste de uma bolsa infl\u00e1vel de ar (air-bag)\u00a0 —\u00a0 essa bolsa \u00e9 automaticamente inflada quando o autom\u00f3vel sofre uma desacelera\u00e7\u00e3o s\u00fabita, de modo que a cabe\u00e7a e o t\u00f3rax do motorista, em vez de colidirem com o volante, colidem com a bolsa.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Veja pela express\u00e3o F.\u2206t=m.\u2206V, que\u00a0m\u00a0e\u00a0\u2206V\u00a0s\u00e3o constantes (mesmo motorista e mesma varia\u00e7\u00e3o de velocidade) e assim, teremos\u00a0 —\u00a0 F. \u2206t=constante, portanto se voc\u00ea aumentar o intervalo de tempo do choque, estar\u00e1 diminuindo a intensidade da for\u00e7a m\u00e1xima que o motorista troca com o volante.<\/b><\/span><\/span><\/span><\/p>\n

Assim, a bolsa aumenta o intervalo de tempo da desacelera\u00e7\u00e3o da cabe\u00e7a do motorista, diminuindo, portanto, a for\u00e7a m\u00e1xima que atua sobre a cabe\u00e7a e o t\u00f3rax\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A.<\/b><\/span><\/span><\/span><\/p>\n

42-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como o sistema \u00e9 isolado, a quantidade de movimento do sistema antes da colis\u00e3o (\"\") deve ser igual \u00e0 quantidade de movimento do sistema depois da colis\u00e3o (\"\")\u00a0 —\u00a0 a quantidade de movimento <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

do sistema antes da colis\u00e3o tem<\/b><\/span><\/span><\/span> dire\u00e7\u00e3o horizontal e sentido para a direita, sendo a do pr\u00f3prio f\u00f3ton, pois a do el\u00e9tron \u00e9 nula (figura)\u00a0 —\u00a0 a quantidade de movimento do sistema depois da colis\u00e3o tamb\u00e9m deve ter dire\u00e7\u00e3o horizontal e sentido para a <\/b><\/span><\/span>direita, pois\u00a0\"\"=\"\"\u00a0 —\u00a0 ent\u00e3o a soma vetorial de\u00a0\"\"com\u00a0\"\"\u00a0deve fornecer\u00a0\"\"=\"\"\u00a0(veja figura)\u00a0 —\u00a0\u00a0R- A.<\/b><\/span><\/span><\/p>\n

43-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A energia mec\u00e2nica antes do disparo que vale Ema=3,75J e \u00e9 a soma das energia cin\u00e9tica Ec=mV2\/2 com a potencial el\u00e1stica armazenada na mola (Epel)\u00a0 —\u00a0 Ema\u00a0= Ec\u00a0+ Epel\u00a0 —\u00a0 3,75=mV2\/2 + Epel\u00a0 —\u00a0 3,75 = (3 + 3).12\/2 + Epel\u00a0 —\u00a0\u00a0Epel=0,75J\u00a0 —\u00a0 agora voc\u00ea deve utilizar o teorema da conserva\u00e7\u00e3o da quantidade de movimento\u00a0 —\u00a0 antes do disparo do gatilho\u00a0 —\u00a0 Qsa=mA.VA\u00a0+ mB.VB=3.1 + 3.1=3kgm\/s\u00a0 —\u00a0 depois do disparo do gatilho\u00a0 —\u00a0 Qsd=mAVA\u00a0+ mBVB=3VA\u00a0+ 3.1,5=3VA\u00a0+ 4,5\u00a0 —\u00a0 Qsa\u00a0= Qsd\u00a0 —\u00a0 3=3VA\u00a0+ 4,5\u00a0 —\u00a0\u00a0VA=1,5\/3=0,5m\/s.<\/b><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Colis\u00f5es mec\u00e2nicas<\/b><\/span><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span> <\/b>01-\u00a0a) Falsa\u00a0 —\u00a0 quantidade de movimento (Q)=massa (m)xvelocidade (V)\u00a0 —\u00a0 bola de v\u00f4lei\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>=(0,27).(30)\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>=8,1kg.m\/s\u00a0 —\u00a0 bola de golfe\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=(0,045).(60)\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=2,7kg.m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\u00a0< Q<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

b) Falsa\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 bola de v\u00f4lei\u00a0 — E<\/b><\/span><\/span><\/span>cV<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=(0,27).(30)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>cV<\/b><\/span><\/span><\/sub><\/span>=121,5J\u00a0 —\u00a0 bola de golfe\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>cG<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=(0,045).(60)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>cG<\/b><\/span><\/span><\/sub><\/span>=81J\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>cG<\/b><\/span><\/span><\/sub><\/span>\u00a0< E<\/b><\/span><\/span><\/span>cV<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

c) Falsa\u00a0 —\u00a0 em todo choque inel\u00e1stico o coeficiente de restitui\u00e7\u00e3o \u00e9 nulo o que implica que, os corpos ap\u00f3s o choque, possuem a mesma velocidade e, como a massa da pessoa \u00e9 bem maior que a das bolas, cada uma delas ter\u00e1 velocidade final nula\u00a0\u00a0 — V<\/b><\/span><\/span><\/span>fV<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 e V<\/b><\/span><\/span><\/span>fG<\/b><\/span><\/span><\/sub><\/span>=0.<\/b><\/span><\/span><\/span><\/p>\n

d) Falsa\u00a0 —\u00a0 “O impulso da for\u00e7a resultante de um sistema de for\u00e7as que age sobre um corpo \u00e9 igual \u00e0 varia\u00e7\u00e3o da quantidade de movimento do corpo”<\/b><\/span><\/span><\/span><\/p>\n

