{"id":3128,"date":"2017-08-08T00:06:37","date_gmt":"2017-08-08T00:06:37","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3128"},"modified":"2024-09-04T14:29:11","modified_gmt":"2024-09-04T14:29:11","slug":"resolucao-dinamica","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/mecanica\/dinamica\/resolucao-dinamica\/","title":{"rendered":"Resolu\u00e7\u00e3o Din\u00e2mica"},"content":{"rendered":"

Resolu\u00e7\u00f5es-Vestibulares Recentes por assunto<\/span><\/span><\/span><\/p>\n

2017, 2016 e 2015<\/span><\/span><\/span><\/p>\n

APLICA\u00c7\u00d5ES DAS TR\u00caS LEIS DE NEWTON<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>Nesse caso o objeto tem uma\u00a0<\/b><\/span><\/span>velocidade constante<\/b><\/span><\/span>, logo\u00a0<\/b><\/span><\/span>sua acelera\u00e7\u00e3o \u00e9 0<\/b><\/span><\/span>. Como sua acelera\u00e7\u00e3o \u00e9 0,\u00a0<\/b><\/span><\/span>a for\u00e7a resultante tamb\u00e9m s\u00f3 pode ser 0<\/b><\/span><\/span>, com isso\u00a0<\/b><\/span><\/span>a alternativa correta \u00e9 a D<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Vamos analisar as outras, tamb\u00e9m:<\/b><\/span><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>velocidade tem sempre a mesma dire\u00e7\u00e3o e sentido da for\u00e7a resultante<\/b><\/span><\/span>, pois\u00a0<\/b><\/span><\/span>ela \u00e9 uma consequ\u00eancia da acelera\u00e7\u00e3o<\/b><\/span><\/span>\u00a0que vai ser exercida por essa for\u00e7a.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Se\u00a0<\/b><\/span><\/span>a resultante fosse igual ao peso, o objeto estaria em queda livre<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

A alternativa C pode causar confus\u00e3o, porque\u00a0<\/b><\/span><\/span>realmente ela \u00e9 verdadeira, mas no caso aonde h\u00e1 uma for\u00e7a resultante<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Sempre\u00a0<\/b><\/span><\/span>h\u00e1 for\u00e7as agindo<\/b><\/span><\/span>, por\u00e9m\u00a0<\/b><\/span><\/span>a soma delas vai ser igual a 0<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R \u2013 D<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

03-<\/b><\/span><\/span>Denominando\u00a0<\/b><\/span><\/span>de ml\u00a0a massa de cada cubo laranja e de ma\u00a0a massa de cada cubo azul\u00a0<\/b><\/span><\/span>e, estando as\u00a0<\/b><\/span><\/span>balan\u00e7as em equil\u00edbrio,\u00a0<\/b><\/span><\/span>voc\u00ea ter\u00e1:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Balan\u00e7a\u00a0<\/b><\/span><\/span>da esquerda\"\"2ma\u00a0+ ml\u00a0= 2 (I)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Balan\u00e7a\u00a0<\/b><\/span><\/span>da direita\u00a0\"\"ma\u00a0+ 3 = 2ml\u00a0(II)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Isolando ma\u00a0em (II) e substituindo em (I<\/b><\/span><\/span>)\u00a0\"\"<\/b><\/span><\/span>ma\u00a0= 2ml\u00a0\u2013 3\u00a0\"\"2(2ml\u00a0\u2013 3) + ml\u00a0= 2\u00a0\"\"<\/b><\/span><\/span>\u00a04ml\u00a0\u2013 6 + ml\u00a0= 2\u00a0\"\"5ml\u00a0= 8\u00a0\"\"<\/b><\/span><\/span>ml<\/b><\/span><\/span>= 8\/5\u00a0<\/b><\/span><\/span>= 1,6 kg.<\/b><\/span><\/span><\/span><\/span><\/p>\n

2ma\u00a0+ 1,6 =2\u00a0\"\"<\/b><\/span><\/span>\u00a0ma\u00a0= 0,4\/2\u00a0\"\"<\/b><\/span><\/span>ma\u00a0= 0,2 kg.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Assim, a\u00a0<\/b><\/span><\/span>massa de um cubo laranja supera a de um cubo azul\u00a0<\/b><\/span><\/span>em\u00a0<\/b><\/span><\/span>ml\u00a0\u2013 ma<\/b><\/span><\/span>= 1,6 \u2013 0,2\u00a0<\/b><\/span><\/span>= 1,4 kg.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

05-<\/b><\/span><\/span><\/span>a)\u00a0<\/b><\/span><\/span>Em\u00a0<\/b><\/span><\/span>todo gr\u00e1fico Vxt o deslocamento\u00a0\"\"\u00a0<\/b><\/span><\/span>do corpo \u00e9\u00a0<\/b><\/span><\/span>numericamente igual \u00e1 \u00e1rea<\/b><\/span><\/span>,\u00a0<\/b><\/span><\/span>nesse caso,\u00a0<\/b><\/span><\/span>entre\u00a0<\/b><\/span><\/span>os\u00a0<\/b><\/span><\/span>instantes 0 e 10 s que \u00e9 a \u00e1rea de um trap\u00e9zio:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Mas, o\u00a0<\/b><\/span><\/span>enunciado n\u00e3o pede o deslocamento entre 0s e 10s,\u00a0<\/b><\/span><\/span>mas sim a<\/b><\/span><\/span>posi\u00e7\u00e3o no instante t = 10 s<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

e,\u00a0<\/b><\/span><\/span>pelo gr\u00e1fico\u00a0<\/b><\/span><\/span>Sxt voc\u00ea observa\u00a0<\/b><\/span><\/span>que ele\u00a0<\/b><\/span><\/span>partiu da origem So\u00a0= 0,\u00a0<\/b><\/span><\/span>ent\u00e3o<\/b><\/span><\/span>quando t = 10s o espa\u00e7o S ser\u00e1\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0= S \u2013 So\u00a0\"\"\u00a026 = S \u2013 0\u00a0\"\"\u00a0<\/b><\/span><\/span>S = 26 m.<\/b><\/span><\/span><\/span><\/span><\/p>\n

b)\u00a0<\/b><\/span><\/span>Observe no\u00a0<\/b><\/span><\/span>gr\u00e1fico da velocidade\u00a0<\/b><\/span><\/span>que, entre\u00a0<\/b><\/span><\/span>5s e 10 s o gr\u00e1fico \u00e9 uma mesma reta inclinada\u00a0<\/b><\/span><\/span>o que\u00a0<\/b><\/span><\/span>significa que nesse intervalo de tempo a acelera\u00e7\u00e3o tem o mesmo m\u00f3dulo<\/b><\/span><\/span>\u00a0que pode ser\u00a0<\/b><\/span><\/span>calculado<\/b><\/span><\/span>\u00a0por\u00a0<\/b><\/span><\/span>a =\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>=<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>a =\u00a0<\/b><\/span><\/span>–\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>em m\u00f3dulo\u00a0\"\"\u00a0a = 0,8 m\/s2.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Pelo\u00a0<\/b><\/span><\/span>Princ\u00edpio Fundamental da Din\u00e2mica<\/b><\/span><\/span>\u00a0ou\u00a0<\/b><\/span><\/span>Segunda lei de Newton\u00a0\u00a0\"\"\u00a0FR= m.a\u00a0<\/b><\/span><\/span>= 8.0,8\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

FR\u00a0= 6,4 N.<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

06-<\/b><\/span><\/span> <\/b>Dados:<\/b><\/span><\/span>\u00a0<\/b><\/span>massa do reboque<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>m =<\/b><\/span><\/span>\u00a016.103\u00a0kg\u00a0\"\"\u00a0<\/b><\/span><\/span>acelera\u00e7\u00e3o do reboque\u00a0\"\"\u00a0a =\u00a0<\/b><\/span><\/span>200 cm\/s<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\u00a0= 200.10<\/b><\/span><\/span>-2<\/b><\/span><\/sup><\/span> m\/s<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span> \"\"<\/b><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span>a = 2 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

07-<\/b><\/span><\/span><\/span>No in\u00edcio,\u00a0<\/b><\/span><\/span>quando\u00a0<\/b><\/span><\/span>ele sobe acelerando,\u00a0<\/b><\/span><\/span>a\u00a0<\/b><\/span><\/span>for\u00e7a exercida pelo cabo\u00a0<\/b><\/span><\/span>na cabine tem\u00a0<\/b><\/span><\/span>intensidade maior que a do peso, pois a for\u00e7a resultante \u00e9 para cima FI\u00a0> P.<\/b><\/span><\/span><\/span><\/span><\/p>\n

No final, antes da parada,\u00a0<\/b><\/span><\/span>ele est\u00e1\u00a0<\/b><\/span><\/span>freando e desacelerando com a for\u00e7a resultante sendo para cima\u00a0<\/b><\/span><\/span>at\u00e9\u00a0<\/b><\/span><\/span>se tornar nula\u00a0<\/b><\/span><\/span>quando\u00a0<\/b><\/span><\/span>FF\u00a0= P\u00a0<\/b><\/span><\/span>(ele para)<\/b><\/span><\/span><\/p>\n

