{"id":1207,"date":"2015-08-25T23:58:17","date_gmt":"2015-08-25T23:58:17","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1207"},"modified":"2024-08-23T12:54:02","modified_gmt":"2024-08-23T12:54:02","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-aplicacoes-das-leis-de-newton-em-blocos-apoiados-ou-suspensos","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/aplicacoes-das-leis-de-newton-em-blocos-apoiados-ou-suspensos\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-aplicacoes-das-leis-de-newton-em-blocos-apoiados-ou-suspensos\/","title":{"rendered":"Aplica\u00e7\u00f5es das Leis de Newton em blocos apoiados ou suspensos – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span> <\/b>Aplica\u00e7\u00f5es das Leis de Newton em blocos apoiados ou suspensos<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

 <\/p>\n

01<\/b><\/span><\/span>– a) Como n\u00e3o existe atrito, por menor que seja a massa do sistema e a for\u00e7a aplicada o sistema sempre se mover\u00e1 no caso, para a direita. Colocando as for\u00e7as que agem sobre cada bloco apenas na dire\u00e7\u00e3o do movimento (for\u00e7as horizontais).<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

N<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>\u00a0\u2013 intensidade da for\u00e7a que A aplica em B.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 N<\/b><\/span><\/span>BA<\/b><\/span><\/span><\/sub>\u00a0\u2013 intensidade da for\u00e7a com que B reage em A. Como N<\/b><\/span><\/span>\u00a0AB<\/b><\/span><\/span><\/sub>\u00a0e N<\/b><\/span><\/span>BA<\/b><\/span><\/span><\/sub>\u00a0constituem par a\u00e7\u00e3o e rea\u00e7\u00e3o elas tem a mesma intensidade que chamaremos de N.<\/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>20 \u2013 N=3.a I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>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<\/b><\/span><\/span>\u00a0N=1.a II<\/b><\/span><\/span><\/p>\n

Somando I com II, obtemos\u00a0 —\u00a0 20=4a\u00a0 —\u00a0 a=5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(\u00e9 a mesma para cada bloco, pois se movem juntos).<\/b><\/span><\/span><\/p>\n

b) Substituindo a=5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0em I ou em II obtemos\u00a0 —\u00a0 N=5N.<\/b><\/span><\/span><\/p>\n

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

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

Observe nas figuras acima que F<\/b><\/span><\/span>RA<\/b><\/span><\/span><\/sub>=20 \u2013 5=15N e que F<\/b><\/span><\/span>RB<\/b><\/span><\/span><\/sub>=5N.<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span>\u00a0Colocando as for\u00e7as que influem no movimento:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Bloco P\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F \u2013 N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=6.a I<\/b><\/span><\/span>\u00a0 —\u00a0 bloco Q\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=4.a\u00a0 II\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>bloco R\u00a0 —\u00a0 N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= m<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 2.a\u00a0 III\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>somando I com II com III, obtemos 48=12\u00aa\u00a0 —\u00a0 a=4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>. A for\u00e7a pedida tem intensidade N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2.a\u00a0 —<\/b><\/span><\/span><\/p>\n

N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2.4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=8N<\/b><\/span><\/span><\/p>\n

03<\/b><\/span><\/span>– Vamos achar a intensidade de\u00a0\"\", considerando os 4 blocos como um s\u00f3, de massa M=12kg e aplicar a lei fundamental\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 F=12.2\u00a0 —\u00a0 F=24N<\/b><\/span><\/span><\/p>\n

Colocando as for\u00e7as apenas sobre o bloco M que s\u00e3o F=24N e a rea\u00e7\u00e3o a entre N e M que \u00e9 N:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Bloco M\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 24 \u2013 N = 3.2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N =18N<\/b><\/span><\/span><\/p>\n

04<\/b><\/span><\/span>– Em fios a for\u00e7a \u00e9 de tra\u00e7\u00e3o\u00a0\"\":<\/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=10.a I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>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 F \u2013 T=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>60 \u2013 T=20.a II<\/b><\/span><\/span>\u00a0 —\u00a0 Somando I com II\u00a0 —<\/b><\/span><\/span><\/p>\n

60 = 30.a\u00a0 —\u00a0 a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, que, substitu\u00eddo em I ou II, fornecer\u00e1\u00a0 —\u00a0 T=20N\u00a0\u00a0\u00a0\u00a0\u00a0 R- E<\/b><\/span><\/span><\/p>\n

05<\/b><\/span><\/span>– Bloco A\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1.10\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0– T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Bloco B\u00a0 —\u00a0\u00a0 FR=m.a\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 T<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=2.10\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 T<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=20<\/b><\/span><\/span><\/p>\n

Bloco C\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 T<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=3.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=30N<\/b><\/span><\/span>, que substitu\u00eddo em T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 T<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=20\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>-30=20\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=50N<\/b><\/span><\/span><\/p>\n

T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=10\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 50=10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=60N<\/b><\/span><\/span><\/p>\n

06<\/b><\/span><\/span>– (a) Colocando as for\u00e7as que influem no movimento:<\/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=3.a I<\/b><\/span><\/span>\u00a0\u00a0 —\u00a0\u00a0 Bloco B\u00a0\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a\u00a0\u00a0 —\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 —\u00a0\u00a0 m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.g \u2013 T=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a\u00a0\u00a0 —\u00a0\u00a0 1.10 \u2013 T=1.a<\/b><\/span><\/span><\/p>\n

