{"id":1374,"date":"2015-09-24T11:36:22","date_gmt":"2015-09-24T11:36:22","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1374"},"modified":"2024-08-23T13:01:25","modified_gmt":"2024-08-23T13:01:25","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-energia-mecanica","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/energia-mecanica\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-energia-mecanica\/","title":{"rendered":"Resolu\u00e7\u00e3o Energia Mec\u00e2nica"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

 <\/p>\n

\u00a0<\/span>01<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>Como n\u00e3o existe atrito a energia mec\u00e2nica se conserva e onde a energia cin\u00e9tica \u00e9 m\u00e1xima (ponto R), a velocidade \u00e9 m\u00e1xima e a energia potencial gravitacional \u00e9 m\u00ednima (altura m\u00ednima). Np ponto P a energia cin\u00e9tica \u00e9 m\u00ednima (menor velocidade) e a energia potencial gravitacional \u00e9 m\u00e1xima (maior altura).<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>Como a energia mec\u00e2nica se conserva, a energia mec\u00e2nica em P, que \u00e9 a cin\u00e9tica (m.v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2) mais a potencial gravitacional (m.g.h) \u00e9 a mesma que chega \u00e0 turbina<\/b><\/span><\/span><\/span>. R- A<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>V=720km\/h\/3,6=200m\/s — h=3.103m — Em=70.106J — Em=m.v2\/2 + m.g.h — 70.106=m.(200)2\/2 + m.10.3.103 — 70.106=m.4.104\/2 + m.3.104 — m=70.106\/5.104 —\u00a0<\/b><\/span><\/span><\/span>m=1.400kg<\/b><\/span><\/span><\/span>\u00a0—<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0C\u00e1lculo da velocidade com que ela chega ao solo\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ g.t=0 + 10.4\u00a0 —\u00a0 V=40m\/s\u00a0 —\u00a0 como se despreza a resist\u00eancia do ar, o sistema \u00e9 conservativo e a energia mec\u00e2nica \u00e9 a mesma em A e em B.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.h=m.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.10.h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=10.m.h\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.10.0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=m.40<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=<\/b><\/span><\/span><\/span>800m<\/b><\/span><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>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<\/b><\/span><\/span><\/span><\/p>\n

10mh=800m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>h=80m R- C<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0Se a velocidade \u00e9 constante, a energia cin\u00e9tica \u00e9 constante. Como a altura diminui a energia potencial gravitacional tamb\u00e9m diminui\u00a0 R- B<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0Observe na figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 +mgh=m(29\/3,6)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 0=\u00a0<\/b><\/span><\/span><\/span>32,45m<\/b><\/span><\/span><\/span>\u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB\u00a0<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh=0 + m.10.h=<\/b><\/span><\/span><\/span>10mh<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mB\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>—\u00a0 32,45m=10mh\u00a0 —\u00a0 h=32,45\/10\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 h=3,245m\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0A menor velocidade que ele deve ter no ponto A deve somente ocorrer se ele chegar em B com velocidade nula.<\/b><\/span><\/span><\/span><\/p>\n

Pela conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pA<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>cA<\/b><\/span><\/span><\/sub><\/span>=mgh + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=m.10.8 + m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=<\/b><\/span><\/span><\/span>80m + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0= mgh=m.10.13=<\/b><\/span><\/span><\/span>130m<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mB\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 80m + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=130m\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=50\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V=10m\/s\u00a0<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0antes de atingir a mola \u2013 E<\/b><\/span><\/span><\/span>ma<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 depois de comprimir a mola e parar (v=0), quando a compress\u00e3o \u00e9 m\u00e1xima \u2013 E<\/b><\/span><\/span><\/span>md<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 –<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>ma<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>md<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 4.100\/2=1.000.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>x=0,4m=40cm\u00a0 R- E<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0a) 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 —\u00a0<\/b><\/span><\/span><\/span>V=4,0m\/s<\/b><\/span><\/span><\/span><\/p>\n

b) mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=mgh + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 —\u00a0 8=10h + 4\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>h=0,6m<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0Colocando os pontos A, B e C na figura:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

Andr\u00e9<\/b><\/span><\/span><\/span>\u00a0\u2013 ponto A \u2013 como V=0 s\u00f3 tem energia potencial gravitacional\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pA<\/b><\/span><\/span><\/sub><\/span>=m.g.h\u00a0 —\u00a0 ponto C \u2013 como h=0 s\u00f3 tem energia cin\u00e9tica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>= E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 sistema conservativo \u2013 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span>=m.g.h\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Daniel\u00a0<\/b><\/span><\/span><\/span>– \u2013 ponto B \u2013 como V=0 s\u00f3 tem energia potencial gravitacional\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pB<\/b><\/span><\/span><\/sub><\/span>=2m.g.h\/2=m.g.h\u00a0 —\u00a0 ponto C \u2013 como h=0 s\u00f3 tem energia cin\u00e9tica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>= E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 sistema conservativo \u2013 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span>=m.g.h\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Portanto, ao chegarem ao solo ambos ter\u00e3o<\/b><\/span><\/span><\/span>\u00a0a mesma energia cin\u00e9tica<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

A energia cin\u00e9tica E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0com que ambos chegam ao solo \u00e9 a mesma, assim:<\/b><\/span><\/span><\/span><\/p>\n

Andr\u00e9\u00a0<\/b><\/span><\/span><\/span>\u2013 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=\u00d62E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\/m<\/b><\/span><\/span><\/span><\/p>\n

Daniel\u00a0<\/b><\/span><\/span><\/span><\/span>– \u2013 E<\/b><\/span><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub><\/span>=2mV<\/b><\/span><\/span><\/span><\/span>D<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2 \u2013\u00a0<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/span>D<\/b><\/span><\/span><\/span><\/sub><\/span>=\u00d6E<\/b><\/span><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub><\/span>\/m<\/b><\/span><\/span><\/span><\/span><\/p>\n

Portanto, ao chegarem ao solo ter\u00e3o velocidades diferentes\u00a0 —\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0Observe a figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

No ponto A s\u00f3 existe\u00a0 energia potencial gravitacional, pois a cin\u00e9tica \u00e9 zero\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>= E<\/b><\/span><\/span><\/span>pA<\/b><\/span><\/span><\/sub><\/span>=m.g.(h +\u00a0Dx)=0,2.10.(h + 0,1)=2h + 0,2<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=2h + 0,2<\/b><\/span><\/span><\/span>\u00a0 — \u00a0no ponto B (n\u00edvel zero de altura), a energia potencial gravitacional (h=0) e a energia cin\u00e9tica (V=0) s\u00e3o nulas, existindo a\u00ed apenas energia potencial el\u00e1stica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=k(Dx)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=1.240.(0,1)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=6,2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=6,2J<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 como o sistema \u00e9 conservativo, E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>\u00a0= E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2h + 0,2=6,2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0h=3m<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0Vamos descrever todas as caracter\u00edsticas deste movimento, desde o in\u00edcio, at\u00e9 quando a velocidade \u00e9 m\u00e1xima. (desprezando-se os atritos):<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0<\/span>I\u00a0<\/b><\/span><\/span><\/span>\u2013 a esfera \u00e9 abandonada do repouso (V\u00ado\u00ad=0) e cai sujeita apenas \u00e0 for\u00e7a peso, com velocidade de intensidade V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, que vai aumentando devido apenas \u00e0 acelera\u00e7\u00e3o da gravidade, at\u00e9 atingir a mola.<\/b><\/span><\/span><\/span><\/p>\n

II<\/b><\/span><\/span><\/span>\u00a0\u2013 ao atingir a mola, come\u00e7ando a comprimi-la, surge a for\u00e7a el\u00e1stica (\"\") que se op\u00f5e ao peso (\"\"). A intensidade de\u00a0\u00a0\"\"\u00a0vai aumentando, diminuindo a intensidade da for\u00e7a resultante que \u00e9 para baixo, fazendo com que V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>aumente cada vez menos<\/b><\/span><\/span><\/span>, at\u00e9 chegar \u00e0 situa\u00e7\u00e3o III.<\/b><\/span><\/span><\/span><\/p>\n