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

08-\"\"<\/b><\/span><\/span><\/span><\/p>\n

09-\"\"<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

 <\/p>\n

10-<\/b><\/span><\/span><\/span>Se a\u00a0<\/b><\/span><\/span>velocidade de queda \u00e9 constante<\/b><\/span><\/span>, ele dever\u00e1\u00a0<\/b><\/span><\/span>percorrer dist\u00e2ncias iguais em intervalos de tempo<\/b><\/span><\/span>\u00a0<\/b><\/span>iguais.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

FOR\u00c7A EL\u00c1STICA<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>a) Observe na figura abaixo<\/b><\/span><\/span>\u00a0que, depois que o objeto foi preso \u00e0 mola que estava em seu\u00a0<\/b><\/span><\/span>comprimento natural de\u00a0<\/b><\/span><\/span>20cm<\/b><\/span><\/span>,<\/b><\/span><\/span>\u00a0ela foi\u00a0<\/b><\/span><\/span>deformada<\/b><\/span><\/span>\u00a0com seu <\/b><\/span><\/span>comprimento passando para\u00a0<\/b><\/span><\/span>25cm<\/b><\/span><\/span>\u00a0e<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

sofrendo uma\u00a0<\/b><\/span><\/span>deforma\u00e7\u00e3o de \u2206x = 5cm = 0,05m<\/b><\/span><\/span>, quando atingiu a\u00a0<\/b><\/span><\/span>situa\u00e7\u00e3o de equil\u00edbrio.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Na\u00a0<\/b><\/span><\/span>situa\u00e7\u00e3o de equil\u00edbrio<\/b><\/span><\/span>, as\u00a0<\/b><\/span><\/span>for\u00e7as<\/b><\/span><\/span>\u00a0que agem sobre o objeto s\u00e3o, seu <\/b><\/span><\/span>peso<\/b><\/span><\/span>\u00a0(vertical e para baixo) e a\u00a0<\/b><\/span><\/span>for\u00e7a el\u00e1stica<\/b><\/span><\/span>\u00a0(vertical e para cima), que sustenta o objeto, que est\u00e1 em\u00a0<\/b><\/span><\/span>equil\u00edbrio, mas n\u00e3o em repouso (for\u00e7a resultante sobre ele \u00e9 nula).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b)<\/b><\/span><\/span>\u00a0<\/b><\/span>Suponha que o\u00a0<\/b><\/span><\/span>objeto seja elevado\u00a0<\/b><\/span><\/span>at\u00e9 a\u00a0<\/b><\/span><\/span>altura em que o comprimento da mola volta a ser\u00a0<\/b><\/span><\/span>20 cm<\/b><\/span><\/span>\u00a0<\/b><\/span>e, em seguida,\u00a0<\/b><\/span><\/span>solto<\/b><\/span><\/span>\u00a0<\/b><\/span>a partir do repouso. <\/b><\/span><\/span>Determine a\u00a0<\/b><\/span><\/span>energia cin\u00e9tica<\/b><\/span><\/span>\u00a0do objeto, em\u00a0<\/b><\/span><\/span>joules<\/b><\/span><\/span>, no instante em que <\/b><\/span><\/span>passa pelo ponto em que o comprimento da mola \u00e9 de 25 cm<\/b><\/span><\/span>. Despreze qualquer perda de energia mec\u00e2nica.<\/b><\/span><\/span><\/span><\/span><\/p>\n

b) C\u00e1lculo da\u00a0<\/b><\/span><\/span>energia mec\u00e2nica no ponto\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>\u00a0onde, pelo enunciado\u00a0<\/b><\/span><\/span>partiu do repouso\u00a0<\/b><\/span><\/span>VP\u00a0=\u00a0<\/b><\/span><\/span>0<\/b><\/span><\/span>\u00a0e se encontra na\u00a0<\/b><\/span><\/span>altura de h=0,05m<\/b><\/span><\/span>\u00a0em rela\u00e7\u00e3o ao ponto Q tomado como referencial. Nesse ponto, a\u00a0<\/b><\/span><\/span>mola est\u00e1 em sua posi\u00e7\u00e3o natural<\/b><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span>n\u00e3o existe energia potencial el\u00e1stica armazenada<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

02-<\/b><\/span><\/span><\/span> <\/b>For\u00e7as\u00a0<\/b><\/span><\/span>que agem\u00a0<\/b><\/span><\/span>sobre cada bloco\u00a0<\/b><\/span><\/span>supondo que a mola est\u00e1\u00a0<\/b><\/span><\/span>sendo esticada\u00a0<\/b><\/span><\/span>e, quando solta, se\u00a0<\/b><\/span><\/span>mover\u00e1 para a esquerda:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

03-<\/b><\/span><\/span><\/span>Veja as\u00a0<\/b><\/span><\/span>caracter\u00edsticas\u00a0<\/b><\/span><\/span>das\u00a0<\/b><\/span><\/span>associa\u00e7\u00f5es s\u00e9rie e paralelo\u00a0<\/b><\/span><\/span>de molas abaixo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Associa\u00e7\u00e3o em paralelo<\/b><\/span><\/span><\/p>\n

Nesse caso a\u00a0<\/b><\/span><\/span>deforma\u00e7\u00e3o\u00a0x<\/b><\/span><\/span>\u00a0sofrida por\u00a0<\/b><\/span><\/span>cada uma das molas \u00e9 a mesma.<\/b><\/span><\/span><\/p>\n

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

Associa\u00e7\u00e3o em s\u00e9rie<\/b><\/span><\/span><\/p>\n

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

Nesse caso as\u00a0<\/b><\/span><\/span>molas 1 e 2 est\u00e3o sujeitas \u00e0 mesma for\u00e7a F<\/b><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span>sofrem deforma\u00e7\u00f5es diferentes<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>x1\u00a0e x2.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe que a\u00a0<\/b><\/span><\/span>constante el\u00e1stica (coeficiente el\u00e1stico)\u00a0<\/b><\/span><\/span>da\u00a0<\/b><\/span><\/span>associa\u00e7\u00e3o paralelo \u00e9 maior do que a da s\u00e9rie.<\/b><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

05-<\/b><\/span><\/span><\/span>Podemos descrever a situa\u00e7\u00e3o\u00a0<\/b><\/span><\/span>conforme o esquema abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

06-<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span>express\u00e3o da frequ\u00eancia de oscila\u00e7\u00e3o de um sistema massa mola\u00a0<\/b><\/span><\/span>\u00e9 \"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>(se voc\u00ea n\u00e3o domina a teoria, a demonstra\u00e7\u00e3o est\u00e1 em fisicaevestibular.com.br \u2013 mec\u00e2nica \u2013 din\u00e2mica \u2013 MHS \u2013 sistema massa-mola).<\/b><\/span><\/span><\/p>\n

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

P\u00caNDULOS SIMPLES<\/b><\/span><\/span><\/span><\/p>\n

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

P\u00eandulo Simples<\/b><\/span><\/span><\/span><\/p>\n

P\u00eandulo Simples<\/b><\/span><\/span>\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span>consta de uma\u00a0<\/b><\/span><\/span>massa m,<\/b><\/span><\/span>\u00a0presa na\u00a0<\/b><\/span><\/span>extremidade inferior de um fio ideal,\u00a0fixada verticalmente\u00a0<\/b><\/span><\/span>na sua extremidade superior (figura)\u00a0.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Se o\u00a0p\u00eandulo simples<\/b><\/span><\/span>\u00a0oscilar, com oscila\u00e7\u00f5es de pequena abertura\u00a0<\/b><\/span><\/span>(no m\u00e1ximo 15o), ele descreve um\u00a0movimento circular de raio R=L, sendo L o comprimento do fio.<\/b><\/span><\/span><\/p>\n

Seu\u00a0per\u00edodo (T),\u00a0que \u00e9 o\u00a0tempo que ele demora para efetuar um \u201cvai e vem\u201d completo\u00a0\u00e9<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>fornecido pela express\u00e3o:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

onde\u00a0<\/b><\/span><\/span><\/span>g \u00e9 a acelera\u00e7\u00e3o da gravidade local.<\/b><\/span><\/span><\/span><\/p>\n

Observe na\u00a0<\/b><\/span><\/span>figura abaixo<\/b><\/span><\/span>\u00a0a sequ\u00eancia da situa\u00e7\u00e3o descrita desde\u00a0<\/b><\/span><\/span>to\u00a0= 0\u00a0<\/b><\/span><\/span>at\u00e9 a esfera efetuar uma\u00a0<\/b><\/span><\/span>oscila\u00e7\u00e3o completa com per\u00edodo T.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