10 \u2013 T=a II<\/b><\/span><\/span>\u00a0\u00a0 —\u00a0\u00a0 somando I com II\u00a0\u00a0 —\u00a0\u00a0 10=4.a\u00a0\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>a=2,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

T=3.a\u00a0\u00a0 —\u00a0\u00a0 T=3.2,5\u00a0\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>T=7,5N<\/b><\/span><\/span><\/p>\n

b) Sobre o bloco A deixa de existir a for\u00e7a de tra\u00e7\u00e3o\u00a0\"\", a for\u00e7a resultante sobre ele torna-se nula, ele fica em equil\u00edbrio din\u00e2mico<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

e segue em MRU com velocidade constante\u00a0\"\", at\u00e9 se chocar com a polia.<\/b><\/span><\/span><\/p>\n

Sobre o bloco B deixa de existir a for\u00e7a de tra\u00e7\u00e3o\u00a0\"\"e a for\u00e7a resultante sobre ele fica sendo apenas seu peso\u00a0 —\u00a0 F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>\u00a0=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a = m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 a = g\u00a0 —\u00a0 ele cai em queda livre com acelera\u00e7\u00e3o da gravidade.<\/b><\/span><\/span><\/p>\n

07<\/b><\/span><\/span>– Colocando o peso do bloco m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u00a0 —\u00a0\u00a0 P=1.10\u00a0 —P=10N<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

O sistema se move no sentido anti-hor\u00e1rio, pois F>P, ou seja, m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0sobe e m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0se desloca para a esquerda.<\/b><\/span><\/span><\/p>\n

Bloco m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0\u00a0 —\u00a0\u00a0 F \u2013 T=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.a\u00a0\u00a0 —\u00a0\u00a0 40 \u2013 T=1.a\u00a0\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>40 \u2013 T=a I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>\u00a0Bloco m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u00a0 —\u00a0\u00a0 T \u2013 P=m.a\u00a0\u00a0 —\u00a0\u00a0 T \u2013 10=1.a\u00a0\u00a0 —<\/b><\/span><\/span><\/p>\n

T \u2013 10=a II\u00a0\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>somando I com II\u00a0\u00a0 —\u00a0\u00a0 40 \u2013 10=2.a\u00a0\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>a=15m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0 —\u00a0\u00a0 T \u2013 10=1.a\u00a0\u00a0 —\u00a0\u00a0 T=10 + 15\u00a0\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>T=25N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

08<\/b><\/span><\/span>– a)Colocando as for\u00e7as:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

bloco 1\u00a0\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.a\u00a0\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=40.a I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>bloco 2<\/b><\/span><\/span>\u00a0\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.a\u00a0\u00a0\u00a0<\/b><\/span><\/span>—\u00a0\u00a0 T<\/b><\/span><\/span>B\u00ad\u00a0<\/b><\/span><\/span><\/sub>\u00a0– T<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=20.a II<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 bloco 3\u00a0\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>.a\u00a0\u00a0 —<\/b><\/span><\/span><\/p>\n

T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 P=m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>.a\u00a0\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 600=60.a III\u00a0\u00a0\u00a0<\/b><\/span><\/span>—\u00a0\u00a0 somando I com II com III\u00a0\u00a0 —\u00a0\u00a0 600=120.a\u00a0\u00a0 —\u00a0\u00a0 a=5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, que, substitu\u00eddo em III nos fornece\u00a0\u00a0 —\u00a0\u00a0 T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=600 + 60.5\u00a0\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=900N<\/b><\/span><\/span><\/p>\n

09<\/b><\/span><\/span>– a) P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.g\u00a0\u00a0 —\u00a0\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=6.10\u00a0\u00a0 —\u00a0\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=60N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.g\u00a0\u00a0 —\u00a0\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=10.10\u00a0\u00a0 —\u00a0\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=100N<\/b><\/span><\/span><\/p>\n

Colocando as for\u00e7as que agem sobre cada bloco e tirando o dinam\u00f4metro, pois sua massa \u00e9 desprez\u00edvel.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe que, como P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>>P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>, o sistema se move no sentido anti-hor\u00e1rio (A sobe e B desce).<\/b><\/span><\/span><\/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 —<\/b><\/span><\/span>\u00a0 T \u2013 60=6.a I<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\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\u00a0<\/b><\/span><\/span>100 \u2013 T=10.a II\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>somando I com II\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

100 \u2013 60=16.a\u00a0 —\u00a0 a=40\/16\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=2,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

b) o dinam\u00f4metro indica a intensidade da for\u00e7a de tra\u00e7\u00e3o no fio no qual ele est\u00e1 inserido, ou seja, indica T\u00a0 —\u00a0 T \u2013 60=6.a\u00a0 —<\/b><\/span><\/span><\/p>\n

T=60 + 6.2,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=75N<\/b><\/span><\/span><\/p>\n