III<\/b><\/span><\/span><\/span>– nesta situa\u00e7\u00e3o as intensidades das for\u00e7as el\u00e1stica e peso tornam-se iguais (P=F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>), quando a velocidade de queda torna-se m\u00e1xima<\/b><\/span><\/span><\/span><\/p>\n

\u00a0 Veja na figura abaixo as condi\u00e7\u00f5es do exerc\u00edcio:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na situa\u00e7\u00e3o final\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>k.x=m.g\u00a0 —\u00a0 200.x=4.10\u00a0 —\u00a0 x=0,2m\u00a0 —\u00a0 toda energia mec\u00e2nica do in\u00edcio (E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=mgh) \u00e9 transformada em E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>cmax<\/b><\/span><\/span><\/sub><\/span>\u00a0na situa\u00e7\u00e3o final\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 mgh=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mV<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —4.10.0,9=200.(0,2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 4.V<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

36= 4 + 2V<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=16\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V=4m\/s\u00a0 R- B<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0Observe a figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Tomando o ponto C como n\u00edvel zero de refer\u00eancia para as alturas, a energia mec\u00e2nica em A, onde a energia cin\u00e9tica \u00e9 zero, ser\u00e1\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=m.g.(2 + x)=0,5.10.(2 + x)=<\/b><\/span><\/span><\/span>10 + 5x\u00a0<\/b><\/span><\/span><\/span>—\u00a0 considerando o sistema conservativo, a energia mec\u00e2nica em C, onde a energia potencial gravitacional \u00e9 nula (h=0) e a cin\u00e9tica tamb\u00e9m (menor comprimento,V=0) ser\u00e1 somente a energia potencial el\u00e1stica\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=k.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=100.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=<\/b><\/span><\/span><\/span>50x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 10 + 5x=50x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 10x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013x -10=0\u00a0 —\u00a0\u00a0D=B<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 4.A.C=1 + 80=81\u00a0 —\u00a0\u00a0\u00d6D=9\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

X=1\u00b19\/20 \u2013 x<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=10\/20=0,5m \u2013 x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 negativo e n\u00e3o satisfaz\u00a0 —\u00a0 o menor comprimento da mola \u00e9 y=0,8 \u2013 x=0,8-0,5\u00a0\u00a0<\/b><\/span><\/span><\/span>y=0,3m<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0Esse sistema massa-mola constitui um oscilador harm\u00f4nico simples, de caracter\u00edsticas:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

* A energia mec\u00e2nica \u00e9 sempre constante no MHS e vale\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>= kA<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 ou\u00a0 E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0 ou\u00a0\u00a0 E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span><\/p>\n

*\u00a0<\/b><\/span><\/span><\/span>Nos extremos<\/b><\/span><\/span><\/span>\u00a0onde v=0 e o m\u00f3dulo de x \u00e9 A, temos que\u00a0 — E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0 — E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>= 0 + k.A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=k.A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = constante<\/b><\/span><\/span><\/span><\/p>\n

*\u00a0<\/b><\/span><\/span><\/span>No ponto m\u00e9dio 0<\/b><\/span><\/span><\/span>, onde o m\u00f3dulo de v \u00e9 m\u00e1ximo e x=0, temos que\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>p<\/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 + 0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=mv<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>max<\/b><\/span><\/span><\/sub><\/span>\/2=const.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da energia mec\u00e2nica num dos extremos\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0E<\/b><\/span><\/span><\/span>me<\/b><\/span><\/span><\/sub><\/span>=k.A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=800.(0,5)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=<\/b><\/span><\/span><\/span>100J<\/b><\/span><\/span><\/span>\u00a0 (constante)\u00a0 —\u00a0 no ponto P, onde E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>me<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 100=E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=2E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=2.mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.1.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 100=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V=10m\/s R- A\u00a0<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0Observe a figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=m.g.h=m.10.1,6=16m\u00a0 —\u00a0 como o sistema \u00e9 conservativo, E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0tamb\u00e9m vale E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=16m\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0=m.g.(L-h) + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=m.10.0,35 + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=3,5m + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\u00a0\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC\u00a0<\/b><\/span><\/span><\/sub><\/span>=\u00a0 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 16m=3,5m + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 12,5=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V=5m\/s<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0a) Energia mec\u00e2nica em A\u00a0 —\u00a0 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=2V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.0=10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=100J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>energia mec\u00e2ncica em B\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0=m.g.h + mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.10.R + 2.V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=40 + V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=20 + V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>\u00a0= E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 100=40 + V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=\u221a60m\/s\u00a0 —\u00a0 acelera\u00e7\u00e3o centr\u00edpeta \u2013 a<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=(\u221a60)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 a<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=30m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

b) Energia mec\u00e2nica em C \u2013 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=m.g.2R + m.V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.10.4 + 2V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=80 + V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=80 + V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 100=80 + V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=\u221a20m\/s\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Quando o carrinho passa por C com velocidade de\u00a0\u00d620m\/s, duas for\u00e7as para baixo agem sobre ele, seu peso\u00a0\"\"\u00a0e a rea\u00e7\u00e3o da parte superior da superf\u00edcie\u00a0\"\"\u00a0\u00a0— for\u00e7a resultante centr\u00edpeta em C \u2013 F<\/b><\/span><\/span><\/span>RC<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 N + P= m.V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 N + 2.10=2V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

N + 20=(\u221a20)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 N=20 \u2013 20\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 N=0<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span> a) Energia mec\u00e2nica no in\u00edcio do movimento E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.h=0 + 0,2.10.0,4\u00a0 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=0,8J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>Energia mec\u00e2nica no ponto X\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mX<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pX<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>cX<\/b><\/span><\/span><\/sub><\/span>=0 + E<\/b><\/span><\/span><\/span>cX<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>cX<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0 E\u00ad<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mX<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0,8=E<\/b><\/span><\/span><\/span>cX<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>cX<\/b><\/span><\/span><\/sub><\/span>=0,8J<\/b><\/span><\/span><\/span><\/p>\n

b) Trabalho como varia\u00e7\u00e3o de energia potencial gravitacional \u2013 W<\/b><\/span><\/span><\/span>XY<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pX<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>pY<\/b><\/span><\/span><\/sub><\/span>=0 \u2013m.g.h= – 0,2.10.0,1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>XY<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>– 0,2J<\/b><\/span><\/span><\/span><\/p>\n

c) Essa velocidade m\u00ednima ocorre quando o carrinho passa por Z estando em contato com a pista, mas sem pression\u00e1-la, ou seja, quando a for\u00e7a de compress\u00e3o\u00a0\"\"\u00a0for nula.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Nesse caso a for\u00e7a resultante centr\u00edpeta \u00e9 o peso\u00a0\"\"\u00a0\u00a0—\u00a0 F<\/b><\/span><\/span><\/span>RC<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>Z<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 P= m.V<\/b><\/span><\/span><\/span>Z<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 0,2.10=0,2.V<\/b><\/span><\/span><\/span>Z<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/0,1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>Z<\/b><\/span><\/span><\/sub><\/span>=1m\/s<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>1- Correta. E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh=0 + mgh\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=mgh\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m.h.0\u00a0 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 sistema conservativo \u2013 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span><\/p>\n

2-Errada, pois se existem for\u00e7as dissipativas, parte da energia mec\u00e2nica \u00e9 \u201cperdida\u201d<\/b><\/span><\/span><\/span><\/p>\n

3- Correta. A for\u00e7a resultante centr\u00edpeta em B, j\u00e1 que a for\u00e7a de compress\u00e3o a\u00ed \u00e9 zero, \u00e9 somente o peso P=mg\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>RC<\/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 mg= mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=\u00d6Rg e esse \u00e9 o valor m\u00ednimo de V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0e para\u00a0 qualquer valor acima dele,\u00a0 sempre haver\u00e1 for\u00e7a de compress\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