 <\/p>\n

02-<\/b><\/span><\/span><\/span>P\u00eandulo Simples:<\/b><\/span><\/span>\u00a0consta de uma\u00a0<\/b><\/span><\/span>massa m,<\/b><\/span><\/span>\u00a0<\/b><\/span>presa na extremidade inferior de um fio ideal<\/b><\/span><\/span>,\u00a0<\/b><\/span><\/span>fixada verticalmente<\/b><\/span><\/span>\u00a0na sua\u00a0<\/b><\/span><\/span>extremidade superior<\/b><\/span><\/span>(figura)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Se o\u00a0<\/b><\/span><\/span>p\u00eandulo simples<\/b><\/span><\/span>\u00a0oscilar, com oscila\u00e7\u00f5es de pequena abertura\u00a0<\/b><\/span><\/span>(no m\u00e1ximo 15o)<\/b><\/span><\/span>, ele descreve um\u00a0<\/b><\/span><\/span>movimento circular de raio R=L, sendo L o comprimento do fio.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Seu\u00a0<\/b><\/span><\/span>per\u00edodo (T)<\/b><\/span><\/span>, que \u00e9 o\u00a0<\/b><\/span><\/span>tempo que ele demora para efetuar um \u201cvai e vem\u201d completo \u00e9 fornecido pela express\u00e3o:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

onde\u00a0<\/b><\/span><\/span>g \u00e9 a acelera\u00e7\u00e3o da gravidade local.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Observe que:<\/b><\/span><\/span><\/span><\/span><\/p>\n

* a\u00a0<\/b><\/span><\/span>massa pendular m<\/b><\/span><\/span>\u00a0n\u00e3o\u00a0<\/b><\/span><\/span>influi no per\u00edodo T do movimento<\/b><\/span><\/span>\u00a0<\/b><\/span>influindo apenas o comprimento L do<\/b><\/span><\/span>\u00a0<\/b><\/span>p\u00eandulo, e a acelera\u00e7\u00e3o da gravidade local.<\/b><\/span><\/span>Assim,\u00a0<\/b><\/span><\/span>dois p\u00eandulos de\u00a0<\/b><\/span><\/span>mesmo comprimento L<\/b><\/span><\/span>, mas\u00a0<\/b><\/span><\/span>de massas diferentes M e m, apresentam o mesmo per\u00edodo T.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>* O\u00a0<\/b><\/span><\/span>per\u00edodo\u00a0<\/b><\/span><\/span>de um p\u00eandulo simples\u00a0<\/b><\/span><\/span>independe da amplitude<\/b><\/span><\/span>, ou seja, da<\/b><\/span><\/span>altura em que m \u00e9<\/b><\/span><\/span>\u00a0<\/b><\/span>abandonada<\/b><\/span><\/span>, assim, os\u00a0<\/b><\/span><\/span>p\u00eandulos da figura abaixo, tanto na situa\u00e7\u00e3o 1 como na 2<\/b><\/span><\/span>,\u00a0<\/b><\/span><\/span>demoram o mesmo tempo para ir de\u00a0<\/b><\/span><\/span>A at\u00e9 B, de B at\u00e9 C, de C at\u00e9 B e de B at\u00e9 A.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

PLANO INCLINADO<\/b><\/span><\/span><\/span><\/p>\n

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

As\u00a0<\/b><\/span><\/span>duas esferas possuem a mesma massa\u00a0<\/b><\/span><\/span>e s\u00e3o\u00a0<\/b><\/span><\/span>abandonadas do repouso da mesma altura h.<\/b><\/span><\/span>\"\"<\/p>\n

A\u00a0<\/b><\/span><\/span>esfera A cai em queda livre\u00a0<\/b><\/span><\/span>sujeita apenas ao\u00a0<\/b><\/span><\/span>seu peso de intensidade P = m.g\u00a0<\/b><\/span><\/span>que \u00e9 a\u00a0<\/b><\/span><\/span>for\u00e7a resultante\u00a0<\/b><\/span><\/span>sobre ele de modo que <\/b><\/span><\/span>FR\u00a0= m.aA<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>FR\u00a0= P\u00a0\"\"\u00a0<\/b><\/span><\/span>m.aA\u00a0= m.g\u00a0\"\"\u00a0<\/b><\/span><\/span>aA\u00a0= g (I)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

C\u00e1lculo\u00a0<\/b><\/span><\/span>da intensidade de PP\u00a0e PN<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>utilizando o\u00a0<\/b><\/span><\/span>tri\u00e2ngulo ret\u00e2ngulo\u00a0<\/b><\/span><\/span>da figura abaixo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

P\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio no\u00a0<\/b><\/span><\/span>trecho BC existe atrito com o Fat<\/b><\/span><\/span>\u00a0<\/b><\/span>valendo\u00a0<\/b><\/span><\/span>Fat\u00a0=\u00a0\"\"mgcos30o.=\u00a0\"\".mg\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>Fat\u00a0=\u00a0\"\"<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

No trecho com atrito\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>sen30o\u00a0= h2\/BC\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0= h2\/BC\u00a0\"\"\u00a0<\/b><\/span><\/span>BC = 2h2<\/b><\/span><\/span><\/span><\/span><\/p>\n

Energia dissipada pelo atrito no<\/b><\/span><\/span>\u00a0<\/b><\/span>trecho BC\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>WFat\u00a0= Fat.BC.cos30o<\/b><\/span><\/span>\u00a0=\u00a0\"\"<\/b><\/span><\/span>.<\/b><\/span><\/span>2h2\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>WFat\u00a0=\u00a0\"\"<\/b><\/span><\/span>.<\/b><\/span><\/span>h2.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Energia mec\u00e2nica no ponto A\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>EmA\u00a0= EcA\u00a0+ EpA<\/b><\/span><\/span>\u00a0<\/b><\/span>= mVA2\/2 + mg(h1\u00a0+ h2) = 0 + mg(h1\u00a0+ h2)\u00a0\"\"\u00a0<\/b><\/span><\/span>EmA\u00a0=mg.(h1\u00a0+ h2).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Energia mec\u00e2nica no ponto C\u00a0<\/b><\/span><\/span>onde, pelo enunciado,<\/b><\/span><\/span>\u00a0chega com velocidade nula\u00a0\"\"\u00a0EmC\u00a0= mVC2\/2 + mgh =\u00a0<\/b><\/span><\/span>m.02\/2 + mg.0\u00a0\"\"\u00a0<\/b><\/span><\/span>EmC\u00a0= 0.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Pelo\u00a0<\/b><\/span><\/span>teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0\"\"\u00a0EmA\u00a0= WFat\u00a0+ EmC\u00a0\"\"\u00a0<\/b><\/span><\/span>mg.(h1\u00a0+ h2) =\u00a0\"\"<\/b><\/span><\/span>.<\/b><\/span><\/span>h2\u00a0+ 0\u00a0\"\"\u00a0(h1\u00a0+ h2)\/h2\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>h1\/h2\u00a0+ 1 =<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>h1\/h2\u00a0=\u00a0\"\"\u00a0\u2013 1 = (3-2)\/2\u00a0\"\"<\/b><\/span><\/span>\u00a0h1\/h2\u00a0=\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

POLIAS E ROLDANAS<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

01-<\/b><\/span><\/span><\/span>C\u00e1lculo da\u00a0<\/b><\/span>intensidade da for\u00e7a\"\"necess\u00e1ria<\/b><\/span>\u00a0para deixar o\u00a0<\/b><\/span>navio na imin\u00eancia do escorregamento<\/b><\/span>\u00a0que deve ser\u00a0<\/b><\/span>maior que o m\u00f3dulo da for\u00e7a de atrito est\u00e1tico m\u00e1xima\"\"\"\"<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

No caso do exerc\u00edcio,\u00a0sendo a praia horizontal N = P\"\"Fate\u00a0=\u00a0\"\"e.N =\u00a0\"\"e.P =\u00a0\"\"e.mg = 0,8x3000x10\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

Fate\u00a0= 24000 N.<\/b><\/span><\/span><\/span><\/span><\/p>\n

A intensidade da for\u00e7a de tra\u00e7\u00e3o\u00a0\"\"\u00a0que puxa o navio tem que ser maior que 24000 N\"\"\u00a0T > 24000 N.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Analise<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>teoria abaixo<\/b><\/span><\/span>\u00a0com\u00a0<\/b><\/span><\/span>uma\u00a0polia fixa\u00a0e\u00a0v\u00e1rias polias m\u00f3veis<\/b><\/span><\/span>\u00a0(talha exponencial):<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

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

Vamos analisar as for\u00e7as desse exerc\u00edcio:<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

A\u00a0<\/b><\/span><\/span>seta vermelha\u00a0<\/b><\/span><\/span>representa a\u00a0<\/b><\/span><\/span>for\u00e7a resultante<\/b><\/span><\/span>, a\u00a0<\/b><\/span><\/span>azul<\/b><\/span><\/span>\u00a0representa a\u00a0<\/b><\/span><\/span>for\u00e7a que a caixa inferior exerce na superior<\/b><\/span><\/span>\u00a0(precisamos descobrir o valor) e a<\/b><\/span><\/span>verde<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>for\u00e7a que a superior exerce na inferior<\/b><\/span><\/span>. A\u00a0<\/b><\/span><\/span>for\u00e7a resultante deve ser igual \u00e0 soma das outras for\u00e7as<\/b><\/span><\/span>, ent\u00e3o:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