\"\"\"\"\"\"\"\"<\/p>\n

\u2206t<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/sub><\/span>= 2\u2206t<\/b><\/span><\/span><\/span><\/span>G<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/sub><\/span>=0,27kg\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span><\/span>G<\/b><\/span><\/span><\/span><\/sub><\/span>=0,045kg\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span><\/span>G<\/b><\/span><\/span><\/span><\/sub><\/span>=60m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/sub><\/span>=30m\/s\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span><\/span>G<\/b><\/span><\/span><\/span><\/sub><\/span>. \u2206t<\/b><\/span><\/span><\/span><\/span>G<\/b><\/span><\/span><\/span><\/sub><\/span>=0,045×60=2,7 (I)\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/sub><\/span>.2\u2206t<\/b><\/span><\/span><\/span><\/span>G<\/b><\/span><\/span><\/span><\/sub><\/span>=0,27×30=8,1 (II)\u00a0 —\u00a0 (I)\/(II)\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span><\/span>G<\/b><\/span><\/span><\/span><\/sub><\/span>.\u2206t<\/b><\/span><\/span><\/span><\/span>G<\/b><\/span><\/span><\/span><\/sub><\/span>\/2F<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/sub><\/span>.\u2206t<\/b><\/span><\/span><\/span><\/span>G<\/b><\/span><\/span><\/span><\/sub><\/span>=2,7\/8,1\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span><\/span>G<\/b><\/span><\/span><\/span><\/sub><\/span>\/2F<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/sub><\/span>=1\/3\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span><\/span>G<\/b><\/span><\/span><\/span><\/sub><\/span>=2F<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/sub><\/span>\/3\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span><\/span>G<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0< F<\/b><\/span><\/span><\/span><\/span>V.<\/b><\/span><\/span><\/span><\/sub><\/span><\/p>\n

e) Correta\u00a0 —\u00a0 Press\u00e3o=for\u00e7a\/\u00e1rea\u00a0 —\u00a0 P=F\/S\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=2F<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>\/3 (I)\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>=10S<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>\/10S<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\u00a0(II)\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

(I)\/(II)\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\/P<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>=2F<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>\/3S<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>x10S<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\/F<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\/P<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>=(20\/3)P<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\u00a0> P<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

R- E.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>50-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo a colis\u00e3o perfeitamente el\u00e1stica, a energia mec\u00e2nica (no caso, cin\u00e9tica) do sistema, antes <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

e depois da colis\u00e3o \u00e9 a mesma\u00a0 —\u00a0 antes\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>ca<\/b><\/span><\/span><\/sub><\/span>=m.2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>ca<\/b><\/span><\/span><\/sub><\/span>=2m\u00a0 —\u00a0 depois\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>cd<\/b><\/span><\/span><\/sub><\/span>=mV\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=m.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>cd<\/b><\/span><\/span><\/sub><\/span>=0,5m + mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>ca<\/b><\/span><\/span><\/sub><\/span>\u00a0= E<\/b><\/span><\/span><\/span>cd<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2m = 0,5m + mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=3\u00a0\u00a0 —\u00a0 raz\u00e3o pedida\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/sup><\/span>=mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=m.3\/2 (I)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=m.4\/2=2m (II)\u00a0 —\u00a0 (I)\/(II)\u00a0 —\u00a0 E’<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/E<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= (3m\/2)\/2m=(3m\/2)x(1\/2m)\u00a0 —\u00a0 E’<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/E<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=3\/4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\u00a051-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Ap\u00f3s o primeiro choque\u00a0 —\u00a0 h<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0,8×5=4,0m\u00a0 —\u00a0 ap\u00f3s 0 segundo choque\u00a0 —\u00a0 h<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0,8×4,0=3,2m\u00a0 —\u00a0 ap\u00f3s o terceiro choque\u00a0 —\u00a0 h<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=0,8×3,2=2,56m\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 R- C.<\/b><\/span><\/span><\/span><\/p>\n

52-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>No primeiro choque ela chega ao solo com\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=mgh=E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 m.10.5=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 v=10m\/s\u00a0 —\u00a0 80%de 10m\/s=8m\/s\u00a0 —\u00a0 aos o primeiro choque ela se solta do solo com\u00a0 —\u00a0\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=mgh=m.10.4\u00a0 —\u00a0 V=\u221a80=8,9ms\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D.<\/b><\/span><\/span><\/span><\/p>\n

53-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Intensidade da for\u00e7a sobre a bola 1\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=3m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=3m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.3=9m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 intensidade da for\u00e7a sobre a bola 2\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mas, pelo princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o essas for\u00e7as tem a mesma intensidade\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 9m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>==\u00a0\u00a0 a<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=9m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

R- B.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES \u2013 2014 – 2013 \u00a0As tr\u00eas leis de Newton 01-\u00a0Se ele est\u00e1 com\u00a0velocidade constante, independente da inclina\u00e7\u00e3o, ele est\u00e1 em\u00a0MRU na vertical e na horizontale, pelo princ\u00edpio da in\u00e9rcia, a\u00a0resultante das for\u00e7as que agem sobre ele \u00e9 nula (equil\u00edbrio din\u00e2mico) —\u00a0R- D 02-\u00a0A for\u00e7a resultante sobre o bloco na dire\u00e7\u00e3o do movimento \u00e9 a for\u00e7a de atrito Fat=\u03bcN=\u03bcP= \u03bcmg=<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3249,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-3251","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3251","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=3251"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3251\/revisions"}],"predecessor-version":[{"id":11082,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3251\/revisions\/11082"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3249"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3251"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}