10<\/b><\/span><\/span>– R- B (veja teoria \u00a0<\/b><\/span><\/span><\/p>\n

11<\/b><\/span><\/span>– primeiro carrinho\u00a0 —\u00a0 F \u2013 T=40.0,5\u00a0 —\u00a0 F \u2013 T=20\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 segundo carrinho\u00a0 —\u00a0 T=100.0,5\u00a0 —\u00a0 T=50N\u00a0 —\u00a0 F \u2013 50=20\u00a0 —\u00a0 F=70N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 R- C<\/b><\/span><\/span><\/p>\n

12<\/b><\/span><\/span>– Considerando o princ\u00edpio fundamental da Din\u00e2mica, F(resultante) = massa x acelera\u00e7\u00e3o temos:\u00a0 —\u00a0 Na descida: mg – T = ma<\/b><\/span><\/span><\/p>\n

Na subida: T’ – mg = ma\u00a0 —\u00a0 Isolando as tra\u00e7\u00f5es\u00a0 —\u00a0 T = mg – ma = m(g – a)\u00a0 —\u00a0 T’ = mg + ma = m(g + a)<\/b><\/span><\/span><\/p>\n

Ent\u00e3o\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>T’\/T = (g + a)\/(g – a)<\/b><\/span><\/span><\/p>\n

13<\/b><\/span><\/span>– A acelera\u00e7\u00e3o de cada\u00a0 bloco em cada caso \u00e9 a mesma, pois F \u00e9 a mesma e a massa do sistema (3m) \u00e9 a mesma.<\/b><\/span><\/span><\/p>\n

Primeira situa\u00e7\u00e3o\u00a0 —\u00a0 bloco m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=m.a I<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 segunda situa\u00e7\u00e3o\u00a0 —\u00a0 bloco2 m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2m.a II<\/b><\/span><\/span><\/p>\n

comparando I com II\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 2T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/p>\n

14<\/b><\/span><\/span>– A propor\u00e7\u00e3o de comprimento \u00e9 v\u00e1lida tamb\u00e9m para massa. Separando as fra\u00e7\u00f5es:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Considerando o sistema todo\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=(m\/3 + 2m\/3).a\u00a0 —\u00a0 F=(3m\/3).a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=m.a I<\/b><\/span><\/span><\/p>\n

Sobre a massa m\/3\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m\/3.a\u00a0 —\u00a0 F \u2013 N=m\/3.a\u00a0 —\u00a0 3F \u2013 3N=m.a\u00a0 —\u00a0\u00a0 veja em I que F=m.a\u00a0 —\u00a0 3F \u2013 3N=F\u00a0 —\u00a0 3N=2F\u00a0 —<\/b><\/span><\/span><\/p>\n

N=2F\/3\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

bloco3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=ma I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>bloco 2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=ma II\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>bloco 1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=ma III\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>somando I, II e III\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=3ma\u00a0 —\u00a0 mg=3ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=g\/3<\/b><\/span><\/span><\/p>\n

Quando o fio que une 2 e 3 se rompe, teremos:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

bloco 2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=ma\u2019 I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>bloco 1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 T=ma\u2019 II\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>somando I com II\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2ma\u2019\u00a0 —\u00a0 mg=2ma\u2019\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a\u2019=2g<\/b><\/span><\/span><\/p>\n

dividindo membro a membro a=g\/3 por a\u2019=2g, obtemos\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a\u2019\/a=3\/2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

16<\/b><\/span><\/span>– A acelera\u00e7\u00e3o \u00e9 a mesma nas duas situa\u00e7\u00f5es, pois as massas e a for\u00e7a aplicada s\u00e3o as mesmas.. A for\u00e7a de contato ser\u00e1 maior na situa\u00e7\u00e3o do conjunto 1, pois o bloco da esquerda ter\u00e1 que empurrar uma massa maior.<\/b><\/span><\/span><\/p>\n

17<\/b><\/span><\/span>– a) As for\u00e7as que atuam no balde s\u00e3o a tra\u00e7\u00e3o do fio, T, e o peso P. Quando o balde est\u00e1 em repouso, temos T = P = 100 N. Como P = mg\u00a0 —\u00a0 100=m.10\u00a0 —\u00a0 m=10kg. . J\u00e1 quando o dinam\u00f4metro acusa T = 120 N, temos, F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0\u00a0 T – P = ma, ou seja, a = (120 – 100)\/10 = 2 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

b) N\u00e3o \u00e9 poss\u00edvel concluir, pois s\u00f3 conhecemos a acelera\u00e7\u00e3o, e n\u00e3o a velocidade. Apenas sabemos que T>P, ou seja, ele pode estar subindo acelerado ou descendo retardado.<\/b><\/span><\/span><\/p>\n

18<\/b><\/span><\/span>–<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 T=2.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=4.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

19<\/b><\/span><\/span>– Vamos calcular a acelera\u00e7\u00e3o do sistema (barco + pessoa + pacote)\u00a0 —\u00a0 Pela segunda lei de Newton, F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\u00a0= m.a\u00a0 —\u00a0 \u00a0240 = (100 + 58 + 2).a\u00a0 —\u00a0 240 = 160.a ==> a = 240\/160 = 1,5 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Apenas sobre o pacote de 2 kg<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F = m.a = 2.1,5 =\u00a0<\/b><\/span><\/span>3,0 N<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span>\u00a0O dinam\u00f4metro indica a tra\u00e7\u00e3o no fio que \u00e9 de 40N e as for\u00e7as sobre o sistema est\u00e3o indicadas na figura.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe no bloco A que, como P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>>T, ele deve descer e consequentemente B deve subir.<\/b><\/span><\/span><\/p>\n