FR\u00a0\u00e9 a for\u00e7a resultante (vermelho)<\/b><\/span><\/span><\/span><\/span><\/p>\n

FI\u00a0\u00e9 a for\u00e7a da inferior (azul)<\/b><\/span><\/span><\/span><\/span><\/p>\n

FS\u00a0\u00e9 a for\u00e7a do superior (verde)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Como\u00a0<\/b><\/span><\/span>as for\u00e7as est\u00e3o em dire\u00e7\u00f5es opostas<\/b><\/span><\/span>, precisamos que\u00a0<\/b><\/span><\/span>uma seja negativa<\/b><\/span><\/span>, para facilitar\u00a0<\/b><\/span><\/span>vamos escolher FI<\/b><\/span><\/span>, porque ela\u00a0<\/b><\/span><\/span>est\u00e1 em dire\u00e7\u00e3o oposta em rela\u00e7\u00e3o \u00e0 resultante<\/b><\/span><\/span>, ent\u00e3o:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Isolando FI:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Nesse caso\u00a0<\/b><\/span><\/span>FS\u00a0\u00e9 apenas a for\u00e7a peso da caixa superior<\/b><\/span><\/span>, n\u00f3s j\u00e1 temos ela no enunciado\u00a0<\/b><\/span><\/span>(300 N<\/b><\/span><\/span>). A for\u00e7a resultante \u00e9:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

FR\u00a0\u00e9 a for\u00e7a resultante<\/b><\/span><\/span><\/span><\/span><\/p>\n

m \u00e9 a massa<\/b><\/span><\/span><\/span><\/span><\/p>\n

a \u00e9 acelera\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/span><\/p>\n

No entanto n\u00f3s\u00a0<\/b><\/span><\/span>n\u00e3o sabemos a massa do bloco superior<\/b><\/span><\/span>, podemos<\/b><\/span><\/span>descobrir utilizando a for\u00e7a peso:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

P \u00e9 o peso<\/b><\/span><\/span><\/span><\/span><\/p>\n

m \u00e9 a massa<\/b><\/span><\/span><\/span><\/span><\/p>\n

g \u00e9 a acelera\u00e7\u00e3o da gravidade<\/b><\/span><\/span><\/span><\/span><\/p>\n

Isolando a massa:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

Substituindo na equa\u00e7\u00e3o da for\u00e7a resultante,\u00a0<\/b><\/span><\/span>a massa \u00e9 apenas do bloco superior, porque estudamos as rea\u00e7\u00f5es que acontecem apenas nele<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Agora que temos tudo,\u00a0<\/b><\/span><\/span>podemos substituir para chegarmos ao resultado:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0<\/span>(resposta da a)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Essa \u00e9 a\u00a0<\/b><\/span><\/span>resposta para o exerc\u00edcio a.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Para o b n\u00f3s teremos que\u00a0<\/b><\/span><\/span>fazer exatamente as mesmas contas<\/b><\/span><\/span>, mas dessa vez o\u00a0<\/b><\/span><\/span>peso ser\u00e1 a soma de ambas as caixas e for\u00e7a \u00e9 a que o piso exerce na caixa inferior<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

A rela\u00e7\u00e3o entre as for\u00e7as ainda pode ser escrita dessa forma, por\u00e9m o peso que antes era apenas 300, agora \u00e9 300 + 500 = 800 N.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Calculando a massa da caixa inferior:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

Sendo assim\u00a0<\/b><\/span><\/span>a resultante fica<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

\"\"\u00a0<\/span>(resposta da b)<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

01<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>Verdadeiro\u00a0\"\"<\/b><\/span><\/span>\u00a0pelo fato de o\u00a0<\/b><\/span><\/span>sistema ser constitu\u00eddo por duas roldanas\u00a0<\/b><\/span><\/span>acontece o\u00a0<\/b><\/span><\/span>seguinte equil\u00edbrio de for\u00e7as:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Onde as\u00a0<\/b><\/span><\/span>setas de mesma cor t\u00eam a mesma intensidade.<\/b><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>seta azul representa a for\u00e7a exercida pelo professor (350N).\u00a0<\/b><\/span><\/span>Por ser uma<\/b><\/span><\/span>roldana e pelas propriedades de a\u00e7\u00e3o e rea\u00e7\u00e3o uma outra for\u00e7a igual e de mesma dire\u00e7\u00e3o \u00e9 aplicada no outro fio.<\/b><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>soma dessas duas for\u00e7as\u00a0<\/b><\/span><\/span>resulta na for\u00e7a representada pela\u00a0<\/b><\/span><\/span>seta vermelha,\u00a0<\/b><\/span><\/span>que ter\u00e1\u00a0<\/b><\/span><\/span>intensidade de 350 + 350 = 700N.<\/b><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span>mesmo racioc\u00ednio \u00e9 v\u00e1lido para se calcular a for\u00e7a representada pela seta amarela,\u00a0<\/b><\/span><\/span>s\u00f3 que nesse caso\u00a0<\/b><\/span><\/span>com for\u00e7as de 700N, ou seja, 700 + 700 = 1400N.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A\u00a0<\/b><\/span><\/span>seta vermelha \u00e9 a for\u00e7a normal,\u00a0<\/b><\/span><\/span>que\u00a0<\/b><\/span><\/span>nesse caso tem mesma intensidade da for\u00e7a peso, j\u00e1 que possuem sentidos opostos e n\u00e3o existem outras for\u00e7as agindo verticalmente no bloco<\/b><\/span><\/span>:<\/b><\/span><\/span><\/p>\n

N = P\u00a0\"\"\u00a0N = 4000N.<\/b><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>seta amarela \u00e9 a tra\u00e7\u00e3o, calculada no (01), logo tem intensidade de F=1400N.<\/b><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>seta verde representa a for\u00e7a de atrito<\/b><\/span><\/span>, nesse caso estudaremos o movimento, portanto\u00a0<\/b><\/span><\/span>deveremos considerar o coeficiente de atrito cin\u00e9tico (\u03bcc\u00a0= 0,25).\u00a0<\/b><\/span><\/span>A\u00a0<\/b><\/span><\/span>for\u00e7a de atrito cin\u00e9tico (Fc)\u00a0<\/b><\/span><\/span>pode ser\u00a0<\/b><\/span><\/span>calculada por\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>Fc\u00a0= N.\u03bcc\u00a0\"\"\u00a0Fc = 4000.0,25\u00a0\"\"\u00a0Fc = 1000N.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Essa\u00a0<\/b><\/span><\/span>for\u00e7a ter\u00e1 sentido para a direita,\u00a0<\/b><\/span><\/span>pois\u00a0<\/b><\/span><\/span>a soma das for\u00e7as exercidas horizontalmente no lado direito do corpo ter\u00e3o maior intensidade que as do esquerdo.\u00a0<\/b><\/span><\/span>A\u00a0<\/b><\/span><\/span>intensidade dessa for\u00e7a\u00a0<\/b><\/span><\/span>pode ser calculada por\u00a0\"\"\u00a0<\/b><\/span><\/span>Fr\u00a0= F \u2013 Fc\u00a0\"\"\u00a0<\/b><\/span><\/span>Fr\u00a0= 1400 \u2013 1000\u00a0\"\"\u00a0<\/b><\/span><\/span>Fr\u00a0= 400 N.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

04\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>Falso\u00a0\"\"<\/b><\/span><\/span>\u00a0Como estudado no item\u00a0<\/b><\/span><\/span>anterior (02),\u00a0<\/b><\/span><\/span>existe uma\u00a0<\/b><\/span><\/span>outra for\u00e7a al\u00e9m das citadas, a for\u00e7a normal, que surge do contato entre o corpo e o ch\u00e3o.<\/b><\/span><\/span><\/p>\n

08\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>Verdadeiro\u00a0\"\"\u00a0Tamb\u00e9m demonstrado no item (02),\u00a0<\/b><\/span><\/span>a\u00a0<\/b><\/span><\/span>for\u00e7a resultante \u00e9 de 400N.<\/b><\/span><\/span><\/p>\n

16<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>Falso<\/b><\/span><\/span>\u00a0\"\"\u00a0Nesse caso\u00a0<\/b><\/span><\/span>vamos estudar o objeto antes do movimento<\/b><\/span><\/span>, com isso utilizaremos o\u00a0<\/b><\/span><\/span>coeficiente de atrito est\u00e1tico (\u03bce),<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>for\u00e7a exercida para que o bloco se movimente deve ser maior que a for\u00e7a de atrito est\u00e1tico (Fe), calculada a seguir\u00a0\"\"\u00a0Fe\u00a0= N.\u03bce<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0Fe = 4000.0.3<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/p>\n

