{"id":1365,"date":"2015-09-20T15:28:22","date_gmt":"2015-09-20T15:28:22","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1365"},"modified":"2024-08-23T13:58:05","modified_gmt":"2024-08-23T13:58:05","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-tipos-de-energia","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/tipos-de-energia\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-tipos-de-energia\/","title":{"rendered":"Tipos de Energia – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

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

 <\/p>\n

01<\/b><\/span><\/span><\/span>– V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=36km\/h\/3,6 \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=10m\/s\u00a0 —\u00a0 a=-4m\/s<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>(freando)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>V=0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

a) V=V<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0+ a.t\u00a0 —\u00a0 0=10 -4.t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>t=2,5s<\/b><\/span><\/span><\/span><\/span><\/p>\n

b)\u00a0DS=V<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>.t + a.t<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2=10.2,5 \u2013 4.(2,5)<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>DS=12,5m<\/b><\/span><\/span><\/span><\/span><\/p>\n

c) Como as forces peso\u00a0\"\"\u00a0e a normal\u00a0\"\"\u00a0se anulam, a for\u00e7a resultante sobre o carro \u00e9 a for\u00e7a aplicada pelo freio\u00a0\"\"\u00a0\u00a0e contr\u00e1ria ao movimento<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

W=E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 m.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=800.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 800.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0 \u2013 40.000\u00a0 —\u00a0<\/b><\/span><\/span><\/span>W=- 40.000J<\/b><\/span><\/span><\/span>\u00a0(negativa, pois o carro est\u00e1 freando e dissipando energia cin\u00e9tica)<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0W=\u00e1rea=b.h\/2=10.20\/2\u00a0 —\u00a0 W=100J\u00a0 —\u00a0 W=E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 m.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 0\u00a0 —\u00a0 100=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V=10m\/s<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0W=E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>=1.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 1.8<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W= -32J\u00a0 R-\u00a0 D<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0W=0,2.(0,6))<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>\u2013 0,2.(0,3)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0,036 \u2013 0,009=0,027\u00a0 — \u00a0W=2,7.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>J\u00a0 R- A<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0R- B (veja teoria)<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=30m\/s\u00a0\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span>V=0\u00a0 —\u00a0 W=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 m.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0 \u2013 1000.900\/2= -900.000\/2=-450.000J\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0m=50g=50.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>=5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>kg \u2013\u00a0\u00a0DS=0,5cm=5.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=5.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>kg – Vo=2.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>m\/s \u2013 V=0\u00a0 —\u00a0 O trabalho realizado pela fibra de kevlar \u00e9 igual ao trabalho realizado pela for\u00e7a resultante resistiva da fibra que se op\u00f5e \u00e0 bala, parando-a.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

W=0 \u2013 5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.(2.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.4.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\/2=20.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 W=1.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 W=F.d.cosa\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=F.5.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.cos180<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=F.5.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.(-1)\u00a0 —\u00a0 F= – 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\/5.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>= – 10.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\/5\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 F= – 2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>N\u00a0 R-\u00a0 A<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=108km\/h=30m\/s \u2013 V=0 (p\u00e1ra) \u2013 m=80kg\u00a0 —\u00a0 W=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 m.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2= 0 \u2013 80.(30)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W= – 36.000J\u00a0 R- C<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0m=20g=20.10<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/sup><\/span>3=2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>kg \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=240m\/s \u2013 V=0<\/b><\/span><\/span><\/span><\/p>\n

a) E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.(240)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.10\u00a0<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.5,76.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>=576J\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 W=E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>=0 \u2013 576\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W= – 576J<\/b><\/span><\/span><\/span><\/p>\n

b)\u00a0DS=d=18cm=18.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 W=F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.d.cos180<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 -576=F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.18.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.(-1)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=3.200N<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0R- A (veja teoria<\/b><\/span><\/span><\/span><\/p>\n

11-R- A\u00a0<\/b><\/span><\/span><\/span>\u00a0(veja teoria)<\/b><\/span><\/span><\/span><\/p>\n