4- Errado. V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 constante e vale\u00a0\u00d6Rg, mas V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 fornecida por E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mgh=mV<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 — V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=\u00d62gh (depende de h)<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0E<\/b><\/span><\/span><\/span>mD<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=m.20<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mD<\/b><\/span><\/span><\/sub><\/span>=200m\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=m.g.H=m.10.H\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mD<\/b><\/span><\/span><\/sub><\/span>=E\u00ad<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 200m=10.m.H\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>H=20m<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=m.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.h=m.100\/2 + m.g.h\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=50m + m.10.h\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mD<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 200m=50m + m.10.h\u00a0 —\u00a0 h=150\/10\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 h=15m\u00a0 R- E<\/b><\/span><\/span><\/span><\/p>\n

21<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0a) Sistema conservativo \u2013 E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mgh=mv<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 — 10.20=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>=20m\/s<\/b><\/span><\/span><\/span><\/p>\n

b) Como n\u00e3o existe atrito, o sistema chega em R com velocidade de 20m\/s (V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=20m\/s) e p\u00e1ra (V=0) ap\u00f3s percorrer\u00a0DS=1,5m\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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+ 2.a.DS\u00a0 —\u00a0 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.a.1,5\u00a0 —\u00a0 a= – 400\/3m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=m.a=90.(-400\/3)\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=- 12.000N em m\u00f3dulo\u00a0<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=12.000N ou F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=12kN<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0Antes do salto s\u00f3 tem energia cin\u00e9tica (com o n\u00edvel zero de altura no ponto 0,8m)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>ma<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span>\u00a0\u00a0 — na altura m\u00e1xima s\u00f3 tem energia potencial gravitacional\u00a0 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mam<\/b><\/span><\/span><\/sub><\/span>=m.g.h=m.10.3,2=<\/b><\/span><\/span><\/span>32m\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 como n\u00e3o h\u00e1 dissipa\u00e7\u00e3o de energia<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=32m\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

V=8m\/s R- D<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0Na situa\u00e7\u00e3o inicial s\u00f3 tem energia potencial gravitacional\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=m.g.4R\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 em x \u2013<\/b><\/span><\/span><\/span>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 + m.g.2R<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 4mgR=<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>m.V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.2R\u00a0 — 2mgR=mV<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=\u00d64gR\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=2\u00d6gR\u00a0<\/b><\/span><\/span><\/span><\/p>\n

No ponto x, agem sobre a esfera duas for\u00e7as verticais e para baixo que s\u00e3o o peso P e a rea\u00e7\u00e3o do trilho f<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 for\u00e7a resultante centr\u00edpeta \u2013 F<\/b><\/span><\/span><\/span>RC<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 P + f<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>= mV<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 mg +f<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=m(2\u00d6Rg)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=m.4Rg\/R \u2013 mg\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=3mg<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0a) Vamos supor que seja no ponto X que a formiga perca contato com a bola. Nesse ponto, a rea\u00e7\u00e3o normal\u00a0\"\"\u00a0deixa de existir agindo sobre a formiga somente seu peso\u00a0\"\", que \u00e9 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.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe no tri\u00e2ngulo menor – cosa=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>=P.cosa=mgcosa\u00a0 —\u00a0\u00a0 mgcosa=mV<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0<\/b><\/span><\/span><\/span>—\u00a0 gcosa=V<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R I\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>observe no tri\u00e2ngulo maior –\u00a0<\/b><\/span><\/span><\/span>cosa=h\/R<\/b><\/span><\/span><\/span>, 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\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=g.h\u00a0<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

b) Chamando a posi\u00e7\u00e3o inicial da formiga de Y e considerando a horizontal que passa por X como n\u00edvel zero de altura:<\/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<\/b><\/span><\/span><\/sub><\/span>=0 + mg.(R-h)\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 E<\/b><\/span><\/span><\/span>mY<\/b><\/span><\/span><\/sub><\/span>=m.g.(R-h)\u00a0\u00a0<\/b><\/span><\/span><\/span>—\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<\/b><\/span><\/span><\/span>—\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<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mY<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>X\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2.g.(R-h)<\/b><\/span><\/span><\/span>\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<\/b><\/span><\/span><\/span>—\u00a0 h=2R\/3\u00a0<\/b><\/span><\/span><\/span><\/p>\n

25<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0No instante em que o bloco deixa de estar em contato com os trilhos no ponto A, a for\u00e7a de compress\u00e3o a\u00ed torna-se nula (N=0) e a for\u00e7a resultante centr\u00edpeta (F<\/b><\/span><\/span><\/span>RC<\/b><\/span><\/span><\/sub><\/span>) que atua sobre o bloco \u00e9 seu peso P\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>RC<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 — P=m.V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 m.g=m.V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=R.g\u00a0 \u00a0<\/b><\/span><\/span><\/span>—\u00a0 sem atrito (sistema conservativo) \u2013 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 m.g.h=m.g.R +\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>m.V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 g.h=g.R +g.R\/2\u00a0 —\u00a0 h=R + R\/2\u00a0 — <\/b><\/span><\/span><\/span>h=3R\/2\u00a0 R- C\u00a0<\/b><\/span><\/span><\/span><\/p>\n

26<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0a) Em A s\u00f3 tem energia cin\u00e9tica \u2013 E<\/b><\/span><\/span><\/span>cA<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=300.144\/2\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=21.600J\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 em C tem energia cin\u00e9tica e energia potencial\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.h=300.V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 300.10.5,4=<\/b><\/span><\/span><\/span>150.V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 16.200\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 Em<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 21.600=150.V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 16.200—\u00a0 V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=\u00d636<\/b><\/span><\/span><\/span><\/p>\n

—\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=6,0m\/s<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

b) No ponto C, a acelera\u00e7\u00e3o do carrinho \u00e9 a acelera\u00e7\u00e3o centr\u00edpeta dada por a<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=36\/5,4\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=6.7m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

c) No ponto C, as for\u00e7as que agem sobre o carrinho s\u00e3o seu peso para baixo e a rea\u00e7\u00e3o dos trilhos para cima, cuja for\u00e7a resultante centr\u00edpeta vale\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>RC<\/b><\/span><\/span><\/sub><\/span>=P \u2013 N=m.V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 m.g –\u00a0 Nm.V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 3000 \u2013 N=300.36\/5,4\u00a0 —\u00a0 N=3.000 \u2013 2.000\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 N=1.000N<\/b><\/span><\/span><\/span><\/p>\n

27<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>a) Observe no gr\u00e1fico que a energia potencial da bola varia com a altura sendo ambas fun\u00e7\u00e3o do tempo.<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>(t)=m.g.H(t)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>(t)=0,6.10.H(t)\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>(t)=6.H(t)<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 colocando valores para H, constru\u00edmos o gr\u00e1fico E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0X t:<\/b><\/span><\/span><\/span><\/p>\n

t=0 \u2013 H=1,6m \u2013 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=6.1,6=9,6J\u00a0 —\u00a0 t=5 – H=0,6m \u2013 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=6.5=3J\u00a0 —\u00a0 t=t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 H=0 \u2013 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 t=9 \u2013 H=0,4 \u2013 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=6.0,4=2,4J e assim por diante.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) Como a energia total, que \u00e9 a energia mec\u00e2nica da bola se conserva no intervalo entre os choques, nesses intervalos E<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 constante e igual \u00e0 energia potencial gravitacional nas alturas m\u00e1ximas, onde a energia cin\u00e9tica \u00e9 nula (V=0).<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Nos pontos mais altos – E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=constante<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

c) Para o choque que ocorre no instante t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, a energia mec\u00e2nica na altura m\u00e1xima E<\/b><\/span><\/span><\/span>p1<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>m1<\/b><\/span><\/span><\/sub><\/span>=m.g.H<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0,6.10.1,6\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>m1<\/b><\/span><\/span><\/sub><\/span>=9,6J \u00e9 igual \u00e0 energia cin\u00e9tica no choque com o solo\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>m1<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>c1<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0,6.V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>m1<\/b><\/span><\/span><\/sub><\/span>=0,3.V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 igualando, pois durante a queda a energia mec\u00e2nica se conserva\u00a0 —\u00a0 9,6=0,3V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=\u00d632\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=4\u00d62m\/s<\/b><\/span><\/span><\/span><\/p>\n