Fe\u00a0= 1200 N.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

FOR\u00c7AS NO MCU<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>Veja a\u00a0<\/b><\/span><\/span>teoria abaixo que est\u00e1 relacionada com o exerc\u00edcio\u00a0<\/b><\/span><\/span>onde o <\/b><\/span><\/span>avi\u00e3o (ou o piloto) comportam-se como a moto e a normal\u00a0\"\"\u00a0que a moto troca com o piso do globo pela for\u00e7a que o avi\u00e3o troca com o ar.<\/b><\/span><\/span><\/p>\n

Globo da morte<\/b><\/span><\/span><\/p>\n

Suponha um\u00a0<\/b><\/span><\/span>motociclista de massa total\u00a0m\u00a0(massa dele + massa da moto)<\/b><\/span><\/span>\u00a0descrevendo voltas segundo um\u00a0<\/b><\/span><\/span>plano vertical,<\/b><\/span><\/span>\u00a0com\u00a0<\/b><\/span><\/span>velocidade escalar constante\u00a0\"\"\u00a0no interior de uma esfera met\u00e1lica oca de raio R \u201cglobo da morte\u201d.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

02-<\/b><\/span><\/span><\/span>Quando a\u00a0<\/b><\/span><\/span>corda atinge a tens\u00e3o m\u00e1xima no ponto A (figura),\u00a0<\/b><\/span><\/span>a ela se <\/b><\/span><\/span>rompe<\/b><\/span><\/span>\u00a0e a part\u00edcula com\u00a0<\/b><\/span><\/span>velocidade VA\u00a0sai pela tangente<\/b><\/span><\/span>\u00a0e executa um <\/b><\/span><\/span>lan\u00e7amento horizontal com velocidade VA, com trajet\u00f3ria parab\u00f3lica at\u00e9 atingir o solo percorrendo a dist\u00e2ncia horizontal 4R e\u00a0<\/b><\/span><\/span>vertical 2R.<\/b><\/span><\/span><\/p>\n

Decomp\u00f5e-se o movimento horizontal em\u00a0<\/b><\/span><\/span>duas parcelas:<\/b><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Segundo o eixo Y<\/b><\/span><\/span>\u00a0\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span>trata-se de um\u00a0<\/b><\/span><\/span>movimento uniformemente variado com velocidade inicial\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>Voy\u00a0= 0,<\/b><\/span><\/span>\u00a0ou seja, \u00e9 uma\u00a0<\/b><\/span><\/span>queda livre com o corpo abandonado da origem, sujeito<\/b><\/span><\/span><\/p>\n

apenas \u00e0 acelera\u00e7\u00e3o da<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>gravidade, de intensidade\u00a0g, dire\u00e7\u00e3o vertical e sentido para baixo<\/b><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/p>\n

Equa\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/p>\n

S = So\u00a0+ Vo.t + at2\/2 \u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>Y= 0 + 0.t + gt2\/2 \u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Y = g.t2\/2<\/b><\/span><\/span><\/span><\/p>\n

Vy\u00a0= Voy\u00a0+ a.t\u00a0<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span><\/span>Vy= 0 + g.t\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>Vy\u00a0= g.t<\/b><\/span><\/span><\/span><\/p>\n

Na\u00a0<\/b><\/span><\/span>vertical,<\/b><\/span><\/span>\u00a0a part\u00edcula em\u00a0<\/b><\/span><\/span>queda livre<\/b><\/span><\/span>, sai do\u00a0<\/b><\/span><\/span>ponto A com Vo\u00a0= 0\u00a0<\/b><\/span><\/span>e percorre a\u00a0<\/b><\/span><\/span>dist\u00e2ncia vertical Y =<\/b><\/span><\/span>\"\"<\/p>\n

2R\u00a0<\/b><\/span><\/span>at\u00e9\u00a0<\/b><\/span><\/span>chegar ao solo e demora um tempo\u00a0<\/b><\/span><\/span>t,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>tal que\u00a0<\/b><\/span><\/span>Y = g.t2\/2\u00a0\"\"\u00a0<\/b><\/span><\/span>2R = g.\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>t2\u00a0=\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>t = 2\"\"<\/b><\/span><\/span><\/p>\n

(tempo de queda na vertical\u00a0<\/b><\/span><\/span>que \u00e9\u00a0<\/b><\/span><\/span>o mesmo tempo\u00a0<\/b><\/span><\/span>que a<\/b><\/span><\/span>part\u00edcula demora para percorrer a dist\u00e2ncia d = 4R na horizontal)<\/b><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Segundo o eixo X\u00a0\"\"<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span>trata-se de um\u00a0<\/b><\/span><\/span>movimento horizontal uniforme com<\/b><\/span><\/span>velocidade constante\u00a0<\/b><\/span><\/span>de<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>intensidade\u00a0VA\u00a0<\/b><\/span><\/span>, que \u00e9 a\u00a0<\/b><\/span><\/span>velocidade de lan\u00e7amento\u00a0<\/b><\/span><\/span>\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span>S = So\u00a0+ V.t \u00a0\"\"<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span>X= 0 + VA.t \u00a0\"\"<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span>X = VA.t.<\/b><\/span><\/span><\/p>\n

O tempo t que ela demora para percorrer x = 4R na horizontal<\/b><\/span><\/span>\u00a0\u00e9 o\u00a0<\/b><\/span><\/span>mesmo que demorou na queda livre<\/b><\/span><\/span>\"\"<\/p>\n

\"\"<\/p>\n

\"\"R \u2013 C<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

P \u2013 N =\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0como a\u00a0<\/b><\/span><\/span>for\u00e7a exercida pela pista sobre o carro tem intensidade N tal que\u00a0\"\"\u00a0<\/b><\/span><\/span>– N =\u00a0\"\"\u00a0\u2013 P\u00a0\"\"\u00a0<\/b><\/span><\/span>N = Mg \u2013\u00a0\"\".<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

TRABALHO E ENERGIA<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

C\u00e1lculo da\u00a0<\/b><\/span><\/span>pot\u00eancia te\u00f3rica (Pt)<\/b><\/span><\/span>\u00a0de cada\u00a0<\/b><\/span><\/span>usina<\/b><\/span><\/span>\u00a0(unidade geradora) que leva em conta\u00a0<\/b><\/span><\/span>a altura (h) da massa m de \u00e1gua represada pela barragem,\u00a0a gravidade local (g = 10 m\/s2) e a densidade da \u00e1gua (da= 1 000 kg\/m3)\u00a0\"\"Pt\u00a0= W<\/b><\/span><\/span>\u00a0(energia potencial gravitacional armazenada pela massa m de \u00e1gua na<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sendo a vaz\u00e3o nominal da \u00e1gua de Z = 690 m3\/s = Volume\/intervalo de tempo (1 s)\u00a0\"\"\u00a0Z = V\/\u2206t\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

690 = V\/1\u00a0\"\"\u00a0V = 690 m3\u00a0(volume de \u00e1gua que cai em cada 1s).<\/b><\/span><\/span><\/span><\/span><\/p>\n

d\u00e1gua\u00a0=\u00a0\"\"\"\"<\/b><\/span><\/span>1000 =\u00a0\"\"\"\"<\/b><\/span><\/span>m =69.104\u00a0kg (massa de \u00e1gua).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Energia potencial gravitacional<\/b><\/span><\/span>\u00a0liberada por\u00a0<\/b><\/span><\/span>m = 69.104\u00a0kg<\/b><\/span><\/span>\u00a0de \u00e1gua numa<\/b><\/span><\/span>altura de h = 118,4<\/b><\/span><\/span>\u00a0m num local onde\u00a0<\/b><\/span><\/span>g = 10 m\/s2\u00a0\"\"W = m.g.h<\/b><\/span><\/span>= 69.104.10.118,4 = 8169,6.105\u00a0<\/b><\/span><\/span>= 816,96.106\u00a0J.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Pot\u00eancia te\u00f3rica (Pt)<\/b><\/span><\/span>\u00a0de cada\u00a0<\/b><\/span><\/span>usina<\/b><\/span><\/span>\u00a0(unidade geradora)\u00a0\"\"Pt\u00a0=\u00a0\"\"=\"\"\"\"<\/b><\/span><\/span>Pt\u00a0=816,96.106\u00a0W<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"Pt = 816,96 MW (pot\u00eancia te\u00f3rica).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Pot\u00eancia n\u00e3o aproveitada em cada unidade geradora de Itaipu = pot\u00eancia te\u00f3rica \u2013 pot\u00eancia instalada<\/b><\/span><\/span><\/span><\/span><\/p>\n

Pperdida<\/b><\/span><\/span>= 816,96MW \u2013\u00a0700MW\u00a0\"\"<\/b><\/span><\/span>Pperdida<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>=\u00a0116,96 MW
\n<\/b><\/span><\/span>R \u2013 C<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

