{"id":1217,"date":"2015-08-26T00:09:19","date_gmt":"2015-08-26T00:09:19","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1217"},"modified":"2024-08-23T13:40:54","modified_gmt":"2024-08-23T13:40:54","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-plano-inclinado-sem-atrito","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/plano-inclinado-sem-atrito\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-plano-inclinado-sem-atrito\/","title":{"rendered":"Plano inclinado sem atrito – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

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

 <\/p>\n

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

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

\"\"<\/p>\n

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

\"\"<\/p>\n

F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=Psenq\u00a0=mgsenq\u00a0 —\u00a0\u00a0 dire\u00e7\u00e3o paralela ao plano, no sentido para baixo (oposto ao do lan\u00e7amento) que retarda o bloco na subida e o acelera na descida.<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span>\u00a0R- C\u00a0 — \u00a0veja a teoria<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span>\u00a0For\u00e7as que agem sobre o balc\u00e3o na dire\u00e7\u00e3o do movimento:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span>P\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0 parcela do peso que tenta fazer o bloco descer o plano inclinado\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=m.g.sen30<\/b><\/span><\/span>o\u00a0\u00a0<\/b><\/span><\/span><\/sup>—\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=200.10.0,5\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=1.000N<\/b><\/span><\/span><\/p>\n

F\u00a0 — intensidade da for\u00e7a aplicada pelo empregado sobre o balc\u00e3o fazendo-o descer com velocidade constante, ou seja, est\u00e1 em equil\u00edbrio est\u00e1tico (F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0). Assim,\u00a0<\/b><\/span><\/span>F=P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=1.000N<\/b><\/span><\/span>.<\/b><\/span><\/span><\/sub><\/p>\n

04-<\/b><\/span><\/span>\u00a0P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=m.g.senq\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=5.10.0,8\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=40N<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como ele sobe com acelera\u00e7\u00e3o de 3m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 F \u2013 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 F \u2013 40=5.3\u00a0 —\u00a0<\/b><\/span><\/span>F=55N R- D<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span>\u00a0A for\u00e7a que reduz a velocidade do bloco \u00e9 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=m.g.senq\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 m.g.senq= m.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=g.senq<\/b><\/span><\/span><\/p>\n

Como ele p\u00e1ra em C \u2013 V=0\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.DS\u00a0 —\u00a0 0<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.(-g.senq).BC\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>BC = V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2.g.senq<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span>\u00a0C\u00e1lculo da acelera\u00e7\u00e3o de subida e de descida do carrinho\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 m.g.sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0= m.a\u00a0 —\u00a0 a=g.sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —<\/b><\/span><\/span><\/p>\n

a=10.0,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

1<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sup>\u00a0etapa\u00a0 —\u00a0 tempo de subida com a velocidade variando de V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=10m\/s a V=0 e com a=-5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

—\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a.t\u00a0 —\u00a0 0=10 \u2013 5t\u00a0 — t=2s (tempo de subida)\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

—\u00a0 dist\u00e2ncia percorrida na subida\u00a0 —\u00a0 d=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t + a.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 d=10.2 \u2013 5.2<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 d=10m\u00a0<\/b><\/span><\/span><\/p>\n

2<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sup>\u00a0etapa\u00a0 —\u00a0 c\u00e1lculo de d\u2019\u00a0 —\u00a0 sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=15\/d\u2019\u00a0 —\u00a0 d\u2019=15\/0,5\u00a0 —\u00a0 d\u2019=30m\u00a0 —\u00a0 tempo que o carrinho demora para ir do topo do plano inclinado (V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0) at\u00e9 sua base, percorrendoDS=40m com a=5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