12- a) h=5.0,3=1,5m\u00a0 — E<\/b><\/span><\/span><\/span>pf<\/b><\/span><\/span><\/sub><\/span>=m.g.h=70.10.1,5=1.050J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>pf<\/b><\/span><\/span><\/sub><\/span>=1.050J<\/b><\/span><\/span><\/span><\/p>\n

b) W=E<\/b><\/span><\/span><\/span>pi<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>pf<\/b><\/span><\/span><\/sub><\/span>=0 \u2013 1.050\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W= – 1.050J<\/b><\/span><\/span><\/span><\/p>\n

c) N\u00e3o, seria o mesmo, pois o trabalho da for\u00e7a peso n\u00e3o depende da trajet\u00f3ria (a for\u00e7a peso \u00e9 uma for\u00e7a conservativa).<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span><\/span>\u00a0a) W(A)=E<\/b><\/span><\/span><\/span><\/span>pi<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span><\/span>pf<\/b><\/span><\/span><\/span><\/sub><\/span>=50.10.0 \u2013 50.10.16= 0 \u2013 8.00\u00a0 —\u00a0 W(A)= – 8.000J\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0 W(B)=E<\/b><\/span><\/span><\/span><\/span>pi<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span><\/span>pf<\/b><\/span><\/span><\/span><\/sub><\/span>=50.10.0 \u2013 50.10.16= 0 \u2013 8.00\u00a0 —\u00a0 W(B)= – 8.000J<\/b><\/span><\/span><\/span><\/span><\/p>\n

W(A)\/W(B)=1<\/b><\/span><\/span><\/span><\/span><\/p>\n

b) Como as velocidades inicial e final s\u00e3o nulas, a varia\u00e7\u00e3o de energia cin\u00e9tica tamb\u00e9m ser\u00e1 nula e a varia\u00e7\u00e3o de energia mec\u00e2nica ser\u00e1 igual \u00e0 varia\u00e7\u00e3o de energia potencial gravitacional que \u00e9 de\u00a0<\/b><\/span><\/span><\/span>8.000J, tanto para Abelardo, quanto para Helo\u00edsa.<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0Todas est\u00e3o corretas – soma=15<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0W=E<\/b><\/span><\/span><\/span>pi<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>pf<\/b><\/span><\/span><\/sub><\/span>= 0 \u2013 450.10.3,2= – 2,5.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>J (aproximadamente)<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0A energia mec\u00e2nica \u00e9 igual a 24\/80% = 24\/0,8 = 30 J<\/b><\/span><\/span><\/span><\/p>\n

Esta energia \u00e9 a energia potencial gravitacional do corpo, E = mgh, ent\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

mgh = 30<\/b><\/span><\/span><\/span><\/p>\n

2.10.h = 30<\/b><\/span><\/span><\/span><\/p>\n

20h = 30<\/b><\/span><\/span><\/span><\/p>\n

h = 30\/20\u00a0\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0h= 1,5 m<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0a) Falsa \u2013W=m.g.h=80.10.5=4.000J<\/b><\/span><\/span><\/span><\/p>\n

b) Falsa, basta apenas ela subir com qualquer tipo de movimento<\/b><\/span><\/span><\/span><\/p>\n

c) Verdadeira \u2013 W=m.<\/b><\/span><\/span><\/span>g<\/b><\/span><\/span><\/span>.h<\/b><\/span><\/span><\/span><\/p>\n

d) Falsa, existe trabalho, pois houve varia\u00e7\u00e3o de energia<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0a)\u00a0 d=1g\/cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg\/m<\/b><\/span><\/span><\/span>3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>V=50m<\/b><\/span><\/span><\/span>3 por segundo\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>—\u00a0\u00a0 d=m\/V\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=m\/50\u00a0 —\u00a0<\/b><\/span><\/span><\/span>m=5.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>kg por segundo<\/b><\/span><\/span><\/span><\/p>\n

b) A energia potencial gravitacional armazenada por m=5.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>kg de \u00e1gua numa altura de 160m em 1s, num local onde g=10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00e9 dada por\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=m.g.h=5.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.10.160\u00a0 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=8.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>J<\/b><\/span><\/span><\/span><\/p>\n