\"\"<\/p>\n

Energia mec\u00e2nica no ponto R\u00a0<\/b><\/span><\/span>onde a mola est\u00e1\u00a0<\/b><\/span><\/span>deformada de x = 10 m, h= 0\u00a0<\/b><\/span><\/span>e o corpo tem\u00a0<\/b><\/span><\/span>velocidade<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>E<\/b><\/span><\/span>mR<\/b><\/span><\/span>\u00a0= m.\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>+ mgh + k.\"\"<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mR<\/b><\/span><\/span><\/span>\u00a0= 50.\"\"<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>+ 50.10.0 + 250.\"\"<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mR<\/b><\/span><\/span><\/span>\u00a0= 25.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>+ 0 + 125.100\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mR<\/b><\/span><\/span><\/span>\u00a0= 25.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span>+12500.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

Sistema conservativo<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>E<\/b><\/span><\/span>mP<\/b><\/span><\/span>\u00a0= E<\/b><\/span><\/span>mR<\/b><\/span><\/span>\u00a0\"\"\u00a012500 = 25.V<\/b><\/span><\/span>2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>+12500\u00a0\"\"\u00a025V<\/b><\/span><\/span>2<\/b><\/span><\/span>\u00a0= 0\u00a0\"\"\u00a0<\/b><\/span><\/span>V = 0 e este \u00e9 o ponto onde a deforma\u00e7\u00e3o da mola \u00e9 m\u00e1xima.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R \u2013 A<\/b><\/span><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>Como em qualquer choque\u00a0<\/b><\/span><\/span>a quantidade de movimento \u00e9 conservada<\/b><\/span><\/span>, j\u00e1 podemos excluir algumas alternativas. Agora s\u00f3 falta resolvermos\u00a0<\/b><\/span><\/span>o problema da energia cin\u00e9tica<\/b><\/span><\/span>, o texto deixa bem claro que\u00a0<\/b><\/span><\/span>as velocidades n\u00e3o mudam, antes e depois do choque<\/b><\/span><\/span>, como a\u00a0<\/b><\/span><\/span>massa tamb\u00e9m se mant\u00e9m<\/b><\/span><\/span>, podemos concluir que\u00a0<\/b><\/span><\/span>a energia cin\u00e9tica n\u00e3o muda<\/b><\/span><\/span>, portanto,\u00a0<\/b><\/span><\/span>a alternativa E \u00e9 a correta<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R \u2013 E<\/b><\/span><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>a)\u00a0<\/b><\/span><\/span>Considerando o sistema \u201cproj\u00e9til, saco de areia e carrinho\u201d<\/b><\/span><\/span>\u00a0isolado de\u00a0<\/b><\/span><\/span>for\u00e7as externas<\/b><\/span><\/span>voc\u00ea pode aplicar o\u00a0<\/b><\/span><\/span>teorema da conserva\u00e7\u00e3o da quantidade de movimento.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Q<\/b><\/span><\/span>sa<\/b><\/span>\u00a0= Q<\/b><\/span><\/span>sd<\/b><\/span>\u00a0\"\"\u00a010 = 1,6 + 100 V\u00a0\"\"\u00a0V =\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>V = 8,4.10<\/b><\/span><\/span>-2<\/b><\/span>\u00a0m\/s<\/b><\/span><\/span><\/p>\n

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

Teorema da energia cin\u00e9tica:<\/b><\/span><\/span><\/p>\n

\u201c\u00a0<\/span>O<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>trabalho da resultant<\/b><\/span><\/span>e<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>de<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>todas as for\u00e7as<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>que agem sobre um corpo<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00e9 igual \u00e0 varia\u00e7\u00e3o da energia cin\u00e9tica sofrida pelo corpo \u201c<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

W<\/b><\/span><\/span>FR<\/b><\/span>\u00a0= \u2206E<\/b><\/span><\/span>c<\/b><\/span>\u00a0= E<\/b><\/span><\/span>cd<\/b><\/span>\u00a0\u2013 E<\/b><\/span><\/span>ca<\/b><\/span>\u00a0= m<\/b><\/span><\/span>p<\/b><\/span>.(V<\/b><\/span><\/span>pd<\/b><\/span>)<\/b><\/span><\/span>2<\/b><\/sup><\/span>\/<\/b><\/span>2 \u2013 m<\/b><\/span><\/span>p<\/b><\/span>.(V<\/b><\/span><\/span>pa<\/b><\/span>)<\/b><\/span><\/span>2<\/b><\/sup><\/span>\/2<\/b><\/span><\/span>\u00a0\"\"\u00a0W<\/b><\/span><\/span>FR<\/b><\/span>\u00a0= 20.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span> <\/b>.80<\/b><\/span><\/span>2<\/b><\/sup><\/span>\/<\/b><\/span>\u00a02 \u2013 20.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span> <\/b>.500<\/b><\/span><\/span>2<\/b><\/sup><\/span>\/<\/b><\/span>\u00a02 = 64 \u2013 2500\u00a0\"\"\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>FR<\/b><\/span>\u00a0= \u2013 2435 J<\/b><\/span><\/span>\u00a0(o sinal negativo significa que a energia cin\u00e9tica do proj\u00e9til foi dissipada).<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>Pelo enunciado o\u00a0<\/b><\/span><\/span>aproveitamento<\/b><\/span><\/span>\u00a0da energia e\u00f3lica ocorre por\u00a0<\/b><\/span><\/span>meio da convers\u00e3o da\u00a0<\/b><\/span><\/span>energia cin\u00e9tica\u00a0<\/b><\/span><\/span>(E<\/b><\/span><\/span>c<\/b><\/span><\/span>) de transla\u00e7\u00e3o em\u00a0<\/b><\/span><\/span>energia cin\u00e9tica <\/b><\/span><\/span>de<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>rota\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>an\u00e1lise das turbinas e\u00f3licas<\/b><\/span><\/span>\u00a0\u00e9\u00a0<\/b><\/span><\/span>complexa porque envolve aerodin\u00e2mica e mec\u00e2nica dos fluidos.<\/b><\/span><\/span><\/p>\n

Contudo, os\u00a0<\/b><\/span><\/span>conceitos b\u00e1sicos podem ser obtidos a partir de simplifica\u00e7\u00f5es na an\u00e1lise.<\/b><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>energia contida no vento horizontal \u00e9 basicamente sua\u00a0<\/b><\/span><\/span>energia cin\u00e9tica, <\/b><\/span><\/span>uma vez que\u00a0<\/b><\/span><\/span>n\u00e3o h\u00e1<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>varia\u00e7\u00e3o de press\u00e3o, varia\u00e7\u00e3o de altitude,<\/b><\/span><\/span>\u00a0e, pelo enunciado, a\u00a0<\/b><\/span><\/span>densidade do ar permanece<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>constante<\/b><\/span><\/span>\u00a0o que implica que a<\/b><\/span><\/span>massa de ar\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>\u00a0que atravessa os permanece constante<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

A<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>energia cin\u00e9tica<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00e9 uma<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>grandeza escalar,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>n\u00e3o tem dire\u00e7\u00e3o nem sentido,<\/b><\/span><\/span>apenas intensidade, fornecida pela express\u00e3o:<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

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

Propriedade do gr\u00e1fico for\u00e7axdeslocamento<\/b><\/span><\/span><\/p>\n

Em\u00a0<\/b><\/span><\/span>todo gr\u00e1fico da for\u00e7a\u00a0F\u00a0em fun\u00e7\u00e3o do deslocamento\u00a0d, o\u00a0trabalho realizado pela for\u00e7a F \u00e9<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>numericamente\u00a0 igual \u00e0 \u00e1rea<\/b><\/span><\/span>\u00a0compreendida entre a\u00a0<\/b><\/span><\/span>reta representativa (linha cheia) e o eixo do deslocamento.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Se, ao final desse\u00a0<\/b><\/span><\/span>deslocamento de d= 50 m\u00a0<\/b><\/span><\/span>a intensidade dessa\u00a0<\/b><\/span><\/span>for\u00e7a F (suposta resultante) fosse mantida constante (F = 310 N), o trabalho <\/b><\/span><\/span>realizado por ela seria\u00a0\"\"<\/b><\/span><\/span>\u00a0W = F.d\u00a0<\/b><\/span><\/span>= 310.50\u00a0\"\"<\/b><\/span><\/span><\/p>\n

W = 15 500 J.<\/b><\/span><\/span><\/p>\n

Supondo que essa for\u00e7a F seja a for\u00e7a resultante voc\u00ea pode aplicar o teorema do Trabalho como varia\u00e7\u00e3o da energia cin\u00e9tica:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

WF\u00a0= Ecf\u00a0\u2013 Eci\u00a0= Ecf\u00a0\u2013 mVo<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\/2\u00a0<\/b><\/span><\/span>= mV<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\/2 \u2013 m0<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\/2 = Ecf\u00a0\u2013 0\u00a0\"\"\u00a0<\/b><\/span><\/span>Ecf\u00a0= 15 500 J.<\/b><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