Da mesma maneira, na segunda queda\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>m2<\/b><\/span><\/span><\/sub><\/span>=mgH=0,6.10.0,4=2,4J=m.V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0,3V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0—\u00a0 2,4=0,3V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=2\u00d62m\/s\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

Coeficiente de restitui\u00e7\u00e3o\u00a0 —\u00a0 C<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2\u00d62\/4\u00d62\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 C<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=0,5<\/b><\/span><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0mgh=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 10.0,6=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V\u00bb3,46m\/s\u00a0\u00a0<\/b><\/span><\/span><\/span>R-E<\/b><\/span><\/span><\/span><\/p>\n

29<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0Energia mec\u00e2nica na posi\u00e7\u00e3o inicial\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>\u00a0+ mgh=0,10 + 0,10.10.0,40=0,10 + 0,40\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=0,50J\u00a0 —\u00a0 na altura h em que ela parar, E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=mgh=0,10.10.h=<\/b><\/span><\/span><\/span>h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0,50=h\u00a0 —\u00a0 ele p\u00e1ra na altura 0,5m que est\u00e1 entre 1 e 2, fica oscilando at\u00e9 parar em 1\u00a0 R- A<\/b><\/span><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0a) E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=mgh=60.10.4\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=2.400J<\/b><\/span><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/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.h=60V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 60.10.0,8=\u00a0<\/b><\/span><\/span><\/span>30V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 480<\/b><\/span><\/span><\/span>\u00a0—\u00a0 2.400=30V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 480\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=8m\/s<\/b><\/span><\/span><\/span><\/p>\n

b) Trata-se de um lan\u00e7amento obliquo com velocidade V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>, no ponto B. Nele, a velocidade de lan\u00e7amento (inicial) V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0possui duas componentes V<\/b><\/span><\/span><\/span>Bx<\/b><\/span><\/span><\/sub><\/span>\u00a0horizontal e constante durante todo o movimento e V<\/b><\/span><\/span><\/span>By<\/b><\/span><\/span><\/sub><\/span>, vertical e vari\u00e1vel (lan\u00e7amento vertical). No ponto mais alto da trajet\u00f3ria, ponto C, a componente vertical V<\/b><\/span><\/span><\/span>By<\/b><\/span><\/span><\/sub><\/span>\u00a0se anula (V<\/b><\/span><\/span><\/span>By<\/b><\/span><\/span><\/sub><\/span>=0), restando apenas a componente constante<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

V<\/b><\/span><\/span><\/span>By<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>Bx<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.cosq=8.\u00d6\u00a03\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>By<\/b><\/span><\/span><\/sub><\/span>=4\u00d63m\/s\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>Cy<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>By<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=60.16.3\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>Cy<\/b><\/span><\/span><\/sub><\/span>=1.440J\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

c) E<\/b><\/span><\/span><\/span>mC=\u00a0<\/b><\/span><\/span><\/sub><\/span>E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>Pc<\/b><\/span><\/span><\/sub><\/span>=60.(4\u00d63)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 60.10.h<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2.440=1.440 + 600h<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 h<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>=1,6m<\/b><\/span><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0Energia mec\u00e2nica inicial da mola (ponto P) \u2013 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>Ci<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>Pei<\/b><\/span><\/span><\/sub><\/span>=0 + k.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>= 10.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 observe na figura abaixo que x=(2\u221a2m), comprimento da mola deformada \u2013 (1m), comprimento natural da mola<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

deforma\u00e7\u00e3o inicial da mola \u2013 x=(2\u221a2 \u2013 1)m\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=10.(2\u221a2-1)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=5(2.1,4 \u2013 1)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=5.1,8<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>—\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=16,2J<\/b><\/span><\/span><\/span><\/p>\n

Energia mec\u00e2nica na situa\u00e7\u00e3o final da mola (ponto T) \u2013 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>Pef<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>Cf<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 observe na figura abaixo que a deforma\u00e7\u00e3o da mola na situa\u00e7\u00e3o final \u00e9 x<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=comprimento da mola deformada (2m) \u2013 comprimento natural da mola (1m)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

x<\/b><\/span><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span><\/sub><\/span>=2 \u2013 1=1m\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/span><\/sub><\/span>=10.1<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2 + 1.V<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2=5 + V<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/span><\/sub><\/span>=5 + V<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span><\/span>\u00a0 —\u00a0 16,2=5 + V<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>V\u00bb\u221a22,4\u00a0\u00bb\u00a04,7m\/s<\/b><\/span><\/span><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0Distens\u00e3o m\u00e1xima (10m) + comprimento natural (30m)=40m<\/b><\/span><\/span><\/span><\/p>\n

Emi=Eci + Epi=0 +.750.40=30.000J\u00a0 — situa\u00e7\u00e3o final \u2013 distens\u00e3o m\u00e1xima \u2013 V=0+ – Ecf=0\u00a0 —\u00a0 Emf= 0 + kx2\/2\u00a0 —\u00a0 Emf=50k —\u00a0 30.00\u22480=50k2\u00a0 —\u00a0 k\u224824,5N\/m<\/b><\/span><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0R \u2013 B veja teoria<\/b><\/span><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0R \u2013 A\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span>\u00a0E energia mec\u00e2nica total (E) \u00e9 s\u00f3 a energia cin\u00e9tica, pois o enunciado diz que a energia potencial n\u00e3o variou\u00a0 —\u00a0 o trabalho das for\u00e7as dissipativas corresponde \u00e0 energia total dissipada \u201cperdida\u201d no processo, no caso, entre 0 e 4s\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>fdiss<\/b><\/span><\/span><\/sub><\/span>=E(final) \u2013 E(inicial)=600 \u2013 1800= – 1.200J –\u00a0\u00a0<\/b><\/span><\/span><\/span>em m\u00f3dulo 1.200J\u00a0 R- B<\/b><\/span><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cA<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pA<\/b><\/span><\/span><\/sub><\/span>=0 + m.g.h=2.10.3,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=70J<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pC<\/b><\/span><\/span><\/sub><\/span>=0 + 2.10.3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=60J\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 a energia dissipada corresponde ao trabalho realizado pelas for\u00e7as dissipativas vale\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>fdis<\/b><\/span><\/span><\/sub><\/span>=DE<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=60 \u2013 70= -10J \u2013\u00a0<\/b><\/span><\/span><\/span>em m\u00f3dulo 10J \u2013 R A<\/b><\/span><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0Nas alturas m\u00e1ximas a energia cin\u00e9tica \u00e9 nula e a mec\u00e2nica \u00e9 s\u00f3 a potencial gravitacional\u00a0 —\u00a0\u00a0DE<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=40.000J\u00a0 —\u00a0\u00a0DE<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>m1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>m2<\/b><\/span><\/span><\/sub><\/span>=mgh<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 mgh<\/b><\/span><\/span><\/span>2=<\/b><\/span><\/span><\/sub><\/span>10.10.(h<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 h<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0DE<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=100.(h<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 h<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 40.000=100.\u00a0Dh\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>Dh=400m R- C<\/b><\/span><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/span>\u00a0A energia potencial gravitacional \u00e9 a mesma em N e em P e como h\u00e1 perda de energia mec\u00e2nica, a energia mec\u00e2nica em N \u00e9 maior que a energia mec\u00e2nica em P. Assim, a energia cin\u00e9tica em N \u00e9 maior que a energia cin\u00e9tica em P. Tendo perda de energia a E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>em P \u00e9 maior que a E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0em Q\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span> E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pA<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E\u00ad<\/b><\/span><\/span><\/span>cA<\/b><\/span><\/span><\/sub><\/span>=mgh + 0=4.10.8=<\/b><\/span><\/span><\/span>320J\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pC<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 0=4.100\/2=<\/b><\/span><\/span><\/span>200J\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 regra de tr\u00eas\u00a0 —\u00a0 320J \u2013 100%\u00a0 —\u00a0 120J \u2013 x\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>x=37,5%\u00a0 R- C<\/b><\/span><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span><\/span>\u00a0E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=100.0,001\/2=0,5J\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=mgh=0,1.10.0,3=0,3J\u00a0 —\u00a0\u00a0DE<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=W=0,3 \u2013 0,5=0,2J\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span><\/span>\u00a0a) E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>=mv<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>=1J<\/b><\/span><\/span><\/span><\/p>\n

b) A energia mec\u00e2nica inicial do bloco de massa 5kg \u00e9 igual a 0,8J, pois houve perda de 20%.e toda essa energia \u00e9 transmitida ao bloco quando se encontra no ponto A, tomado como n\u00edvel zero de altura.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=0,8J\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=mgh=5.10h\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=50h\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0,8=50h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>h=0,16m<\/b><\/span><\/span><\/span><\/p>\n

