{"id":3660,"date":"2018-02-13T13:53:23","date_gmt":"2018-02-13T13:53:23","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3660"},"modified":"2024-08-21T13:25:33","modified_gmt":"2024-08-21T13:25:33","slug":"resolucao-comentada-transformacoes-de-energia-energia-mecanica","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/transformacoes-de-energia-energia-mecanica-exercicios-com-caracteristicas-de-enem\/resolucao-comentada-transformacoes-de-energia-energia-mecanica\/","title":{"rendered":"Energia Mec\u00e2nica – Resolu\u00e7\u00e3o"},"content":{"rendered":"

\u00a0<\/span>Resolu\u00e7\u00e3o Comentada \u00a0<\/b><\/span><\/span><\/span><\/p>\n

Transforma\u00e7\u00f5es de Energia \u2013 Energia Mec\u00e2nica<\/b><\/span><\/span><\/span><\/p>\n

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

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que na figura I ele est\u00e1 correndo (energia cin\u00e9tica) no solo horizontal sem utilizar a vara (energia el\u00e1stica) e nem subir (energia gravitacional), ent\u00e3o, em I voc\u00ea identifica apenas energia cin\u00e9tica\u00a0 —\u00a0 na figura II ele est\u00e1 se movendo (energia cin\u00e9tica),<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

a flexibilidade da vara o impulsiona para cima (energia el\u00e1stica) e ele est\u00e1 ganhando altura (energia gravitacional)\u00a0 —\u00a0 na figura 3 ele est\u00e1 se movendo para baixo (energia cin\u00e9tica), perdendo altura (energia gravitacional) e n\u00e3o havendo energia el\u00e1stica (abandonou a vara)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe pelo enunciado que n\u00e3o h\u00e1 perda de energia mec\u00e2nica por atrito, ent\u00e3o o sistema \u00e9 conservativo\u00a0 —\u00a0 no instante em que o carrinho deixa a mola com velocidade v, toda sua energia potencial el\u00e1stica armazenada (kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2) se transforma em energia cin\u00e9tica (mv<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2)\u00a0 —\u00a0\u00a0 kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 8000.(2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0,2.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=32.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\/2.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>=16\u00a0 —\u00a0V=4,0m\/s<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Ao iniciar a subida da rampa o carrinho tem energia cin\u00e9tica E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mv<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0,2.4<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=1,6J\u00a0 — essa energia na altura h onde a velocidade \u00e9 de 2,0m\/s se transforma em cin\u00e9tica e potencial gravitacional\u00a0 —\u00a0 1,6=mv<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh\u00a0 —\u00a0 1,6=0,2.2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 0,2.10.h\u00a0 —\u00a0 1,6 \u2013 0,4=2h\u00a0 —\u00a0 h=0,6m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Antes do salto a \u00fanica energia existente \u00e9 a energia cin\u00e9tica (com o n\u00edvel zero de altura no ponto 0,8m)\u00a0 —\u00a0E<\/b><\/span><\/span><\/span>msolo<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\u00a0 — na altura m\u00e1xima s\u00f3 tem energia potencial gravitacional\u00a0 —\u00a0E<\/b><\/span><\/span><\/span>mhm\u00e1x<\/b><\/span><\/span><\/sub><\/span>=m.g.h=m.10.3,2=32m\u00a0\u00a0—\u00a0 como n\u00e3o h\u00e1 dissipa\u00e7\u00e3o de energia elas s\u00e3o id\u00eanticas\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>msolo<\/b><\/span><\/span><\/sub><\/span>= E<\/b><\/span><\/span><\/span>mhm\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=32m\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=32\u00a0 —\u00a0 V=8m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A energia total do corpo na altura\u00a0 de 10m, que \u00e9 s\u00f3 a gravitacional (a cin\u00e9tica \u00e9 zero, abandonado do repouso) vale E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=mgh=2.10.10\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=200J\u00a0 —\u00a0 ao chegar ao solo (a gravitacional \u00e9 nula, altura nula), a parte de energia que \u00e9 transformada em cin\u00e9tica vale\u00a0 —\u00a0\u00a0 E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=200 \u2013 4\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>= 196J\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\u00a0 —\u00a0 196=2V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V=\u221a196\u00a0 —\u00a0 V=14m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados: r = 80 cm = 0,8 m; h = 2 m; m = 36 kg; H = 6 m e g = 10 m\/s2\u00a0 —\u00a0 como na descida o atrito \u00e9 desprez\u00edvel, o sistema pode ser considerado conservativo\u00a0 —\u00a0 ent\u00e3o, tomando como refer\u00eancia o plano que cont\u00e9m o final do tobo\u00e1gua, pela conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0na altura de 6m quando a crian\u00e7a parte do repouso (V=0, E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=0), s\u00f3 existe energia potencial gravitacional\u00a0 —\u00a0\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=mgH\u00a0 —\u00a0 quando ela chega \u00e0 altura de 2m\u00a0 ela ter\u00e1 energia cin\u00e9tica e energia potencial\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=mgh + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 m g H = m g h + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 10.(6) = 10.(2) + V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u00a0—\u00a0 \u00a0v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 80\u00a0 —\u00a0 a <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