 <\/p>\n

08-<\/b><\/span><\/span><\/span> Observe na\u00a0<\/b><\/span><\/span>figura a\u00a0<\/b><\/span><\/span>que a for\u00e7a\u00a0\"\"\u00a0\u00e9\u00a0<\/b><\/span><\/span>decomposta<\/b><\/span><\/span>\u00a0em\u00a0<\/b><\/span><\/span>duas parcelas\u00a0\"\"e\"\"\u00a0<\/b><\/span><\/span>e que a\u00a0<\/b><\/span><\/span>parcela respons\u00e1vel pela realiza\u00e7\u00e3o do trabalho no deslocamento \u00e9 Fx\u00a0tal que Fx\u00a0< F.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Em\u00a0<\/b><\/span><\/span>b a for\u00e7a que age no deslocamento \u00e9\u00a0\"\".<\/b><\/span><\/span><\/span><\/span><\/p>\n

Como o\u00a0<\/b><\/span><\/span>trabalho\u00a0<\/b><\/span><\/span>em a vale\u00a0<\/b><\/span><\/span>Wa\u00a0=Fx.d\u00a0<\/b><\/span><\/span>e em\u00a0<\/b><\/span><\/span>b<\/b><\/span><\/span>\u00a0vale,\u00a0<\/b><\/span><\/span>Wb\u00a0= F.d\u00a0\"\"\u00a0Wa\u00a0< Wb.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

Essa \u00e9 a\u00a0<\/b><\/span><\/span>express\u00e3o da energia cin\u00e9tica<\/b><\/span><\/span>\u00a0de um corpo de\u00a0<\/b><\/span><\/span>massa m<\/b><\/span><\/span>\u00a0no instante em que ele possui<\/b><\/span><\/span>velocidade<\/b><\/span><\/span>\u00a0de<\/b><\/span><\/span><\/p>\n

intensidade V<\/b><\/span><\/span><\/p>\n

No caso do exerc\u00edcio:<\/b><\/span><\/span><\/p>\n

Massa do navio\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>navio<\/b><\/span>\u00a0= 450.000 kg\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>massa das pessoas\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>pessoas<\/b><\/span>\u00a0= 1000\u00d770\u00a0<\/b><\/span><\/span>= 70.000 kg<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/p>\n

Massa dos carros mcarros\u00a0= 150\u00d71000 = 150.000 kg.<\/b><\/span><\/span><\/p>\n

Massa total do sistema\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>=\u00a0<\/b><\/span><\/span>450000 + 70000 + 150000\u00a0<\/b><\/span><\/span>=<\/b><\/span><\/span><\/p>\n

670.000 kg.\u00a0\"\"
\n\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"\u00a0\"\"\u00a0\"\"\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0301,5 MJ<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>O\u00a0<\/b><\/span><\/span>enunciado afirma<\/b><\/span><\/span>\u00a0que a esfera est\u00e1 em\u00a0<\/b><\/span><\/span>repouso na posi\u00e7\u00e3o x = x1<\/b><\/span><\/span>\u00a0<\/b><\/span>o que implica que,\u00a0<\/b><\/span><\/span>nessa posi\u00e7\u00e3o, sua energia cin\u00e9tica \u00e9 nula<\/b><\/span><\/span>\u00a0(Ec\u00a0= mV2\/2 = m.02\/2=0).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Sendo os\u00a0<\/b><\/span><\/span>efeitos dissipativos desprez\u00edveis\u00a0<\/b><\/span><\/span>voc\u00ea pode aplicar o\u00a0<\/b><\/span><\/span>Princ\u00edpio da Conserva\u00e7\u00e3o da energia Mec\u00e2nica em que, para\u00a0<\/b><\/span><\/span>qualquer posi\u00e7\u00e3o da trajet\u00f3ria, a energia mec\u00e2nica (Em) \u00e9 constante<\/b><\/span><\/span>,<\/b><\/span><\/span>\u00a0ou seja, a\u00a0<\/b><\/span><\/span>soma das energias cin\u00e9tica (E\u00adc) e potencial gravitacional (Ep) em cada ponto fornece sempre o mesmo valor.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Assim, se\u00a0<\/b><\/span><\/span>Em\u00a0= Ec\u00a0+ Ep\u00a0= constante<\/b><\/span><\/span>, ent\u00e3o no ponto\u00a0<\/b><\/span><\/span>x = x1\u00a0onde a cin\u00e9tica \u00e9 nula a\u00a0<\/b><\/span><\/span>potencial ser\u00e1 m\u00e1xima.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Da mesma maneira, se\u00a0<\/b><\/span><\/span>Em\u00a0= Ec\u00a0+ Ep\u00a0= constante<\/b><\/span><\/span>, ent\u00e3o no ponto\u00a0<\/b><\/span><\/span>x = x2onde a energia potencial gravitacional \u00e9 m\u00ednima (veja gr\u00e1fico) e de valor Uo,\u00a0<\/b><\/span><\/span>a cin\u00e9tica ser\u00e1 m\u00e1xima.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

DIN\u00c2MICA IMPULSIVA<\/b><\/span><\/span><\/span><\/p>\n

01<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span>C\u00e1lculo das\u00a0<\/b><\/span><\/span>velocidades dos carrinhos 1 e 2 de massas m1\u00a0= 150 g e m2\u00a0pedida, <\/b><\/span><\/span>antes do choque,<\/b><\/span><\/span>\u00a0registrados nos\u00a0<\/b><\/span><\/span>instantes O e 1 s fornecidos no quadro:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Ap\u00f3s o choque,<\/b><\/span><\/span>\u00a0pelo enunciado eles se\u00a0<\/b><\/span><\/span>movem unidos<\/b><\/span><\/span>\u00a0(choque inel\u00e1stico) com\u00a0<\/b><\/span><\/span>massa (M = 150 +m2),<\/b><\/span><\/span>\u00a0com a\u00a0<\/b><\/span><\/span>mesma velocidade V\u2019<\/b><\/span><\/span>\u00a0fornecida no quadro nos instantes\u00a0<\/b><\/span><\/span>8s e 11 s:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Aplicando o\u00a0<\/b><\/span><\/span>teorema da conserva\u00e7\u00e3o da quantidade de movimento:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Quantidade de mov<\/b><\/span><\/span>imento do sistema\u00a0<\/b><\/span><\/span>antes do choque \"\" Qsa\u00a0= m1.V1\u00a0+ m2.V2<\/b><\/span><\/span>\u00a0= 150.15 + m2.0\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

Qsa\u00a0= 2250 + 0 \"\" <\/b><\/span><\/span>Qsa\u00a0= 2250 g.cm\/s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Quantidade de mov<\/b><\/span><\/span>imento do sistema\u00a0<\/b><\/span><\/span>depois do choque \"\"Qsd\u00a0= M.V\u2019<\/b><\/span><\/span>\u00a0= (150 + m2).5 = 750 + 5m2 \"\" <\/b><\/span><\/span>Qsd\u00a0= 750 + 5m2.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Qsa\u00a0= Qsd \"\" 2250 = 750 + 5m2\"\"<\/b><\/span><\/span>\u00a0m2\u00a0= 1500\/5\u00a0\"\"<\/b><\/span><\/span>\u00a0m2\u00a0= 300 g.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

02-<\/b><\/span><\/span><\/span>As\u00a0<\/b><\/span><\/span>for\u00e7as trocadas\u00a0<\/b><\/span><\/span>entre o\u00a0<\/b><\/span><\/span>trem e a camionete\u00a0<\/b><\/span><\/span>obedecem ao\u00a0<\/b><\/span><\/span>princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o<\/b><\/span><\/span>, ou seja, tem sempre a\u00a0<\/b><\/span><\/span>mesma intensidade, mesma dire\u00e7\u00e3o, mas sentidos contr\u00e1rios.<\/b><\/span><\/span><\/p>\n

Choque inel\u00e1stico<\/b><\/span><\/span><\/p>\n

Neste tipo de choque a\u00a0<\/b><\/span><\/span>dissipa\u00e7\u00e3o de energia \u00e9 m\u00e1xima,<\/b><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span>coeficiente de restitui\u00e7\u00e3o \u00e9 nulo<\/b><\/span><\/span>, e,\u00a0<\/b><\/span><\/span>ap\u00f3s<\/b><\/span><\/span>\u00a0o choque, os\u00a0<\/b><\/span><\/span>corpos obrigatoriamente se juntam e se movem unidos com a mesma velocidade.<\/b><\/span><\/span>\u00a0Lembre-se de que em\u00a0<\/b><\/span><\/span>qualquer tipo\u00a0<\/b><\/span><\/span>de choque a\u00a0<\/b><\/span><\/span>quantidade de movimento sempre se conserva<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