42<\/b><\/span><\/span><\/span>-E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=mgh=25.10.2,4=<\/b><\/span><\/span><\/span>600J<\/b><\/span><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=25V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=<\/b><\/span><\/span><\/span>12,5V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=mN=mPcos37=0,5.250.0,8=100N\u00a0 —\u00a0 \u00a0sen37<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=h\/d\u00a0 —\u00a0 d=4m\u00a0 —\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>.d.cos37<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=100.4.0.(-1<\/b><\/span><\/span><\/span>)=-400J<\/b><\/span><\/span><\/span>\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 — -400=12,5V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 600\u00a0 —\u00a0<\/b><\/span><\/span><\/span>V=4m\/s\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

43-<\/b><\/span><\/span><\/span>\u00a0a) Terra\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=mgh=<\/b><\/span><\/span><\/span>mgH<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mx<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>como na rampa n\u00e3o existe atrito<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>mgH<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= mV<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2gH<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u00a0\u00a0\u00a0\u00a0 Marte\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=mg\/3h=<\/b><\/span><\/span><\/span>mg\/<\/b><\/span><\/span><\/span>3H<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mx<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>como na rampa n\u00e3o existe atrito<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>mg\/3H<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= mV<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2\/3gH<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u00a0 V<\/b><\/span><\/span><\/span>xTerra<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/ V<\/b><\/span><\/span><\/span>xMarte<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2gH<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/2\/3gH<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0\u00a0\u00a0 V<\/b><\/span><\/span><\/span>xTerra<\/b><\/span><\/span><\/sub><\/span>\/ V<\/b><\/span><\/span><\/span>xMarte<\/b><\/span><\/span><\/sub><\/span>=\u00d63<\/b><\/span><\/span><\/span><\/p>\n

b)Terra\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mx<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cx<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mx<\/b><\/span><\/span><\/sub><\/span>=m2gH<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/2=<\/b><\/span><\/span><\/span>mgH<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>my<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=<\/b><\/span><\/span><\/span>mV<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 no trecho horizontal existe for\u00e7a de atrito\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=mN=mP=mmg\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=ma\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>at\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 ma=mmg\u00a0 —\u00a0 a=mg\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2(-a).DS\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2gH<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0–<\/b><\/span><\/span><\/span>\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span><\/p>\n

2mgL\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>= 2g(H<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub><\/span>–\u00a0mL)<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>my<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>mx<\/b><\/span><\/span><\/sub><\/span>\u00a0 — W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 mV<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 — W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>2g(H<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub><\/span>–\u00a0mL)<\/b><\/span><\/span><\/span>\/2 \u2013 m(2gH<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>)\/2\u00a0\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

m<\/b><\/span><\/span><\/span>g(H<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub><\/span>–\u00a0mL)<\/b><\/span><\/span><\/span>\u00a0–\u00a0 mgH<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>fatTerra<\/b><\/span><\/span><\/sub><\/span>= – mgm<\/b><\/span><\/span><\/span>L\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\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\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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Marte\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>fatMarte<\/b><\/span><\/span><\/sub><\/span>= – mg\/3mL<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>fatTerra<\/b><\/span><\/span><\/sub><\/span>\/ W<\/b><\/span><\/span><\/span>fatMarte<\/b><\/span><\/span><\/sub><\/span>= – mgmL X \u2013 3\/mgmL\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>fatTerra<\/b><\/span><\/span><\/sub><\/span>\/ W<\/b><\/span><\/span><\/span>fatMarte<\/b><\/span><\/span><\/sub><\/span>= 3<\/b><\/span><\/span><\/span><\/p>\n

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

44-<\/b><\/span><\/span><\/span>\u00a0Calculando a energia mec\u00e2nica na situa\u00e7\u00e3o inicial<\/b><\/span><\/span><\/span>\u00a0i<\/b><\/span><\/span><\/span>\u00a0e na final\u00a0<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=mgH=mg20\u00d63\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=20\u00d63mg I<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=mg2R + mV<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 a condi\u00e7\u00e3o para que ele fa\u00e7a o loop em f sem perder contato com a pista \u00e9 V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=\u00d6Rg\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=mg2R + m(Rg)\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=2,5mRg II<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo do trabalho das for\u00e7as dissipativas (F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>):<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Trabalho do F<\/b><\/span><\/span><\/span>at\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0na rempa<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sup><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=mN=mPcos60<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sup><\/span>=mmgcos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=1\/2.mg.1\/2\u00a0 —\u00a0 \u00a0F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sup><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=mg\/4\u00a0 —\u00a0 sen60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=H\/d\u00a0 —\u00a0\u00d63\/2=20\u00d63\/d\u00a0 —\u00a0 d=40m\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sup><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sup><\/span>fat<\/b><\/span><\/span><\/sub><\/span>.d.cos180<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=mg\/4.40.(-1)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sup><\/span>fat<\/b><\/span><\/span><\/sub><\/span>= – 10mg<\/b><\/span><\/span><\/span><\/p>\n

Trabalho do F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0no plano horizontal:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span><\/span>PH<\/b><\/span><\/span><\/sup><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=mN=mP=mmg=1\/2.mg\u00a0 — F<\/b><\/span><\/span><\/span>PH<\/b><\/span><\/span><\/sup><\/span>fat<\/b><\/span><\/span><\/sub><\/span>= mg\/2\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>PH<\/b><\/span><\/span><\/sup><\/span>fat\u00a0<\/b><\/span><\/span><\/sub><\/span>= F<\/b><\/span><\/span><\/span>PH<\/b><\/span><\/span><\/sup><\/span>fat<\/b><\/span><\/span><\/sub><\/span>.d.cos180<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=mg\/2.20.(-1)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>PH<\/b><\/span><\/span><\/sup><\/span>fat<\/b><\/span><\/span><\/sub><\/span>= – 10mg<\/b><\/span><\/span><\/span><\/p>\n

Trabalho total do F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>= -10mg \u2013 10mg\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>= – 20mg III<\/b><\/span><\/span><\/span><\/p>\n

O trabalho das for\u00e7as dissipativas (F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>) \u00e9 igual \u00e0 varia\u00e7\u00e3o de energia mec\u00e2nica\u00a0 —\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>\u00a0 IV\u00a0 —\u00a0\u00a0\u00a0 substituindo I, II e III em IV\u00a0 —\u00a0 -20mg=2,5mgR – 20\u00d63mg\u00a0 —\u00a0 R=(20\u00d63 \u2013 20)\/2,5\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 R=8(\u00d63 \u2013 1)m\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

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

Descida\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span>=mgh<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=1.10.2=20J\u00a0 — —\u00a0 E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span>=20J\u00a0 —\u00a0 subida\u00a0 —\u00a0 perde 12J, sobram 8J\u2013 E<\/b><\/span><\/span><\/span>mS\u00a0<\/b><\/span><\/span><\/sub><\/span>=8J\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mR<\/b><\/span><\/span><\/sub><\/span>=mgh<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1.10.h<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=10h<\/b><\/span><\/span><\/span>2\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>—\u00a0 E<\/b><\/span><\/span><\/span>mS\u00a0<\/b><\/span><\/span><\/sub><\/span>= E<\/b><\/span><\/span><\/span>mR<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 10h<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=8\u00a0 —\u00a0<\/b><\/span><\/span><\/span>h2=0,8m=80cm<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