c) P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/Dt=4,2.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\/60\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=7.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>W<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0R- A \u2013 Observe que o gr\u00e1fico \u00e9 de E em fun\u00e7\u00e3o de\u00a0<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>pois se fosse em fun\u00e7\u00e3o de x seria a alternativa b<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0Quando x=1m, F=500N\u00a0 —\u00a0 F=K.x\u00a0 —\u00a0 500=K.1\u00a0 —\u00a0 K=500N\/m\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>=K.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = 500.(0,6)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>=90J<\/b><\/span><\/span><\/span>\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>=K.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=10.0,04\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>=0,2J\u00a0 R- B<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0E<\/b><\/span><\/span><\/span>pei<\/b><\/span><\/span><\/sub><\/span>=K.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=150.(2,0.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\/2=150.4,010<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\/2= 300.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>pef<\/b><\/span><\/span><\/sub><\/span>=150.(2,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=468,75.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 W=DE<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pei<\/b><\/span><\/span><\/sub><\/span>\u00a0– E<\/b><\/span><\/span><\/span>pef<\/b><\/span><\/span><\/sub><\/span>= 300.10<\/b><\/span><\/span><\/span>-4\u00a0<\/b><\/span><\/span><\/sup><\/span>\u2013 468,75.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>= – 168,75.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>= – 1,6875.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>J (negativo, pois o deslocamento \u00e9 for\u00e7ado por uma for\u00e7a externa) ou, graficamente\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

quando x=2,0.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m \u2013 F=K.x=150.2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=<\/b><\/span><\/span><\/span>3N<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 quando x=2,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m \u2013 F=K.x=150.2,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=<\/b><\/span><\/span><\/span>3,75m<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 colocando esses valores no gr\u00e1fico F X x, teremos:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Energia potencial armazenada quando x=2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 W=E<\/b><\/span><\/span><\/span>pei<\/b><\/span><\/span><\/sub><\/span>=\u00e1rea do tri\u00e2ngulo=b.h\/2=2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.3\/2=3,0.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>J<\/b><\/span><\/span><\/span><\/p>\n

Energia potencial armazenada quando x=2,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 W=E<\/b><\/span><\/span><\/span>pei<\/b><\/span><\/span><\/sub><\/span>=\u00e1rea do tri\u00e2ngulo=b.h\/2=2,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.3,75\/2=4,6875.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>J<\/b><\/span><\/span><\/span><\/p>\n

\u2206E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>= W=\u2206E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pei<\/b><\/span><\/span><\/sub><\/span>\u00a0– E<\/b><\/span><\/span><\/span>pef<\/b><\/span><\/span><\/sub><\/span>=3,0.10\u00a0<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/sup><\/span>\u00a04,6875.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0= –<\/b><\/span><\/span><\/span>1,6875.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>ou ainda, pela \u00e1rea do trap\u00e9zio que representa o trabalho realizado (varia\u00e7\u00e3o de energia potencial el\u00e1stica)\u00a0 —\u00a0 W=\u2206E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>=(B + b).h\/2=(3,75 + 3,0).0,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\/2=<\/b><\/span><\/span><\/span>1,6875.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>J\u00a0\u00a0 R-E<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0a) Do enunciado e do gr\u00e1fico\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

W<\/b><\/span><\/span><\/span>de B para A<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pB<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>pA<\/b><\/span><\/span><\/sub><\/span>=24.000J\u00a0 —\u00a0 24.000= E<\/b><\/span><\/span><\/span>pB<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>pA<\/b><\/span><\/span><\/sub><\/span>=m.g.y<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 m.g.y<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=P.y<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 P.y<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 24.000=P.30 \u2013 P.0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P=800N<\/b><\/span><\/span><\/span><\/p>\n

Observe no gr\u00e1fico que a energia el\u00e1stica s\u00f3 come\u00e7a a surgir a partir de 20m, quando a corda come\u00e7a a ser esticada. Assim,\u00a0<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=20m<\/b><\/span><\/span><\/span><\/p>\n

b) Se o comprimento natural da mola \u00e9 20m e ela foi deformada at\u00e9 30m, a deforma\u00e7\u00e3o \u00e9 x=30 \u2013 20=10m, e para efetuar essa deforma\u00e7\u00e3o a energia potencial el\u00e1stica foi de 24.000J (veja gr\u00e1fico)\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>el\u00e1stica<\/b><\/span><\/span><\/sub><\/span>=K.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 24.000=K.(10)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>K=480N\/m<\/b><\/span><\/span><\/span><\/p>\n