for\u00e7a horizontal de intensidade F<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>\u00a0sobre a crian\u00e7a na altura de 2m, quando V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=80 \u00e9 a for\u00e7a resultante centr\u00edpeta\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=36×80\/0,8\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>=3.600N\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

Fx\u00a0= Fc\u00a0= mV2\/R= 36×80\/08\u00a0 —\u00a0\u00a0Fx\u00a0= 3.600 N. <\/b><\/span><\/span><\/span><\/p>\n

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

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Vamos supor que seja no ponto X que a formiga perca contato com a bola\u00a0 —\u00a0 nesse ponto, a rea\u00e7\u00e3o normal\u00a0\"\"\u00a0deixa de existir agindo sobre a formiga somente seu peso\u00a0\"\", que tem como componente a for\u00e7a resultante centr\u00edpeta, dirigida para o centro da circunfer\u00eancia e de intensidade F<\/b><\/span><\/span><\/span>RC<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 observe no tri\u00e2ngulo menor\u00a0 —\u00a0 cos\u03b1=F<\/b><\/span><\/span><\/span>RC<\/b><\/span><\/span><\/sub><\/span>\/P\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>RC<\/b><\/span><\/span><\/sub><\/span>=Pcos\u03b1=mgcos\u03b1\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

mgcos\u03b1=mV<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0\u00a0—\u00a0 gcos\u03b1=V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R (I)\u00a0 —\u00a0\u00a0observe no tri\u00e2ngulo maior –\u00a0cos\u03b1=h\/R, que substitu\u00eddo em (I)\u00a0 —\u00a0 g.h\/R=V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=g.h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=\u221a(gh).<\/b><\/span><\/span><\/span><\/p>\n

C<\/b><\/span><\/span><\/span>hamando a posi\u00e7\u00e3o inicial da formiga de Y e considerando a horizontal que passa por X como n\u00edvel zero de altura (figura abaixo):<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

E<\/b><\/span><\/span><\/span>mY<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cy<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>py\u00a0<\/b><\/span><\/span><\/sub><\/span>=0 + mg.(R-h)\u00a0\u00a0—\u00a0 E<\/b><\/span><\/span><\/span>my<\/b><\/span><\/span><\/sub><\/span>=m.g.(R-h)\u00a0\u00a0—\u00a0 E<\/b><\/span><\/span><\/span>mx<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cx<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>px<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 0\u00a0\u00a0—\u00a0 E<\/b><\/span><\/span><\/span>mx<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>my<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mx<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 m.g.(R-h)=m.V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2.g.(R-h)\u00a0 —\u00a0 que substitu\u00eddo em V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=g.h\u00a0 —\u00a0 g.h=2.g.R \u2013 2.g.h\u00a0 —\u00a0 3h=2R\u00a0\u00a0—\u00a0<\/b><\/span><\/span><\/span>\u00a0h=2R\/3.<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- A\u00a0<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como o sistema \u00e9 conservativo a energia mec\u00e2nica total \u00e9 constante e diferente de zero (gr\u00e1fico III)\u00a0 —\u00a0 se a energia total \u00e9 constante quando a energia potencial diminui a cin\u00e9tica deve aumentar ou quando Ep = m\u00e1xima, \u00a0Ec =0 (gr\u00e1fico I).\u00a0 \u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura abaixo que mostra os pontos de velocidade nula (A) e de velocidade m\u00e1xima <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(B)\u00a0 —\u00a0 pela conserva\u00e7\u00e3o da \u00a0 energia mec\u00e2nica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 3mgh<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=3mgh<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 3mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u00a0—\u00a0 g(h<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 h<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>)=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=20.(60 \u2013 2)=1.160\u00a0 —\u00a0 V=\u221a(1.160)\u00a0 —\u00a0v\u00a0\u2248\u00a034 m\/s\u00a0\u00a0—\u00a0 a energia cin\u00e9tica m\u00e1xima a que eles ficam submetidos \u00e9 a energia cin\u00e9tica do sistema formado pelos tr\u00eas jovens, no ponto de velocidade m\u00e1xima (B)\u00a0 —\u00a0 Ec=3mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=3x50x1.160\/2\u00a0 —\u00a0\u00a0Ec=87.000J=87kJ \u00a0—\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo o sistema \u00e9 conservativo, em todos os casos a velocidade em B \u00e9 v<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>, que pode ser calculada pelo Teorema da Energia Mec\u00e2nica\u00a0 —\u00a0 fazendo AB = h\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 mgh=mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=\u221a(2gh)\u00a0 —\u00a0 sendo H a altura do solo at\u00e9 B, o tempo de queda (t<\/b><\/span><\/span><\/span>q<\/b><\/span><\/span><\/sub><\/span>) \u00e9 obtido pela express\u00e3o: H = gt<\/b><\/span><\/span><\/span>q<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 tq=\u221a(2H\/g)\u00a0 —\u00a0 na dire\u00e7\u00e3o horizontal, o movimento \u00e9 uniforme com velocidade v<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 a dist\u00e2ncia horizontal percorrida durante o tempo de queda \u00e9: d = v<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0t<\/b><\/span><\/span><\/span>q<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 d =(\u221a2gh).(\u221a2H\/g)\u00a0 —\u00a0 d=2\u221ahH\u00a0 —\u00a0 sendo h e H iguais em todos os casos, a dist\u00e2ncia de B ao solo tamb\u00e9m \u00e9 a mesma para todos eles\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Nos dois casos, a deforma\u00e7\u00e3o da mola \u00e9 a mesma (x), armazenando as duas molas mesma energia potencial el\u00e1stica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>Pel<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 — \u00a0energia potencial gravitacional em rela\u00e7\u00e3o \u00e0 linha da <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