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

46-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Dados\u00a0 —\u00a0 m = 2 kg\u00a0 —\u00a0 h<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 1 m\u00a0 —\u00a0 \u00a0g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 considere que a cada choque a bola perde 50% de sua energia mec\u00e2nica, e que n\u00e3o haja perda de energia mec\u00e2nica enquanto a bola est\u00e1 no ar\u00a0 —\u00a0 assim, ap\u00f3s o primeiro choque, a energia cin\u00e9tica da bola E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 50% ou metade de sua energia potencial gravitacional inicial E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=0,5E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mgh<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/2=2x10x1\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=10J<\/b><\/span><\/span><\/span><\/p>\n

b) Ao atingir o solo pela segunda vez, a bola tem a mesma energia cin\u00e9tica que tinha ap\u00f3s o primeiro choque, calculada no item anterior\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\u00a0 —\u00a0 10=2V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V=\u221a10\u22483,2m\/s<\/b><\/span><\/span><\/span><\/p>\n

c) Dado\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u2019=0,08J \u00a0—\u00a0 ap\u00f3s o n-\u00e9simo choque, a energia cin\u00e9tica da bola \u00e9\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u2019=(1\/2)<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sup><\/span>.mgh<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0,08=(1\/2)<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sup><\/span>.2x10x1\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

(1\/2)<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sup><\/span>=8\/2000\u00a0 —\u00a0 (1\/2)<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sup><\/span>=1\/250\u00a0 —\u00a0 considerando 1\/250\u22481\/256\u00a0 —\u00a0 (1\/2)<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sup><\/span>=(1\/256)\u00a0 —\u00a0 (1\/2)<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sup><\/span>=(1\/2)<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0n=8<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\u00a0 — \u00a0energia potencial gravitacional em rela\u00e7\u00e3o \u00e0 linha da 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>\u00a0+ 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 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=\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\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

48-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observando a tabela dada\u00a0 —\u00a0 k=F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\/x=160\/10=320\/20=480\/30\u00a0 —\u00a0 k=16N\/cm=1.600N\/m\u00a0 —\u00a0 k=1,6kN\/m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A energia potencial el\u00e1stica (Epot) armazenada no sistema \u00e9 igual ao trabalho da for\u00e7a el\u00e1stica W<\/b><\/span><\/span><\/span>fel\u00a0<\/b><\/span><\/span><\/sub><\/span>para provocar essa deforma\u00e7\u00e3o\u00a0 —\u00a0 como a for\u00e7a el\u00e1stica varia com a deforma\u00e7\u00e3o, esse trabalho \u00e9 dado pela \u201c\u00e1rea\u201d entre a linha do gr\u00e1fico e o eixo da deforma\u00e7\u00e3o, como mostra a figura:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

E<\/b><\/span><\/span><\/span>pot<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>fel<\/b><\/span><\/span><\/sub><\/span>=\u201d\u00e1rea\u201d=base x altura\/2=x.F\/2=x.(kx)\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>pot<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pela conserva\u00e7\u00e3o da energia mec\u00e2nica, a energia potencial el\u00e1stica armazenada no sistema \u00e9 transferida para a flecha na forma de energia cin\u00e9tica\u00a0 —\u00a0 \u00a0dados relevantes\u00a0 —\u00a0 \u00a0x = 1 m\u00a0 —\u00a0 m = 10 g = 10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0kg\u00a0 —\u00a0 \u00a0k = 1.600 N\/m\u00a0 —\u00a0 mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 v=x.\u221ak\/m\u00a0 —V=1.\u221a1.600\/10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=\u221a16.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 v=400m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

51-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como 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\u00a0<\/b><\/span><\/span><\/sub><\/span>\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>=\u221a2gh\u00a0 —\u00a0 sendo H a altura do solo at\u00e9 B, o tempo de queda (tq) \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 t<\/b><\/span><\/span><\/span>q<\/b><\/span><\/span><\/sub><\/span>=\u221a2H\/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<\/b><\/span><\/span><\/span><\/p>\n

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

52-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como as for\u00e7as dissipativas s\u00e3o desconsideradas, o sistema \u00e9 conservativo e a energia mec\u00e2nica \u00e9 a mesma em todos os pontos\u00a0 —\u00a0 \u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

53-\u00a0<\/b><\/span><\/span><\/span>Observe a figura abaixo que mostra os pontos de velocidade nula (A) e de velocidade m\u00e1xima (B)\u00a0 —\u00a0 ela conserva\u00e7\u00e3o da energia<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

mec\u00e2nica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mB\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 3mgh<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=3mhh<\/b><\/span><\/span><\/span>B\u00a0<\/b><\/span><\/span><\/sub><\/span>+ 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=\u221a1.160\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>v\u00a0\u2248\u00a034 m\/s\u00a0<\/b><\/span><\/span><\/span>\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 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=3mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=3x50x1.160\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=87.000J=87kJ<\/b><\/span><\/span><\/span><\/p>\n

54-<\/b><\/span><\/span><\/span>\u00a0Observe que o sistema conservativo, a energia mec\u00e2nica inicial, no lan\u00e7amento, \u00e9 igual \u00e0 energia mec\u00e2nica no ponto mais alto, que \u00e9 o mesmo para as tr\u00eas trajet\u00f3rias. Portanto, a energia potencial tamb\u00e9m \u00e9 a mesma. Assim, fica na depend\u00eancia da energia cin\u00e9tica. A partir do ponto mais alto, a trajet\u00f3ria de maior alcance horizontal \u00e9 a III, portanto, a de maior velocidade horizontal e, consequentemente, a de maior energia cin\u00e9tica. Ent\u00e3o, a trajet\u00f3ria III \u00e9 a que apresenta maior energia mec\u00e2nica no ponto mais alto, logo, maior energia mec\u00e2nica no lan\u00e7amento.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

55<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0A menor altura h \u00e9 aquela que faz com que o carrinho passe pela posi\u00e7\u00e3o (2) com a velocidade m\u00ednima permitida para um c\u00edrculo vertical (quase perdendo o contato com a pista), ou seja, a for\u00e7a normal que a pista aplica no carrinho \u00e9 praticamente nula\u00a0 —\u00a0 assim, a intensidade da for\u00e7a resultante centr\u00edpeta nessa posi\u00e7\u00e3o (2) vale\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=N + P\u00a0 —\u00a0 \u00a0F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=0 + P\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>= P \u00a0—\u00a0 F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00e9 o pr\u00f3prio peso do carrinho\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 P=mg\u00a0 —\u00a0 mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=mg\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=Rg (I)\u00a0 —\u00a0 pela conserva\u00e7\u00e3o da energia mec\u00e2nica entre as posi\u00e7\u00f5es (1) e (2)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>p1<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>c2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>p2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mgh=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mg(2R)\u00a0 —\u00a0 2gh=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 4Rg (II)\u00a0 —\u00a0 substituindo (I) em (II)\u00a0 —\u00a0 2gh=Rg + 4Rg\u00a0 —\u00a0 h=5R\/2\u00a0 —\u00a0 h=2,5.(24)\u00a0 —\u00a0 h=60m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

56-<\/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). Se a energia total \u00e9 constante quando a energia potencial diminui a cin\u00e9tica deve aumentar ou quando Ep = m\u00e1xima\u00a0\u00ae\u00a0Ec =0 (gr\u00e1fico I).\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