Assim, considerando que A des\u00e7a acelerado, pelo princ\u00edpio fundamental da din\u00e2mica temos, para o corpo A, que:\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a\u00a0 —<\/b><\/span><\/span><\/p>\n

100 – 40 = 10.a ==> a = 60\/10 = 6 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

Para o corpo B:\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 40 – m.10 = m.6\u00a0 —\u00a0 40 = 16.m\u00a0\u00a0 —\u00a0 m = 40\/16 =\u00a0<\/b><\/span><\/span>2,5 kg<\/b><\/span><\/span><\/p>\n

21<\/b><\/span><\/span>– Bloco A\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><\/p>\n

\"\"<\/p>\n

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>F=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

Bloco B sobre o bloco A<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=(m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>).a\/4 \u00a0—-\u00a0 F==(m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>).a\/4\u00a0 —\u00a0 4F=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>a + m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>a\u00a0 —\u00a0 4F=F + m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>a\u00a0 —\u00a0<\/b><\/span><\/span>3F=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>a II<\/b><\/span><\/span><\/p>\n

Dividindo membro a membro II por I\u00a0 —\u00a0 3m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=1\/3<\/b><\/span><\/span><\/p>\n

22<\/b><\/span><\/span>– Calculando a acelera\u00e7\u00e3o do sistema pelo gr\u00e1fico\u00a0 —\u00a0 a=DV\/Dt\u00a0 —\u00a0 a=24\/6\u00a0 —\u00a0 a=4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

Bloco A\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 T=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>a\u00a0 —\u00a0 10m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 T=4m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=6m<\/b><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub>I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>blobo B\u00a0 —\u00a0 T=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>a\u00a0 —\u00a0<\/b><\/span><\/span>T=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.4 II<\/b><\/span><\/span>\u00a0\u00a0\u00a0 —\u00a0\u00a0 igualando I com II\u00a0 —\u00a0 6m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=4m<\/b><\/span><\/span>B\u00a0\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=2\/3<\/b><\/span><\/span><\/p>\n

23<\/b><\/span><\/span>– a) caminhonete\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 F \u2013 T=2.000X1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F \u2013 T=2.000 I\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>caixote\u00a0 —\u00a0 T \u2013 800=80X1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=880N II<\/b><\/span><\/span><\/p>\n

substituindo II em I\u00a0 —\u00a0 F \u2013 880=2.000\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=2.<\/b><\/span><\/span>880N<\/b><\/span><\/span><\/p>\n

b) Sim, ser\u00e1 poss\u00edvel, pois a tra\u00e7\u00e3o m\u00e1xima que o cabo suporta \u00e9 2000N e a tra\u00e7\u00e3o aplicada \u00e9 de 880N<\/b><\/span><\/span><\/p>\n

 <\/p>\n

24<\/b><\/span><\/span>-1<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sup>\u00a0etapa\u00a0 —\u00a0 queda livre da esfera m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>, com a=g e a velocidade variando de 0 a V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>, numa queda de altura H<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0= 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.g.H<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0=2gH<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>2<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sup>\u00a0etapa\u00a0 —\u00a0 as duas esferas se movem juntas com acelera\u00e7\u00e3o a que vale:<\/b><\/span><\/span><\/p>\n

M=3m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>, se desloca para a direita puxada por T\u00a0 —\u00a0 F\u00ad<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=3m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>a I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>, desce de modo que P \u2013 T=m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>a\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

M<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.g – T=m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.a II<\/b><\/span><\/span>\u00a0—\u00a0 substituindo I em II\u00a0 —\u00a0 m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>g \u2013 3m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>a=m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>a\u00a0 —\u00a0 g=4.a\u00a0 —\u00a0 a=g\/4<\/b><\/span><\/span><\/p>\n

Queda de m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0com acelera\u00e7\u00e3o a=g\/4 e velocidade variando de V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0a V, numa queda de altura H<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/p>\n

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.DS\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.g\/4.H<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=2gH<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ (2gH<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\/4\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+ (V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>)\/4\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=(5V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>)\/4\u00a0 —\u00a0 V=\u00d65V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=2,2.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/p>\n

25<\/b><\/span><\/span>– a) equil\u00edbrio est\u00e1tico (for\u00e7a resultante nula)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0corda B bloco de baixo\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0\u00a0 —\u00a0\u00a0 T=40N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 bloco de cima\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 F = 20 + T\u00a0 —\u00a0 F = 20 + 40\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=60N<\/b><\/span><\/span><\/p>\n

b) os dois blocos sobem com a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0e F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\u00b90<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Bloco de baixo\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 T-40=4.2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=48N (tra\u00e7\u00e3o na corda B<\/b><\/span><\/span>)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 bloco de cima\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 F \u2013 T \u2013 20=2.2\u00a0 —<\/b><\/span><\/span><\/p>\n