DS= v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>.t + a.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 40=0.t + 5.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=16\u00a0 —\u00a0 t=4s\u00a0 —\u00a0 tempo total de subida e descida = 2 + 4=<\/b><\/span><\/span>6s<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span>\u00a0A parcela do peso paralela ao plano inclinado vale\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=mgsenq. A parcela de F (F<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>), paralela ao plano inclinado vale:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

cosq=cateto adjacente\/hipotenusa\u00a0 —\u00a0 cosq=F<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\/F\u00a0 —\u00a0 F<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=F.cosq\u00a0 —\u00a0 sendo F=P=mg\u00a0 —\u00a0 F<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=mgcosq<\/b><\/span><\/span><\/p>\n

Como o bloco sobe com acelera\u00e7\u00e3o a\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0\u2013 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 mgcosq\u00a0– mgsenq=m.a\u00a0 gcosq\u00a0– gsenq\u00a0=a\u00a0 —\u00a0 10.0,8 \u2013 10.0,6=a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span>\u00a0For\u00e7as que agem sobre o corpo na figura 1\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o corpo est\u00e1 em repouso\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 — P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P.sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0= F.cos30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P\/2 = F.\u00d63\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P=F.\u00d63 I\u00a0<\/b><\/span><\/span><\/p>\n

For\u00e7as que agem sobre o corpo na figura 2<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o corpo est\u00e1 em repouso\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P.sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0= F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P\/2=F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>P=2F<\/b><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub>II<\/b><\/span><\/span><\/p>\n

Igualando I com II\u00a0 —\u00a0 F.\u00d63 = 2.F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/F=\u00d63\/2<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span>\u00a0Em A\u00a0 —\u00a0 N=P\u00a0 —\u00a0 N=100N (indica\u00e7\u00e3o da balan\u00e7a)<\/b><\/span><\/span><\/p>\n

Em B\u00a0 —\u00a0 N=P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 N=P.cosa\u00a0 —\u00a0 N=100.cateto adjacente\/hipotenusa\u00a0 —\u00a0 N=100.40\/50\u00a0 —\u00a0 N=100.0,8\u00a0 —\u00a0 N=80N\u00a0 R- D<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span>\u00a0Da figura 2\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=6m\/s\u00a0 —\u00a0 V=0\u00a0 — t=1,2s\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a.t\u00a0 —0=6 + a.1,2\u00a0 —\u00a0 a= – 5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 em m\u00f3dulo a=5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

A for\u00e7a que retarda o bloco \u00e9 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0tal que\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=m.a\u00a0 —m.g.sena=ma\u00a0 —\u00a0 10.sena=5\u00a0 —\u00a0 sena=1\/2\u00a0 —\u00a0\u00a0a=30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/p>\n

Em todo gr\u00e1fico VXt o deslocamento do corpo \u00e9 fornecido pela \u00e1rea\u00a0 —\u00a0 de 0 a 0,4s a \u00e1rea \u00e9:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

DS=\u00e1rea do trap\u00e9zio=(B + b).h\/2\u00a0 —\u00a0\u00a0DS=(6 + 4).0,4\/2\u00a0 —\u00a0\u00a0DS=2m.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=h\/2\u00a0 —\u00a0 0,5=h\/2\u00a0 —\u00a0 h<\/b><\/span><\/span>=1m\u00a0\u00a0 R- B<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span>\u00a0a) Como n\u00e3o tem atrito\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=m.g.sen37<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=1.10.0,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=6N<\/b><\/span><\/span><\/p>\n

b) F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 6=1.a\u00a0 —\u00a0 a=6m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span>S=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>.t + a.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 12=0.t + 6.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 t=<\/b><\/span><\/span><\/span>\u00d6<\/b><\/span><\/span>4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>t=2s<\/b><\/span><\/span><\/span><\/p>\n

12<\/b><\/span><\/span><\/span>–\u00a0<\/b><\/span><\/span><\/span>Colocando as for\u00e7as:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Bloco A\u00a0 —\u00a0 sobe com acelera\u00e7\u00e3o de 2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0tal que\u00a0 —\u00a0 T \u2013 P<\/b><\/span><\/span>PA<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 T \u2013 m.g.sen37<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=1.2\u00a0 —\u00a0 T=1.10.0,6+2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=8N<\/b><\/span><\/span><\/p>\n