mola n\u00e3o deformada\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>pg<\/b><\/span><\/span><\/sub><\/span>= = m g x\u00a0 —\u00a0 pela conserva\u00e7\u00e3o da energia, a velocidade v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0de lan\u00e7amento de um dardo \u00e9\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pel<\/b><\/span><\/span><\/sub><\/span>+ E<\/b><\/span><\/span><\/span>pg<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=mgx + kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2\/m(mgx + kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2)\u00a0 —\u00a0 Vo=\u221a(mgx + kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2)\u00a0 —\u00a0 observe que, como a massa m aparece no denominador, o dardo de maior massa \u00e9 o que tem menor velocidade inicial, ou seja, o dardo I, que tem um peda\u00e7o de chumbo grudado nele.<\/b><\/span><\/span><\/span><\/p>\n

Ap\u00f3s sair dos canos dos brinquedos, desprezando a resist\u00eancia do ar, os dados ficam sujeitos exclusivamente \u00e0 for\u00e7a peso, tendo, portanto, a mesma acelera\u00e7\u00e3o g\u00a0 —\u00a0 por isso os gr\u00e1ficos s\u00e3o retas paralelas, como mostrado na op\u00e7\u00e3o A\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo da energia mec\u00e2nica em A que \u00e9 s\u00f3 a energia potencial gravitacional\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=mgh=25.10.2,4=600J\u00a0 —\u00a0 energia mec\u00e2nica em B, s\u00f3 a energia cin\u00e9tica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=25.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=12,5V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 for\u00e7a de atrito\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcN=\u03bcP<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=\u03bcPcos37<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcmgcos37<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=0,5x250x0,8\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=100N\u00a0 —\u00a0 sen37<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=h\/d\u00a0 —\u00a0 0,6=2,4\/d\u00a0 —\u00a0 d=4m\u00a0 —\u00a0 trabalho da for\u00e7a <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

de atrito\u00a0 —\u00a0W<\/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>=100.4.(-1)\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>Fat<\/b><\/span><\/span><\/sub><\/span>= – 400J\u00a0 —\u00a0 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 —\u00a0 -400=12,5V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 600\u00a0 —\u00a0 V=4m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

\u00a0Resolu\u00e7\u00e3o Comentada \u00a0 Transforma\u00e7\u00f5es de Energia \u2013 Energia Mec\u00e2nica \u00a0 01-\u00a0Observe que na figura I ele est\u00e1 correndo (energia cin\u00e9tica) no solo horizontal sem utilizar a vara (energia el\u00e1stica) e nem subir (energia gravitacional), ent\u00e3o, em I voc\u00ea identifica apenas energia cin\u00e9tica\u00a0 —\u00a0 na figura II ele est\u00e1 se movendo (energia cin\u00e9tica), a flexibilidade da vara o impulsiona para cima<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3658,"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-3660","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3660","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=3660"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3660\/revisions"}],"predecessor-version":[{"id":10711,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3660\/revisions\/10711"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3658"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3660"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}