F \u2013 48 \u2013 20=4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=72N (tra\u00e7\u00e3o na corda A)<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span>\u00a0a) sistema em repouso\u00a0 —\u00a0\u00a0\"\"\u00a0for\u00e7a que o anteparo exerce sobre A, impedindo-o de ir para a direita.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

bloco B\u00a0 —\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=T\u00a0 —\u00a0 T=50N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 bloco A\u00a0 —\u00a0 N=T\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=50N<\/b><\/span><\/span><\/p>\n

b) retirando-se o anteparo C, N deixa de agir sobre A e o sistema se move no sentido hor\u00e1rio com acelera\u00e7\u00e3o a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

bloco A\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=20.a I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>bloco B\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 P<\/b><\/span><\/span>B\u00a0<\/b><\/span><\/span><\/sub>\u2013 T =ma\u00a0 —\u00a0<\/b><\/span><\/span>50 \u2013 T =5.a II<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0 substituindo I em II\u00a0 —<\/b><\/span><\/span><\/p>\n

50 \u2013 20.a = 5.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span>\u00a0Colocando as for\u00e7as sobre cada bloco:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

bloco A \u2013 sobe\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub>\u00a0– 40= 4.a I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>bloco B \u2013 direita\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=11.a II\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>bloco C \u2013 desce\u00a0 —<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 50 \u2013 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>50 \u2013 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=5.a III\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>Somando I, II e III<\/b><\/span><\/span>\u00a0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>10=20.a<\/b><\/span><\/span>\u00a0\u00a0 —\u00a0<\/b><\/span><\/span>a=<\/b><\/span><\/span>\u00a00,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

Bloco A que sobe com acelera\u00e7\u00e3o de 0,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, percorrendo\u00a0DS=25cm=0,25m e tendo V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0 em X e V em Y.<\/b><\/span><\/span><\/p>\n

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 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.0,5.0,25\u00a0 —\u00a0 V=\u00d60,25\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=0,5m\/s<\/b><\/span><\/span><\/p>\n

28<\/b><\/span><\/span>– Colocando as for\u00e7as e calculando a acelera\u00e7\u00e3o do sistema:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

bloco 1\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=6.a I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>bloco 2\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0<\/b><\/span><\/span>\u00a040 \u2013 T=4.a II\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>substituindo I em II\u00a0 —\u00a0 40 \u2013 6.a=4.a\u00a0 —<\/b><\/span><\/span><\/p>\n

a=4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.\u00a0 Observe na figura que os dois blocos se movem juntos enquanto percorrem apenas 0,5m, com acelera\u00e7\u00e3o a=4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0e, no instante em que 2 chega ao solo,eles tem velocidade:<\/b><\/span><\/span><\/p>\n

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 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>= 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.4.0,5\u00a0 —\u00a0 V=2m\/s. A partir da\u00ed, 2 n\u00e3o puxa mais 1 e ele segue em MRU com velocidade constante de 2m\/s, com que chega em B.<\/b><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span>\u00a0(1) bloco A\u00a0 —\u00a0 F \u2013 T=m.a\u00a0 —\u00a0 F \u2013 T=4.2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F \u2013 T=8 I<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 bloco B\u00a0 —\u00a0 T=6.2 \u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span>T=12N II<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0 substituindo II em I\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

F \u2013 12=8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=20N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>est\u00e1 correta<\/b><\/span><\/span><\/p>\n

(2) est\u00e1 correta \u2013 vide (1)<\/b><\/span><\/span><\/p>\n

(3) est\u00e1 correta, pois o dinam\u00f4metro \u00e9 ideal (como se n\u00e3o existisse).<\/b><\/span><\/span><\/p>\n

(4) est\u00e1 correta, o dinam\u00f4metro indica a tra\u00e7\u00e3o no fio.<\/b><\/span><\/span><\/p>\n

Todas corretas<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span>\u00a0colocando as\u00a0 for\u00e7as e, como as massas s\u00e3o iguais, a tra\u00e7\u00e3o (T) em cada fio \u00e9 a mesma.\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na figura acima que o bloco B sobe, pois nele temos 2T para cima. Bloco A\u00a0 —\u00a0 desce\u00a0 — P-T=ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>mg-T=ma I\u00a0\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 bloco B\u00a0 —\u00a0 sobe\u00a0 —\u00a0 2T-P-ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>2T-mg=ma II\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 bloco C\u00a0 —\u00a0 desce\u00a0 —\u00a0 P-T=ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>mg-T=ma III<\/b><\/span><\/span>\u00a0 —\u00a0 somando I, II e III\u00a0 —\u00a0 mg=3ma\u00a0 —\u00a0 a=g\/3\u00a0\u00a0 R- C<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span>\u00a0Colocando as for\u00e7as:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Bloco B\u00a0 —\u00a0 sobe\u00a0 —\u00a0 T \u2013 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0= m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 T \u2013 100 = 10.a I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 bloco A\u00a0 —\u00a0 para a esquerda\u00a0 —\u00a0 F \u2013 T=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