57-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 \u00a0h = 2,4 m\u00a0 —\u00a0 \u00a0v<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\u00a0= 4 m\/s\u00a0 —\u00a0 conserva\u00e7\u00e3o da energia mec\u00e2nica em AB e CD\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mAB<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mCD<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mv<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\/2 +<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

mgh=mv<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>CD<\/b><\/span><\/span><\/sub><\/span>\/2\u00a0 —\u00a0 4<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 10×2,4=v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>CD<\/b><\/span><\/span><\/sub><\/span>\/2\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>CD<\/b><\/span><\/span><\/sub><\/span>=64\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>CD<\/b><\/span><\/span><\/sub><\/span>=8m\/s\u00a0 — conserva\u00e7\u00e3o da energia mec\u00e2nica em CD e E\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mCD<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mE<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mv<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>CD<\/b><\/span><\/span><\/sub><\/span>\/2=mgH\u00a0 —\u00a0 8<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=10H\u00a0 —\u00a0 H=3,2m\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

58-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Dados: r = 80 cm = 0,8 m; h = 2 m; m = 36 kg; H = 6 m e g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\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 —\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 for\u00e7a horizontal (F<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>) sobre a crian\u00e7a durante a descida \u00e9 a resultante centr\u00edpeta.<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\u00a0= F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0= mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R= 36×80\/08\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\u00a0= 3.600 N.<\/b><\/span><\/span><\/span><\/p>\n

b) Dados: v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 10,9 m\/s; v = 0;\u00a0DS = 1,5 m; g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 durante a descida da crian\u00e7a na \u00e1gua da piscina, ela sofre a a\u00e7\u00e3o do peso (\"\") e das for\u00e7as dissipativas exercidas pela \u00e1gua\u00a0\"\", em sentido oposto ao movimento, formando com a velocidade \u00e2ngulo\u00a0a\u00a0= 180\u00b0. Aplicando o teorema da energia cin\u00e9tica\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>FR<\/b><\/span><\/span><\/sub><\/span>=\u0394E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0+ W<\/b><\/span><\/span><\/span>FDM<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0mV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 mg\u0394S + F<\/b><\/span><\/span><\/span>DM<\/b><\/span><\/span><\/sub><\/span>\u0394Scos180<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=0 \u2013 mV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 10x36x1,5 \u2013 F<\/b><\/span><\/span><\/span>DM<\/b><\/span><\/span><\/sub><\/span>x1,5= -36x(10,9)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 1,5F<\/b><\/span><\/span><\/span>DM<\/b><\/span><\/span><\/sub><\/span>=540 + 2.138,6\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0F<\/b><\/span><\/span><\/span>DM<\/b><\/span><\/span><\/sub><\/span>=1.785,7J<\/b><\/span><\/span><\/span><\/p>\n

59-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) No trajeto do ponto P at\u00e9 o ponto B, agem no bloco tr\u00eas for\u00e7as: o peso\u00a0\"\"\u00a0a normal\u00a0\"\"\u00a0e a de atrito\u00a0\"\"\u00a0\u00a0—\u00a0 a for\u00e7a peso realiza trabalho apenas ao longo da descida PA, a normal n\u00e3o realiza trabalho, pois \u00e9 perpendicular \u00e0 trajet\u00f3ria em todo o percurso, e a for\u00e7a de atrito somente realiza trabalho no trecho AB\u00a0 —\u00a0 aplicando, ent\u00e3o, o teorema da energia cin\u00e9tica, notando que a energia cin\u00e9tica final em B \u00e9 nula e que em P \u00e9 1\/20 da energia potencial nesse mesmo ponto, suposta calculada a partir do plano horizontal de lan\u00e7amento\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=\u0394E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>(PA) + W<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>(PB) + W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>(AB) =E<\/b><\/span><\/span><\/span>cB\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0– E<\/b><\/span><\/span><\/span>cP<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mg(H \u2013 h) + 0 \u2013 Fd= 0 \u2013 1\/20(mgH)\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

mg(H \u2013 0,15H) \u2013 \u03bcmg3H=-0,05mgH\u00a0 —\u00a0 \u03bc=0,9\/3\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0\u03bc=0,3<\/b><\/span><\/span><\/span><\/p>\n

b) \u0394E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=\u2502W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>\u2502\/E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span>=\u03bcmgd\/1\/20(mgH) + mgH\u00a0 —\u00a0 \u0394E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=0,3mg3H\/21\/20mgH\u00a0 —\u00a0 \u0394E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=18\/21=0,857\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>(%)=85,7%<\/b><\/span><\/span><\/span><\/p>\n

60-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados: v = 10 m\/s; g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>; h = 10 m; m = 2 kg\u00a0 —\u00a0 enquanto voa, no esquilo agem duas for\u00e7as: o peso\u00a0\"\"\u00a0e a for\u00e7a de sustenta\u00e7\u00e3o\u00a0\"\"\u00a0do ar. Como a velocidade \u00e9 constante, o trabalho da resultante\u00a0\"\"\u00a0\u00e9 nulo. Mas o trabalho da resultante \u00e9 igual ao somat\u00f3rio dos trabalhos das for\u00e7as atuantes\u00a0 —\u00a0\u00a0\"\"\u00a0\u00a0—\u00a0 \u00a0\u00a0\"\"\u00a0\u00a0—\u00a0 \u00a0m g h =\u00a0\"\"\u00a0—\u00a0 \u00a0\"\"\u2013 2 (10) (10) \u00a0—\u00a0<\/b><\/span><\/span><\/span><\/p>\n

W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>= 200 J\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

61-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pelo teorema da energia cin\u00e9tica, o trabalho da resultante W<\/b><\/span><\/span><\/span>R\u00a0<\/b><\/span><\/span><\/sub><\/span>das for\u00e7as que atuam sobre um corpo \u00e9 igual \u00e0 varia\u00e7\u00e3o da energia cin\u00e9tica do corpo\u00a0 —\u00a0 como a velocidade \u00e9 constante, esse trabalho \u00e9 nulo.\u00a0\u00a0 \u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

62-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A energia total do corpo na altura\u00a0 de 10m 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\u00a0 a energia cin\u00e9tica vale 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\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

63-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Velocidade de A imediatamente antes de se chocar com B\u00a0 —\u00a0 conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0 mgh=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 10.0,8=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V=4m\/s\u00a0 —\u00a0 velocidade de B imediatamente ap\u00f3s o choque com A\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 3.4=3.V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=4m\/s<\/b><\/span><\/span><\/span><\/p>\n

b) Como, no o atrito \u00e9 desprez\u00edvel o bloco B incide na mola com velocidade de 4m\/s e a mola \u00e9 comprimida de x, at\u00e9 o bloco B parar (V=0)\u00a0 —\u00a0 conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0 imediatamente antes de se chocar com a mola o bloco s\u00f3 possui energia cin\u00e9tica E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=3.16\/2=24J\u00a0 —\u00a0 quando o bloco B p\u00e1ra ele s\u00f3 possui energia potencial el\u00e1stica armazenada\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=4x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=2x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\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 24=2x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 x=\u221a12\u00a0 —\u00a0 x\u22483,46m<\/b><\/span><\/span><\/span><\/p>\n

c) Agora, com atrito no plano horizontal a \u00fanica for\u00e7a na dire\u00e7\u00e3o do movimento \u00e9 a for\u00e7a de atrito\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcN=\u03bcP=0,4.30\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=ma\u00a0 —\u00a0 12=3.a\u00a0 —\u00a0 a=4m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 como a velocidade do bloco B est\u00e1 diminuindo essa acelera\u00e7\u00e3o \u00e9 negativa\u00a0 —\u00a0 equa\u00e7\u00e3o de Torricelli at\u00e9 o bloco B parar\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+ 2.a.\u0394S\u00a0 —\u00a0 0=4<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.(-4).\u0394S\u00a0 —\u00a0 \u0394S=2m\u00a0 — \u00a0<\/b><\/span><\/span><\/span>como d=3m, o bloco B n\u00e3o comprime a mola parando a 1m da mesma.<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