Bloco B\u00a0 —\u00a0 desce com a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 T=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.10 \u2013 8=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.2\u00a0 — 8m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=8\u00a0 —\u00a0 m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=1kg\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=10N\u00a0 R-D<\/b><\/span><\/span><\/p>\n

13<\/b><\/span><\/span>– Colocando as for\u00e7as:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0Repouso\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 bloco B\u00a0 —\u00a0\u00a0\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=T\u00a0 —\u00a0 T=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 T=8.10\u00a0 —\u00a0 T=80N\u00a0\u00a0\u00a0\u00a0 bloco A\u00a0 —\u00a0 P<\/b><\/span><\/span>PA<\/b><\/span><\/span><\/sub>=T\u00a0 — P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.sen45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=80\u00a0 —<\/b><\/span><\/span><\/p>\n

P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.\u00d62\/2=80\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=80.\u00d62N<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span>\u00a0a)Figura 1: M<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P=10.10\u00a0 —\u00a0 P=100N\u00a0 —\u00a0 M<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=m.g.sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=10.10.1\/2\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=50N<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

O sistema se move no sentido anti-hor\u00e1rio\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>100 \u2013 T=10.a I<\/b><\/span><\/span>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0T \u2013 50=10.a II\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 somando I com II\u00a0 —\u00a0 50 =20.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=2,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0e \u00e9 a mesma para os dois blocos.<\/b><\/span><\/span><\/p>\n

b) figura 2\u00a0 —\u00a0 observe que a polia de baixo \u00e9 m\u00f3vel, ent\u00e3o ela transmite apenas metade do peso de M<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0para M<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0sendo que a outra metade \u00e9 suportada pelo teto em A.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Assim, observe que a for\u00e7a resultante sobre M<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u00e9 nula e\u00a0<\/b><\/span><\/span>a=0<\/b><\/span><\/span>.\u00a0 O mesmo ocorre com M<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span>\u00a0\u00a0\u00a0m\u00a0 —\u00a0 P=10m\u00a0 —\u00a0 M\u00a0 — \u00a0observe na figura abaixo que\u00a0q\u00a0e\u00a0a\u00a0s\u00e3o complementares portanto sena=cosq\u00a0 —\u00a0\u00a0 \u00a0P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=mgsena=mgcosq\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=8.10.\u00d63\/2\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=40\u00d63N<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na figura acima que do peso de M apenas 10m \u00e9 transmitido a m, pois os outros 10m s\u00e3o sustentados pelo teto. Assim,<\/b><\/span><\/span><\/p>\n

10m=40\u00d63\u00a0 —\u00a0 m=4\u00d63kg.\u00a0 R- B<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span>\u00a0Como n\u00e3o existe atrito o sistema se move no sentido hor\u00e1rio. Colocando as for\u00e7as que atuam na dire\u00e7\u00e3o do movimento.<\/b><\/span><\/span><\/p>\n

P<\/b><\/span><\/span>PB<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.g.sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=3.10.0,5\u00a0 —\u00a0 P<\/b><\/span><\/span>PB<\/b><\/span><\/span><\/sub>=15N<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

bloco A\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=2.a I<\/b><\/span><\/span>\u00a0 —\u00a0 bloco B\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 P<\/b><\/span><\/span>PB<\/b><\/span><\/span><\/sub>\u00a0\u2013 T=3.a\u00a0\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>15 \u2013 T=3.a II\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 somando I com II\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

15=5.a\u00a0 —\u00a0 a=3m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 T=2.3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=6N.<\/b><\/span><\/span><\/p>\n

a.<\/b><\/span><\/span><\/p>\n

17-\u00a0<\/b><\/span><\/span>Dados: m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= 0,4 kg; m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= 0,6 kg\u00a0 —\u00a0 Analisando a figura:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