500 \u2013 T=15.a II\u00a0 —\u00a0 somando I com II\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=16m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 T \u2013 100=10.16\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0T=260N\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span>\u00a0a) m=100g=0,1kg\u00a0\u00a0 —\u00a0 F-P=m.a\u00a0 —\u00a0 F \u2013 3=0,3.3\u00a0 —\u00a0 F=3,9N.<\/b><\/span><\/span><\/p>\n

b) s\u00f3 sobre o elo do meio\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0,1.3\u00a0 —\u00a0 F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=0,3N<\/b><\/span><\/span><\/p>\n

c) elo de baixo\u00a0 —\u00a0 T-P=ma\u00a0 —\u00a0 T-1=0,1.3\u00a0 —\u00a0 T=1,3N<\/b><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span>\u00a0C\u00e1lculo da acelera\u00e7\u00e3o do sistema:<\/b><\/span><\/span><\/p>\n

Bloco da esquerda\u00a0 —\u00a0 2mg \u2013 T=2ma\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 bloco da direita\u00a0 —\u00a0 T \u2013 mg=ma\u00a0\u00a0\u00a0 resolvendo o sistema\u00a0 —\u00a0 mg=3ma\u00a0 —\u00a0 a=g\/3<\/b><\/span><\/span><\/p>\n

Torricelli —\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.DS\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>= 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.g\/3.h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=\u00d62.g.h\/3<\/b><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span>\u00a0Quem est\u00e1 subindo \u00e9 o anel do meio mais os tr\u00eas an\u00e9is de baixo, portanto 4 an\u00e9is de peso P=(4X0,2).10\u00a0 —\u00a0 P=8N<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

T \u2013 P=m.a\u00a0 —\u00a0 T \u2013 8=0,8.2\u00a0 —\u00a0 T=9,6N\u00a0 R- B<\/b><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span>Decompondo, na figura I a for\u00e7a de tra\u00e7\u00e3o T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>, observamos que:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

figura I\u00a0 —\u00a0 T<\/b><\/span><\/span>1X<\/b><\/span><\/span><\/sub>\u00a0+ N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=P \u2013 T<\/b><\/span><\/span>1X<\/b><\/span><\/span><\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>figura II\u00a0 —\u00a0 N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 portanto N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>>N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R-B<\/b><\/span><\/span><\/p>\n

36-Colocando as for\u00e7as nas barca\u00e7as:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Barca\u00e7a A\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 T<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>\u00a0=30.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.a\u00a0 —\u00a0\u00a0 T<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>=20.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.a\u00a0 —\u00a0 8.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>=20.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.a<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 a<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0qualquer acelera\u00e7\u00e3o acima desse valor arrebenta o cabo II<\/b><\/span><\/span><\/p>\n

O cabo I puxa as duas barca\u00e7as de massa m<\/b><\/span><\/span>A+B<\/b><\/span><\/span><\/sub>=50.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\u00a0 — F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A+B<\/b><\/span><\/span><\/sub>.a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 — T<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>= m<\/b><\/span><\/span>A+B<\/b><\/span><\/span><\/sub>.a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 6.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=50.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=12m\/s<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0\u00a0—\u00a0 qualquer acelera\u00e7\u00e3o acima desse valor arrebenta o cabo I.<\/b><\/span><\/span><\/p>\n

Como a acelera\u00e7\u00e3o das duas barca\u00e7as deve ser a mesma, para que os dois cabos n\u00e3o arrebentem temos que pegar o menor valor de a, ou seja, a=4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

37-\u00a0<\/b><\/span><\/span>De uma maneira bem simples e pr\u00e1tica\u00a0 —\u00a0 quando o recipiente \u00e9 acelerado para a direita, o l\u00edquido, por in\u00e9rcia, tende a permanecer em repouso, ou seja, fica contr\u00e1rio ao deslocamento dos recipientes\u00a0 —<\/b><\/span><\/span>\u00a0 R- B<\/b><\/span><\/span><\/p>\n

38-\u00a0<\/b><\/span><\/span>O bloco m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0est\u00e1 sujeito a 6 for\u00e7as. Seu pr\u00f3prio peso e a for\u00e7a de a\u00e7\u00e3o F s\u00e3o duas delas\u00a0 —\u00a0 as outras quatro s\u00e3o devidas aos contatos com os outros dois corpos, sendo duas delas para cada corpo\u00a0 —\u00a0 a a\u00e7\u00e3o na dire\u00e7\u00e3o da gravidade em fun\u00e7\u00e3o do peso destes corpos e a\u00e7\u00f5es na dire\u00e7\u00e3o do movimento, mas no sentido oposto, por resist\u00eancia a a\u00e7\u00e3o de F\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

39-\u00a0<\/b><\/span><\/span>Considere o sistema abaixo em repouso em rela\u00e7\u00e3o a um referencial inercial em que o bloco de massa M representa o peso do cabo (aplicado em seu centro de gravidade) e as cordas representam o cabo, e q o \u00e2ngulo que o cabo faz com a horizontal do lugar.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na seq\u00fc\u00eancia das figuras acima que, como o sistema est\u00e1 em equil\u00edbrio, F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 2Tsen\u03b8=Mg\u00a0 —\u00a0 senq = Mg\/2T\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

assim, se q = 0\u00a0 —\u00a0 \u00a0sen(q) = 0 o que significa que as for\u00e7as exercidas pela corda sobre o bloco devem ser infinitas\u00a0 —\u00a0 ou, deve existir uma for\u00e7a de m\u00f3dulo infinito para que, somada ao peso do bloco (aqui modelando a massa do cabo), resulte zero\u00a0 —\u00a0 como isso n\u00e3o \u00e9 poss\u00edvel, n\u00e3o h\u00e1 como se ter q = 0, isto \u00e9, dever\u00e1 sempre existir, num campo gravitacional, a \u201ccurva\u201d observada por Lucinha.<\/b><\/span><\/span><\/p>\n