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

24-<\/b><\/span><\/span><\/span>\u00a001. Correta\u00a0 —\u00a0 trabalho \u00e9 a medida da quantidade de energia transferida por uma for\u00e7a ao longo de um deslocamento\u00a0 —\u00a0 em F\u00edsica, trabalho \u00e9 transfer\u00eancia de energia, provocando deslocamento.<\/b><\/span><\/span><\/span><\/p>\n

02. Correta\u00a0 —\u00a0 observe na express\u00e3o E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 que a energia \u00e9 diretamente proporcional ao quadrado da velocidade e, como V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=4V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0ela ficar\u00e1 4<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=16 vezes maior.<\/b><\/span><\/span><\/span><\/p>\n

04. Correta\u00a0 —\u00a0 energia potencial \u00e9 energia de posi\u00e7\u00e3o, dependendo, portanto, de um referencial.<\/b><\/span><\/span><\/span><\/p>\n

08. Correta\u00a0 —\u00a0 a cada transforma\u00e7\u00e3o, parte da energia transformada \u00e9 dissipada, pois n\u00e3o h\u00e1 transforma\u00e7\u00e3o com 100% de rendimento.<\/b><\/span><\/span><\/span><\/p>\n

16. Correta\u00a0 —\u00a0 \u00a0energia cin\u00e9tica prov\u00e9m de outras formas de energia.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- (01 + 02 + 04 + 08 + 16) = 31<\/b><\/span><\/span><\/span><\/p>\n

25-\u00a0<\/b><\/span><\/span><\/span>(01) Errada. O trabalho da for\u00e7a peso \u00e9 igual a m g h. A for\u00e7a que provoca rota\u00e7\u00e3o \u00e9 resist\u00eancia do ar e n\u00e3o o peso.<\/b><\/span><\/span><\/span><\/p>\n

(02) Correta. Como a folha cai com velocidade constante, pelo princ\u00edpio da in\u00e9rcia, a resultante das for\u00e7as sobre ela \u00e9 nula, ou seja, a for\u00e7a de resist\u00eancia do ar sobre ela e seu peso t\u00eam mesma intensidade.<\/b><\/span><\/span><\/span><\/p>\n

(04) Correta. Se as velocidades de transla\u00e7\u00e3o e de rota\u00e7\u00e3o s\u00e3o constantes, as energias cin\u00e9ticas de transla\u00e7\u00e3o e rota\u00e7\u00e3o tamb\u00e9m o s\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

(08) Correta. Como j\u00e1 afirmado, a for\u00e7a que provoca rota\u00e7\u00e3o \u00e9 a de resist\u00eancia do ar e n\u00e3o o peso.<\/b><\/span><\/span><\/span><\/p>\n

(16) Errada. A folha est\u00e1 perdendo energia potencial. Como a energia cin\u00e9tica total permanece constante, a energia mec\u00e2nica est\u00e1 diminuindo.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- (02+ 04 + 08) = 14<\/b><\/span><\/span><\/span><\/p>\n

26-\u00a0<\/b><\/span><\/span><\/span>Para atingir a altura desejada ele deve ganhar de energia potencial a quantidade \u0394E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>, com N idas e vindas\u00a0 —\u00a0 \u0394E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=mgh-800.3\u00a0 —\u00a0 \u0394E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=2.400J\u00a0 —\u00a0 \u0394E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=NE<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2.400=N.600\u00a0 —\u00a0 N=4\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span><\/p>\n