Como os corpos est\u00e3o em equil\u00edbrio, as for\u00e7as tamb\u00e9m se equilibram em todas as dire\u00e7\u00f5es: Assim:<\/b><\/span><\/span><\/p>\n

T = P<\/b><\/span><\/span>x1<\/b><\/span><\/span><\/sub>\u00a0T = P<\/b><\/span><\/span>x2\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 P<\/b><\/span><\/span>x2<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>x1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>gsen\u03b2=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>gsen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 sen<\/b><\/span><\/span>\u03b2<\/b><\/span><\/span><\/span>=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 sen<\/b><\/span><\/span>\u03b2<\/b><\/span><\/span><\/span>=1\/3\u00a0 —<\/b><\/span><\/span>\u00a0 \u03b2=arc sem 1\/3<\/b><\/span><\/span><\/p>\n

18-\u00a0<\/b><\/span><\/span>Como os atritos s\u00e3o desprezados, o movimento \u00e9 acelerado no plano inclinado e uniforme no plano horizontal:\u00a0<\/b><\/span><\/span><\/p>\n

 <\/p>\n

\"\"<\/p>\n

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

19-\u00a0<\/b><\/span><\/span>Como n\u00e3o existe atrito, a for\u00e7a resultante sobre o bloco na dire\u00e7\u00e3o do deslocamento, tem intensidade\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=mgsen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 pela lei fundamental da din\u00e2mica F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 mgsen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=ma\u00a0 —\u00a0 a=gsen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=10.1\/2\u00a0 —\u00a0 a=5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S=S<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t + at<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 4=0 +<\/b><\/span><\/span><\/p>\n

0.t + 5t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=1,6\u00a0 —\u00a0 t\u22481,26s
\n<\/b><\/span><\/span>20-\u00a0<\/b><\/span><\/span>A intensidade da for\u00e7a resultante sobre o bloco na dire\u00e7\u00e3o do movimento \u00e9 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=mgsen\u03b8=ma\u00a0 —\u00a0 a=gsen\u03b8\u00a0 —\u00a0 quando t=1s \u2013 S=d\u00a0 —\u00a0 d=at<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 d=gsen\u03b8.1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 d=gsen\u03b8\/2\u00a0 —\u00a0 quando t=3s\u00a0 —\u00a0 d\u2019=at<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 d\u2019=gsen\u03b8.9\/2\u00a0 —\u00a0 d\u2019\/d=(9.gsen\u03b8\/2)\/gsen\u03b8\/2\u00a0 —\u00a0 d\u2019=9d\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

21-\u00a0<\/b><\/span><\/span>a) mgsen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=ma\u00a0 —\u00a0 a=gsen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=10.1\/2\u00a0 —\u00a0 a=5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 falsa<\/b><\/span><\/span><\/p>\n

b) c\u00e1lculo do deslocamento \u0394S\u00a0 —\u00a0 equa\u00e7\u00e3o de Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a. \u0394S\u00a0 —\u00a0 6<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>+ 2.5.\u0394S\u00a0 —\u00a0 \u0394S=3,6m\u00a0 —\u00a0 altura h\u00a0 —\u00a0 sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=h\/3,6\u00a0 —\u00a0 h=1,8m\u00a0 —\u00a0 correta<\/b><\/span><\/span><\/p>\n

c) V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at\u00a0 —\u00a0 6=0 + 5t\u00a0 —\u00a0 t=1,2s\u00a0 —\u00a0 correta<\/b><\/span><\/span><\/p>\n

d) Na base\u00a0 —\u00a0 I=m.\u0394V=2.(6 \u2013 0)\u00a0 —\u00a0 I=12N.s\u00a0 —\u00a0 correta<\/b><\/span><\/span><\/p>\n

e) Trabalho na vertical\u00a0 —\u00a0 W=P.h=20.1,8\u00a0 —\u00a0 W=36J\u00a0 —\u00a0 correta<\/b><\/span><\/span><\/p>\n