40-\u00a0<\/b><\/span><\/span>a) A tens\u00e3o (ou tra\u00e7\u00e3o, que \u00e9 o termo mais adequado) na corda corresponde \u00e0 intensidade da for\u00e7a aplicada por Alberto\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

T = 200 N.<\/b><\/span><\/span><\/p>\n

b)\u00a0\"\": for\u00e7a de tra\u00e7\u00e3o no centro da polia, aplicada por Cabral\u00a0 —\u00a0\u00a0\"\": for\u00e7as aplicadas pela corda que passa pela polia\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

c) Como a polia n\u00e3o tem massa (ou seja, sua massa \u00e9 desprez\u00edvel) e, al\u00e9m disso, ela est\u00e1 sendo arrastada quase que estaticamente (ou seja, com velocidade constante\u00a0 —\u00a0 a = 0)\u00a0 —\u00a0\u00a0 princ\u00edpio fundamental\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 F \u2013 2 T = m a\u00a0 —\u00a0 F \u2013 2 T = 0\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

F = 2 T\u00a0 = 2 (200)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0F = 400 N<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

d) A figura a seguir mostra que quando a ponta da corda desloca D (do ponto P at\u00e9 o ponto P\u2019 ), o centro da polia desloca D\/2.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Se corda que Alberto puxa enrola D, essa dist\u00e2ncia \u00e9 distribu\u00edda nos dois bra\u00e7os da polia, fazendo com o seu centro desloque D\/2\u00a0<\/b><\/span><\/span>\u00a0—\u00a0\u00a0 portanto, se Carlos avan\u00e7a 2 m, Alberto recua 4 m.<\/b><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span>\u00a0As for\u00e7as que agem sobre a esfera s\u00e3o seu peso (\"\"), vertical e para baixo e a for\u00e7a de tra\u00e7\u00e3o no fio (\"\"), conforme figura\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

somando vetorialmente\u00a0\"\"\u00a0com\u00a0\"\"\u00a0voc\u00ea obt\u00e9m a for\u00e7a resultante\u00a0\"\"\u00a0\u00a0\u00a0—\u00a0 tg45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0= cateto oposto\/cateto adjacente\u00a0 —\u00a0 tg45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\/P\u00a0 —\u00a0 1=ma\/mg\u00a0 —\u00a0 a=g=9,8m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

42-\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span><\/sub>=KV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 K=F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span><\/sub>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>=m.aV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>=(m.V\/t)\/V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 K=(m\/t)\/V\u00a0 —\u00a0 K=(kg\/s)\/(m\/s)\u00a0 —\u00a0 K=kg\/s x s\/m\u00a0 —\u00a0 K=kg\/m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

43-\u00a0<\/b><\/span><\/span>Como o andaime se encontra parado (equil\u00edbrio est\u00e1tico) a resultante das for\u00e7as que agem sobre ele \u00ea nula\u00a0 —\u00a0 \u00a0observe que os pesos sobre o andaime encontram-se no meio (peso de um oper\u00e1rio + peso do andaime, supondo-o homog\u00eaneo) e na extremidade esquerda (peso do outro oper\u00e1rio)\u00a0 —\u00a0 assim, a tens\u00e3o na corda X \u00e9 maior que a na corda Y\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

44-\u00a0<\/b><\/span><\/span>Colocando as for\u00e7as que agem sobre cada bloco (pesos de A e de B, P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0e P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>, verticais e para baixo; tra\u00e7\u00e3o T em cada bloco,<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

verticais e para cima e o empuxo E sobre o bloco A devido ao l\u00edquido, vertical e para cima)\u00a0 —\u00a0 E=\u03c1<\/b><\/span><\/span>liq<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>liq<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 E= \u03c1<\/b><\/span><\/span>liq<\/b><\/span><\/span><\/sub>.3(V\/2).g\u00a0 —\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=m.g\u00a0 —\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>= \u03c1<\/b><\/span><\/span>bloco<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>bloco<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>= \u03c1<\/b><\/span><\/span>bloco.<\/b><\/span><\/span><\/sub>Vg\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=3mg=3.\u03c1<\/b><\/span><\/span>bloco<\/b><\/span><\/span><\/sub>.V.g\u00a0 —\u00a0 o sistema est\u00e1 em equil\u00edbrio (F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0)\u00a0 —\u00a0 bloco B\u00a0 —\u00a0 T=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T= \u03c1<\/b><\/span><\/span>bloco.<\/b><\/span><\/span><\/sub>Vg\u00a0 —\u00a0 bloco A\u00a0 —\u00a0 V<\/b><\/span><\/span>liq<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>bloco<\/b><\/span><\/span><\/sub>\/2\u00a0 —\u00a0 V=2V\u00a0 —\u00a0 E + T = P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>liq<\/b><\/span><\/span><\/sub>.3(V\/2).g + \u03c1<\/b><\/span><\/span>bloco.<\/b><\/span><\/span><\/sub>Vg = 3.\u03c1<\/b><\/span><\/span>bloco<\/b><\/span><\/span><\/sub>.V.g\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>liq<\/b><\/span><\/span><\/sub>.3(V\/2).g = \u03c1<\/b><\/span><\/span>bloco<\/b><\/span><\/span><\/sub>.2.(2V).g\u00a0 —\u00a0 3\u03c1<\/b><\/span><\/span>liq<\/b><\/span><\/span><\/sub>\/2 = 2.\u03c1<\/b><\/span><\/span>bloco<\/b><\/span><\/span><\/sub>.2\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>liq<\/b><\/span><\/span><\/sub>=4 \u03c1<\/b><\/span><\/span>bloco<\/b><\/span><\/span><\/sub>\/3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