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

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

\u00a0\u00a0a) Quando o bloco da esquerda de massa m<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=2,0kg chega \u00e0 superf\u00edcie rugosa com velocidade de V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=1m\/s, a for\u00e7a de atrito, contr\u00e1ria ao movimento, que \u00e9 a for\u00e7a resultante sobre ele, faz com que surja uma acelera\u00e7\u00e3o de retardamento at\u00e9 que ele pare<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(V=0) ap\u00f3s percorrer a dist\u00e2ncia \u2206S=0,5m\u00a0 —\u00a0 c\u00e1lculo da acelera\u00e7\u00e3o de retardamento\u00a0 —\u00a0 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+2.(-a).\u2206S\u00a0 —\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>-2.a.0,5\u00a0 —\u00a0 a=1m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 na dire\u00e7\u00e3o do movimento a for\u00e7a resultante sobre o bloco \u00e9 a for\u00e7a de atrito F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0lei de Newton\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a=F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=2.1=2N\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcN=\u03bcP=\u03bcmg\u00a0 —\u00a0 2=\u03bc.2.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03bc=0,1<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

b) Utilizando o teorema da conserva\u00e7\u00e3o da quantidade de movimento\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/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>=m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.0 + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.0=0\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>depois<\/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+<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= 2.(-1) + 1.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 2\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—\u00a0 0=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2m\/s<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

c) Utilizando o teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica, enquanto a mola est\u00e1 comprimida a energia total (mec\u00e2nica) armazenada \u00e9 a energia potencial el\u00e1stica que vale\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>Mantes<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 depois que a mola \u00e9 liberada e cada bloco se move<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

em sentidos contr\u00e1rios com velocidades V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=1m\/s e V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2m\/s, toda energia potencial el\u00e1stica sedo sistema se transforma em energia cin\u00e9tica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>Mdepois<\/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>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.(-1)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 1.2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=3J\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>Mantes\u00a0<\/b><\/span><\/span><\/sub><\/span>= E<\/b><\/span><\/span><\/span>Mdepois<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=3\u00a0 —\u00a0 6.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = 3\u00a0 —\u00a0 x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —<\/b><\/span><\/span><\/span>\u00a0x=10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m=1,0cm<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

\u00a0\u00a0Pela conserva\u00e7\u00e3o da quantidade de movimento\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=m.V + 2m.0\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=m.V\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span>=(m + 2m).V\u2019\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span><\/p>\n

\"\"<\/p>\n

=Q<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span>\u00a0 — m.V=3m.V\u2019\u00a0 —\u00a0 V\u2019=V\/3\u00a0 —\u00a0 supondo que a energia mec\u00e2nica do sistema de massa 3m se conserve, como a altura n\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

varia, toda energia energia cin\u00e9tica antes do contato com a mola se transforma em energia potencial el\u00e1stica ap\u00f3s a composi\u00e7\u00e3o parar e a mola for comprimida de x\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=3mV\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 —\u00a0 E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 3mV\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 (3m\/2)\/(V\/3)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 x = v \u00b7\u221a m\/ (3k)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

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

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

\u00a0Como as for\u00e7as externas s\u00e3o desprezadas, o sistema \u00e9 conservativo e a energia mec\u00e2nica \u00e9 constante sendo a soma da energia cin\u00e9tica com a potencial gravitacional:<\/b><\/span><\/span><\/span><\/p>\n

a) Falsa\u00a0 —\u00a0 a energia mec\u00e2nica \u00e9 constante.<\/b><\/span><\/span><\/span><\/p>\n

b) Falsa\u00a0 —\u00a0 quanto maior o afastamento, maior ser\u00e1 altura atingida (ponto B) e, consequentemente maior a energia potencial e claro, tamb\u00e9m a mec\u00e2nica.<\/b><\/span><\/span><\/span><\/p>\n

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

d) Falsa\u00a0 —\u00a0 a energia mec\u00e2nica m\u00e1xima total do sistema depende apenas da altura m\u00e1xima atingida pelo ponto B.<\/b><\/span><\/span><\/span><\/p>\n

e) Falsa\u00a0 —\u00a0 em B, onde V=0 a 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) \u00e9 nula e a energia potencial gravitacional (E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=mgh) \u00e9 m\u00e1xima.<\/b><\/span><\/span><\/span><\/p>\n

R – C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

\u00a0<\/span>68-<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da altura h\u00a0 —\u00a0 sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=h\/1440\u00a0 —\u00a0 h=720m\u00a0 —\u00a0 n\u00e3o havendo atrito o sistema pode ser considerado conservativo\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0 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 + m.g.h = 0 + m.10.720\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=7.200m\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.h = 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 + 0\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\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>\u00a0 — \u00a0mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = 7200m\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=14400\u00a0 — V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=120m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

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

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

Entre A e B n\u00e3o tem atrito e a for\u00e7a na dire\u00e7\u00e3o do movimento \u00e9 a for\u00e7a resultante\u00a0\"\"\u00a0(veja figura)\u00a0 —\u00a0 c\u00e1lculo da<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

acelera\u00e7\u00e3o do movimento\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a.t\u00a0 —\u00a0 50 = 0 + a.2\u00a0 —\u00a0 a= 25m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 intensidade da for\u00e7a resultante\u00a0 —\u00a0 princ\u00edpio fundamental da din\u00e2mica\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a=800.25\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=20 000N\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 entre B e C (come\u00e7o da subida da<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

rampa), houve energia perdida devido aos atritos de valor E<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=190 000J\u00a0 —\u00a0 energia mec\u00e2nica em B\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 +<\/b><\/span><\/span><\/span><\/p>\n

m.g.h=800.50<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 800.10.0\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=1 000 000J\u00a0 —\u00a0 energia mec\u00e2nica em C\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.h=800.5<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 +<\/b><\/span><\/span><\/span><\/p>\n

800.10.h\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=10 000 + 8 000h\u00a0 —\u00a0 observe que E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>\u00a0= E<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 1 000 000 = 190 000 + 10 000 + 8 000h\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

8 000h = 800 000\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>h=100m<\/b><\/span><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Como n\u00e3o existem for\u00e7as dissipativas a energia mec\u00e2nica \u00e9 constante\u00a0 —\u00a0 no ponto fornecido\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cA<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pA<\/b><\/span><\/span><\/sub><\/span>\u00a0 —<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=9 + 9=18J\u00a0 —\u00a0 quando chega ao solo h=0 e sua energia potencial gravitacional \u00e9 nula\u00a0 —\u00a0 no solo\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>ms<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh=4.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 4.10.0\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>ms<\/b><\/span><\/span><\/sub><\/span>=2V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>\u00a0= E<\/b><\/span><\/span><\/span>ms<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 18=2V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V=3m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Como o sistema \u00e9 conservativo toda energia mec\u00e2nica \u00e9 sempre a mesma em qualquer instante\u00a0 —\u00a0 imediatamente antes do salto ele deve ter a maior velocidade poss\u00edvel (EtapaI) e consequentemente maior energia cin\u00e9tica (Ec=m.V2\/2)\u00a0 —\u00a0 essa maior energia cin\u00e9tica, deve se transformar em energia potencial el\u00e1stica (Etapa II) que acontece quando a vara \u00e9 deformada\u00a0 —\u00a0 ent\u00e3o, essa m\u00e1xima energia potencial el\u00e1stica \u00e9 transformada em energia potencial gravitacional (Ep=m.g.h) para que a maior altura atingida (Etapa III) tamb\u00e9m seja m\u00e1xima\u00a0 —\u00a0\u00a0R- C.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Energia Mec\u00e2nica     \u00a001–Como n\u00e3o existe atrito a energia mec\u00e2nica se conserva e onde a energia cin\u00e9tica \u00e9 m\u00e1xima (ponto R), a velocidade \u00e9 m\u00e1xima e a energia potencial gravitacional \u00e9 m\u00ednima (altura m\u00ednima). Np ponto P a energia cin\u00e9tica \u00e9 m\u00ednima (menor velocidade) e a energia potencial gravitacional \u00e9 m\u00e1xima (maior<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1370,"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-1374","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1374","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=1374"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1374\/revisions"}],"predecessor-version":[{"id":10821,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1374\/revisions\/10821"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1370"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1374"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}