R- (F,C,C,C,C)<\/b><\/span><\/span><\/p>\n

22-\u00a0<\/b><\/span><\/span>Os corpos se movimentam nos seguintes planos inclinados:<\/b><\/span><\/span><\/p>\n


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

As componentes dos pesos de\u00a0P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>, ao longo das respectivas trajet\u00f3rias, s\u00e3o dadas por\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=Psen\u03b2\u00a0 —\u00a0 F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=Psen\u03b1\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=sen\u03b2\/sen\u03b1\u00a0 —\u00a0 sen<\/b><\/span><\/span>\u03b2<\/b><\/span><\/span><\/span>=CF\/BF=6\/10\u00a0 —\u00a0 AF \u00e9 a diagonal do paralelep\u00edpedo\u00a0 — \u00a0(AF)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=36 + 64 + 225\u00a0 —\u00a0 AF=5\u221a13\u00a0 –sen<\/b><\/span><\/span>\u03b1<\/b><\/span><\/span><\/span>=CF\/AF=6\/5\u221a13\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=6\/10 x 5\u221a13\/6\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=\u221a13\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Observe que os dois tri\u00e2ngulos s\u00e3o pitag\u00f3ricos, ent\u00e3o a altura de I vale 4m e a de II, 3m\u00a0 —\u00a0 como n\u00e3o existe atrito<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

e a for\u00e7a peso \u00e9 conservativa (independe da trajet\u00f3ria), voc\u00ea pode consider\u00e1-las como em queda livre, com a=g e V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0, I da altura de 4m e II da altura de 3m\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 I\u00a0 —\u00a0 V<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0= V<\/b><\/span><\/span>oI<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2gh<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>=0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.10.4\u00a0 —\u00a0 V<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>\u00a0= \u221a80\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

II\u00a0 —\u00a0 V<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0= V<\/b><\/span><\/span>oII<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2gh<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>=0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.10.3\u00a0 —\u00a0 V<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>\u00a0= \u221a60\u00a0 —\u00a0 V<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>\u00a0= \u221a80\/\u221a60= (\u221a3.\u221a5)\/(2.\u221a5)\u00a0 —\u00a0 V<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>=\u221a3\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

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

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

A parcela do peso respons\u00e1vel pela descida do bloco A \u00e9 dada por P<\/b><\/span><\/span>pA<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>sen\u03b1=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.g.sen\u03b1\u00a0 —\u00a0 a figura mostra todas as for\u00e7as<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0que agem sobre o bloco\u00a0 —\u00a0 como o sistema est\u00e1 em equil\u00edbrio a for\u00e7a resultante sobre cada bloco \u00e9 nula\u00a0 —\u00a0 bloco A\u00a0 —\u00a0 P<\/b><\/span><\/span>pA<\/b><\/span><\/span><\/sub>=T\u00a0 —\u00a0 m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.g.sen\u03b1 =T\u00a0 —\u00a0 bloco B\u00a0 —\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=2T\u00a0 —\u00a0 m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.g<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>=2.m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.g.sen\u03b1\u00a0 —\u00a0 m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=2m<\/b><\/span><\/span>A.<\/b><\/span><\/span><\/sub>g.sen\u03b1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/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 Plano inclinado sem atrito   01- a) b) FR=PP=Psenq\u00a0=mgsenq\u00a0 —\u00a0\u00a0 dire\u00e7\u00e3o paralela ao plano, no sentido para baixo (oposto ao do lan\u00e7amento) que retarda o bloco na subida e o acelera na descida. 02-\u00a0R- C\u00a0 — \u00a0veja a teoria 03-\u00a0For\u00e7as que agem sobre o balc\u00e3o na dire\u00e7\u00e3o do movimento: PP\u00a0\u00a0—\u00a0 parcela do peso<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1213,"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-1217","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1217","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=1217"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1217\/revisions"}],"predecessor-version":[{"id":10851,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1217\/revisions\/10851"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1213"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1217"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}