28-\u00a0<\/b><\/span><\/span><\/span>a) Dados: \u0394x = 6,4 nm = 6,4.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>\u00a0m; d = 100 nm = 100.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>\u00a0m = 10\u20137 m; b\u00ad = 9,6.10<\/b><\/span><\/span><\/span>-39<\/b><\/span><\/span><\/sup><\/span>N.m<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>;
\nF<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=b\/d<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=k \u0394x\u00a0 —\u00a0 k=b\/d<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.\u0394x\u00a0 —\u00a0 k=9,6.10<\/b><\/span><\/span><\/span>-39<\/b><\/span><\/span><\/sup><\/span>\/10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>.6,4.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0k=0.015N\/m<\/b><\/span><\/span><\/span><\/p>\n

b) Dados: E<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0= kB<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>; kB = 1,4 .10<\/b><\/span><\/span><\/span>-23<\/b><\/span><\/span><\/sup><\/span>\u00a0J\/K; T = 300 K; k<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0,21 N\/m = 2,1 10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0N\/m\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 k\u0394x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>2=k<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>T\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u0394x=\u221a(2.1,4.10<\/b><\/span><\/span><\/span>-23<\/b><\/span><\/span><\/sup><\/span>.300)\/2,1.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u0394x=\u221a8,4.10<\/b><\/span><\/span><\/span>-21<\/b><\/span><\/span><\/sup><\/span>\/2,1.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u0394x=\u221a4.10<\/b><\/span><\/span><\/span>-20<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u0394x=2.10<\/b><\/span><\/span><\/span>-10<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394x=0,2nm<\/b><\/span><\/span><\/span><\/p>\n

29-\u00a0<\/b><\/span><\/span><\/span>Observe a figura\u00a0 —\u00a0 como o bloco est\u00e1 em equil\u00edbrio, a resultante das for\u00e7as sobre ele \u00e9 nula. Ou<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

seja, a for\u00e7a el\u00e1stica \u00e9 equilibrada pela componente tangencial do peso\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 kx=Psen\u03b8\u00a0 —\u00a0 x=10.0,8\/100\u00a0 —\u00a0 x=8,0.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 como centro de massa do bloco est\u00e1 abaixo de plano referencial, sua energia potencial gravitacional \u00e9 negativa\u00a0 —\u00a0 soma pedida das energias\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"= 0,32 \u2013 1\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/sub><\/span>= \u2013<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>0,68 J.\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

30-\u00a0<\/b><\/span><\/span><\/span>Em rela\u00e7\u00e3o a Jo\u00e3o, as velocidade de Pedro e Paulo s\u00e3o, respectivamente\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>Pe<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>Pa<\/b><\/span><\/span><\/sub><\/span>=1m\/s\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>cPe<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>cPa<\/b><\/span><\/span><\/sub><\/span>=60.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>cPa<\/b><\/span><\/span><\/sub><\/span>=30J\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

32-\u00a0<\/b><\/span><\/span><\/span>Observe nas figuras abaixo as for\u00e7as que agem em cada uma das partes:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

a) Como as partes se movem em movimento retil\u00edneo e uniforme, a resultante das for\u00e7as em cada uma delas \u00e9 nula\u00a0 —\u00a0\u00a0 corpo A\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 \u00a0N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= 5 N<\/b><\/span><\/span><\/span><\/p>\n

F = A<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F = \u03bc<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2022\u00a0N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u00a0F = 1,5 N\u00a0 —\u00a0\u00a0 corpo B\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= 10 N\u00a0 —\u00a0 T = F + A<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 T = F + \u03bc<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2022\u00a0N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

T = 4,5N<\/b><\/span><\/span><\/span><\/p>\n

b) A for\u00e7a T \u00e9 constante\u00a0 –p-\u00a0 trabalho da for\u00e7a T\u00a0 —\u00a0 W = T\u00a0\u2022\u00a0\u0394s\u00a0\u2022\u00a0cos \u03b1\u00a0 —\u00a0 o \u00e2ngulo entre a for\u00e7a e o deslocamento \u00e9 nulo\u00a0 —\u00a0\u00a0 \u00a0cos \u03b1 = 1\u00a0 —\u00a0 como a velocidade dos corpos tamb\u00e9m \u00e9 constante\u00a0 —\u00a0 \u00a0\u0394s = v\u00a0\u2022\u00a0\u0394t\u00a0 —\u00a0 W = T\u00a0\u2022\u00a0v\u00a0\u2022\u00a0\u0394t\u00a0 —\u00a0 W = 4,5\u00a0\u2022\u00a00,1\u00a0\u2022\u00a0120\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