04-<\/b><\/span><\/span><\/span>Definimos<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>impulso\u00a0da for\u00e7a\u00a0\"\"\u00a0<\/b><\/span><\/span>no\u00a0<\/b><\/span><\/span>intervalo de tempo\u00a0\u2206t<\/b><\/span><\/span>\u00a0como sendo\u00a0a\u00a0<\/b><\/span><\/span>grandeza vetorial:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como\u00a0<\/b><\/span><\/span>\u2206t<\/b><\/span><\/span>\u00a0\u00e9 uma\u00a0<\/b><\/span><\/span>grandeza escalar positiva<\/b><\/span><\/span>, o\u00a0<\/b><\/span><\/span>impulso\u00a0I<\/b><\/span><\/span>\u00a0tem sempre a mesma dire\u00e7\u00e3o e mesmo sentido<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>que a<\/b><\/span><\/span>\u00a0for\u00e7a\u00a0\"\"e, com as\u00a0<\/b><\/span><\/span>caracter\u00edsticas abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe no\u00a0<\/b><\/span><\/span>gr\u00e1fico<\/b><\/span><\/span>\u00a0que, no\u00a0<\/b><\/span><\/span>intervalo de tempo entre 5 s e 105 s,<\/b><\/span><\/span>\u00a0a for\u00e7a de\u00a0<\/b><\/span><\/span>mant\u00e9m constante\u00a0<\/b><\/span><\/span>em\u00a0<\/b><\/span><\/span>100N,\u00a0<\/b><\/span><\/span>assim, em um intervalo de tempo de\u00a0<\/b><\/span><\/span>\u0394t =<\/b><\/span><\/span>105 \u2013 5 =\u00a0<\/b><\/span><\/span>100 s,\u00a0<\/b><\/span><\/span>aplicando a f\u00f3rmula do impulso, em modulo, temos\u00a0\"\"\u00a0<\/b><\/span><\/span>I = F.\u0394t\u00a0<\/b><\/span><\/span>= 100.100\u00a0\"\"\u00a0<\/b><\/span><\/span>I = 10.000 N.s = 104\u00a0N.s.<\/b><\/span><\/span><\/p>\n

OBS:<\/b><\/span><\/span>\u00a0repare que o exerc\u00edcio pede somente a\u00a0<\/b><\/span><\/span>intensidade do impulso<\/b><\/span><\/span>, mesmo ela sendo uma\u00a0<\/b><\/span><\/span>grandeza VETORIAL\u00a0<\/b><\/span><\/span>(possui intensidade, dire\u00e7\u00e3o e sentido).<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>I. Correta\u00a0\"\"\u00a0<\/b><\/span><\/span>Quando temos<\/b><\/span><\/span>\u00a0uma colis\u00e3o perfeitamente el\u00e1stica <\/b><\/span><\/span>significa que<\/b><\/span><\/span>\u00a0o sistema \u00e9 conservativo,\u00a0<\/b><\/span><\/span>isto \u00e9<\/b><\/span><\/span>, a quantidade de movimento e a energia cin\u00e9tica permanecem constantes.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

IV. Errada\u00a0\"\"\u00a0<\/b><\/span><\/span>Analogamente<\/b><\/span><\/span>\u00a0a afirma\u00e7\u00e3o III, como m1\u00a0< m2\u00a0\"\"\u00a0V\u20192\u00a0< V\u20191,<\/b><\/span><\/span>por\u00e9m,<\/b><\/span><\/span>\u00a0V\u20191\u00a0ser\u00e1 menor que V1.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Sendo a\u00a0<\/b><\/span><\/span>colis\u00e3o el\u00e1stica a energia total (cin\u00e9tica do sistema constitu\u00eddo pelas 5 esferas) \u00e9 a mesma<\/b><\/span><\/span>\u00a0<\/b><\/span>antes e depois do choque<\/b><\/span><\/span>\u00a0e o<\/b><\/span><\/span>coeficiente de restitui\u00e7\u00e3o vale e=1<\/b><\/span><\/span>. Como em\u00a0<\/b><\/span><\/span>todo em tipo de choque, a<\/b><\/span><\/span>quantidade de movimento do sistema tamb\u00e9m se conserva.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Se voc\u00ea\u00a0<\/b><\/span><\/span>chamar\u00a0<\/b><\/span><\/span>de m a massa de cada esfera e de V sua respectiva velocidade<\/b><\/span><\/span>, o\u00a0<\/b><\/span><\/span>m\u00f3dulo<\/b><\/span><\/span>\u00a0da<\/b><\/span><\/span>\u00a0<\/b><\/span>quantidade de movimento do sistema antes do choque<\/b><\/span><\/span>\u00a0(com as 3 esferas se movendo para a direita) \u00e9\u00a0<\/b><\/span><\/span>Qsa<\/b><\/span><\/span>\u00a0= mV + mV = mV\u00a0<\/b><\/span><\/span>= 3mV.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Assim, o\u00a0<\/b><\/span><\/span>m\u00f3dulo da quantidade de movimento<\/b><\/span><\/span>\u00a0do sistema\u00a0<\/b><\/span><\/span>depois do choque deve ser tamb\u00e9m Qsd\u00a0=<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

3mV\u00a0<\/b><\/span><\/span>\u00e9 a \u00fanica alternativa que satisfaz \u00e9 a C.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

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

\"\"<\/p>\n

 <\/p>\n

08-<\/b><\/span><\/span><\/span>Supondo a cabe\u00e7a do jogador parada\u00a0<\/b><\/span><\/span>(obst\u00e1culo fixo),<\/b><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span>m\u00f3dulo quantidade de movimento<\/b><\/span><\/span>\u00a0da bola\u00a0<\/b><\/span><\/span>antes do choque, quando incide verticalmente com velocidade de m\u00f3dulo Vvertical<\/b><\/span><\/span>\u00a0ser\u00e1:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Qsa\u00a0= mVvertical, com dire\u00e7\u00e3o vertical e sentido para baixo.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Supondo a cabe\u00e7a do jogador parada\u00a0<\/b><\/span><\/span>(obst\u00e1culo fixo)<\/b><\/span><\/span>, o m\u00f3dulo\u00a0<\/b><\/span><\/span>da quantidade de movimento<\/b><\/span><\/span>\u00a0da bola\u00a0<\/b><\/span><\/span>depois do choque, quando retorna horizontalmente com velocidade de m\u00f3dulo Vhorizontal\u00a0ser\u00e1:<\/b><\/span><\/span><\/p>\n

Qsd\u00a0= mVhorizontal, com dire\u00e7\u00e3o horizontal e sentido para a direita.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

MOVIMENTO HARM\u00d4NICO SIMPLES<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

02-<\/b><\/span><\/span><\/span>O\u00a0<\/b><\/span><\/span>per\u00edodo T<\/b><\/span><\/span>\u00a0(tempo que o oscilador massa-mola demora para efetuar uma oscila\u00e7\u00e3o \u201cvai e vem\u201d completa)\u00a0<\/b><\/span><\/span>\u00e9 fornecido por:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

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

01\u00a0\"\"\u00a0Falso\u00a0\"\"\u00a0O\u00a0<\/b><\/span><\/span>per\u00edodo<\/b><\/span><\/span>\u00a0e a\u00a0<\/b><\/span><\/span>frequ\u00eancia<\/b><\/span><\/span>\u00a0dependem apenas do <\/b><\/span><\/span>n\u00famero de ciclos\u00a0<\/b><\/span><\/span>e do\u00a0<\/b><\/span><\/span>intervalo de tempo<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

02\u00a0\"\"\u00a0Verdadeiro\u00a0\"\"\u00a0A\u00a0<\/b><\/span><\/span>frequ\u00eancia<\/b><\/span><\/span>\u00a0no MHS pode ser calculada como:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Onde T \u00e9 o\u00a0<\/b><\/span><\/span>per\u00edodo<\/b><\/span><\/span>, m a\u00a0<\/b><\/span><\/span>massa desejada<\/b><\/span><\/span>\u00a0e k a\u00a0<\/b><\/span><\/span>constante el\u00e1stica<\/b><\/span><\/span>.Para calcularmos a frequ\u00eancia, precisamos elevar toda a equa\u00e7\u00e3o por -1, pois a frequ\u00eancia \u00e9 o inverso do per\u00edodo, nesse caso apenas vamos inverter as fra\u00e7\u00f5es.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Colocando os valores:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Transformando 25.104\u00a0em 2500.102:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Saindo da raiz:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h2>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00f5es-Vestibulares Recentes por assunto 2017, 2016 e 2015 APLICA\u00c7\u00d5ES DAS TR\u00caS LEIS DE NEWTON 01-Nesse caso o objeto tem uma\u00a0velocidade constante, logo\u00a0sua acelera\u00e7\u00e3o \u00e9 0. Como sua acelera\u00e7\u00e3o \u00e9 0,\u00a0a for\u00e7a resultante tamb\u00e9m s\u00f3 pode ser 0, com isso\u00a0a alternativa correta \u00e9 a D. Vamos analisar as outras, tamb\u00e9m: A\u00a0velocidade tem sempre a mesma dire\u00e7\u00e3o e sentido da for\u00e7a resultante,<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3125,"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-3128","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3128","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=3128"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3128\/revisions"}],"predecessor-version":[{"id":11081,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3128\/revisions\/11081"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3125"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3128"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}