W = 54 J<\/b><\/span><\/span><\/span><\/p>\n

c) Do resultado obtido no item a\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0F = 1,5 N<\/b><\/span><\/span><\/span><\/p>\n

d) Chamando de \u03b4 a deforma\u00e7\u00e3o da mola\u00a0 —\u00a0 F = k\u00a0\u2022\u00a0\u03b4\u00a0 —\u00a0 1,5 = 10\u00a0\u2022\u00a0\u03b4 \u00a0—\u00a0 \u03b4 = 0,15 m\u00a0 —\u00a0 comprimento da mola\u00a0 —\u00a0 x = x<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ \u03b4\u00a0 —\u00a0 x = 0,1 + 0,15\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>x = 0,25 m ou x = 25 cm<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

34-\u00a0<\/b><\/span><\/span><\/span>A energia potencial independe da trajet\u00f3ria\u00a0 —\u00a0 apenas a III est\u00e1 correta\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

35-\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 se a massa cai pela metade a energia cin\u00e9tica de 1 tamb\u00e9m cai pela metade\u00a0 —\u00a0 se V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=4V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 a energia cin\u00e9tica de 1 fica 4<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=16 vezes maior\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>x16\/2=8E\u00ad<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

36-\u00a0<\/b><\/span><\/span><\/span>Como as perdas de energia por dissipa\u00e7\u00e3o s\u00e3o desprezadas, a energia mec\u00e2nica total do sistema estudante + corda el\u00e1stica \u00e9 conservada. No in\u00edcio, n\u00e3o h\u00e1 energia cin\u00e9tica do estudante, que inicia o salto a partir do repouso, nem energia potencial el\u00e1stica (corda sem tens\u00e3o). No ponto mais baixo do salto, tamb\u00e9m n\u00e3o h\u00e1 energia cin\u00e9tica do estudante, que apresenta velocidade nula. Al\u00e9m disso, a energia cin\u00e9tica da corda \u00e9 desprezada. Assim, a energia potencial el\u00e1stica da corda nesse ponto \u00e9 igual ao m\u00f3dulo da varia\u00e7\u00e3o de energia potencial gravitacional do estudante, isto \u00e9, P<\/b><\/span><\/span><\/span>H<\/b><\/span><\/span><\/sub><\/span>\u00a040 = 24.000 J\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\uf020<\/b><\/span><\/span><\/span>= 600.\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

37-\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=mgh\u00a0 —\u00a0 M andares\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=Mmgh\u00a0 —\u00a0 N pessoas\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=NMmgh\u00a0 —\u00a0 duas vezes ao dia\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=2NMmgh\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

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

38-(ENEM-MEC-011)<\/b><\/span><\/span><\/span><\/p>\n

Pode-se definir o Segundo Princ\u00edpio da Termodin\u00e2mica da seguinte maneira: \u201c\u00c9 imposs\u00edvel obter uma m\u00e1quina t\u00e9rmica que, operando em ciclos, seja capaz de transformar totalmente o calor por ela recebido em trabalho\u201d\u00a0 — sempre haver\u00e1 energia dissipada pelo motor\u00a0 —\u00a0\u00a0\u00a0R- C.\u00a0<\/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 Tipos de Energia   01– Vo=36km\/h\/3,6 \u2013 Vo=10m\/s\u00a0 —\u00a0 a=-4m\/s2\u00a0(freando)\u00a0 —\u00a0\u00a0V=0 a) V=Vo\u00a0+ a.t\u00a0 —\u00a0 0=10 -4.t\u00a0 —\u00a0\u00a0t=2,5s b)\u00a0DS=Vo.t + a.t2\/2=10.2,5 \u2013 4.(2,5)2\/2\u00a0 —\u00a0\u00a0DS=12,5m c) Como as forces peso\u00a0\u00a0e a normal\u00a0\u00a0se anulam, a for\u00e7a resultante sobre o carro \u00e9 a for\u00e7a aplicada pelo freio\u00a0\u00a0\u00a0e contr\u00e1ria ao movimento W=Ecf\u00a0\u2013 Eci=mV2\/2 \u2013 m.Vo2\/2=800.02\/2 \u2013 800.102\/2=0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1361,"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-1365","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1365","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=1365"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1365\/revisions"}],"predecessor-version":[{"id":10864,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1365\/revisions\/10864"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1361"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1365"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}