I. Falsa\u00a0 —\u00a0 para que o corpo se desloque \u00e9 necess\u00e1rio que a proje\u00e7\u00e3o de F na horizontal (Fcos\u03b8) seja maior que a for\u00e7a de atrito\u00a0 —\u00a0 Fcos\u03b8 > F<\/b><\/span><\/span>atrito<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

II. Correta\u00a0 —\u00a0 o corpo estar\u00e1 na imin\u00eancia de se mover quando Fcos\u03b8 = F<\/b><\/span><\/span>atrito<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

III. Correta\u00a0 —\u00a0 se F > P e \u03b8=90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>, o bloco girar\u00e1 no sentido anti-hor\u00e1rio at\u00e9 que a dire\u00e7\u00e3o de F seja a mesma que de P e, a partir da\u00ed,<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

como F>P a for\u00e7a resultante ser\u00e1 vertical e para cima, e o corpo subir\u00e1.<\/b><\/span><\/span><\/p>\n

IV- Falsa\u00a0 —\u00a0 o corpo s\u00f3 ser\u00e1 elevado se N + Fsen\u03b8 > P, pois, nesse caso sobrar\u00e1 uma for\u00e7a resultante vertical e para cima.<\/b><\/span><\/span><\/p>\n

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

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

Acelera\u00e7\u00e3o de subida da plataforma A que sobe \u2206S=4,5m em \u2206t=3,0s, a partir do repouso V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0 —\u00a0 \u2206S=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t + a.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 4,5 =<\/b><\/span><\/span><\/p>\n

0.3 + a.3<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 a=4,4\/4,5\u00a0 —\u00a0 a=1,0m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 esta \u00e9 a acelera\u00e7\u00e3o de subida da massa m e de descida da massa M\u00a0 —\u00a0 a figura<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

mostra as for\u00e7as que agem sobre o sistema\u00a0 —\u00a0 equa\u00e7\u00f5es\u00a0 —\u00a0 massa m\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 T \u2013 mg = m.a\u00a0 —\u00a0 T \u2013 225.10 = 225.1\u00a0 —<\/b><\/span><\/span><\/p>\n

T=2 475N\u00a0 —\u00a0 massa M\u00a0 —\u00a0 M.g \u2013 T = M.a\u00a0 —\u00a0 M.10 \u2013 2 475 = M.1\u00a0 —\u00a0 9M=2 475\u00a0 —\u00a0 M=275kg\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

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

a) Se voc\u00ea considerar o sistema carro-\u00f4nibus como um sistema isolado, voc\u00ea pode utilizar o teorema da conserva\u00e7\u00e3o da<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0quantidade de movimento\u00a0 —\u00a0 Q<\/b><\/span><\/span>antes<\/b><\/span><\/span><\/sub>=M.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=9000.80 +1000.0\u00a0 —\u00a0 Q<\/b><\/span><\/span>antes<\/b><\/span><\/span><\/sub>=720000kg.m\/s\u00a0 —\u00a0 Q<\/b><\/span><\/span>depois<\/b><\/span><\/span><\/sub>=(M + m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>).V=10.000.V<\/b><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span>depois<\/b><\/span><\/span><\/sub>=10000V\u00a0 —\u00a0 Q<\/b><\/span><\/span>antes<\/b><\/span><\/span><\/sub>\u00a0= Q<\/b><\/span><\/span>depois<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 720000=10000V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=72km\/h<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

b) Pela figura fornecida voc\u00ea pode determinar a intensidade da for\u00e7a lateral\u00a0\"\"\u00a0\u00a0—\u00a0 sen3<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=F<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\/F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0,05=F<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\/8000\u00a0 —\u00a0 F<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=400N\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A acelera\u00e7\u00e3o lateral\u00a0\"\"\u00a0do carro tem intensidade\u00a0 —\u00a0 F<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=m.a<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 400=1600.a<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0a<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=0,25m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/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 Aplica\u00e7\u00f5es das Leis de Newton em blocos apoiados ou suspensos     01– a) Como n\u00e3o existe atrito, por menor que seja a massa do sistema e a for\u00e7a aplicada o sistema sempre se mover\u00e1 no caso, para a direita. Colocando as for\u00e7as que agem sobre cada bloco apenas na dire\u00e7\u00e3o do movimento<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1203,"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-1207","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1207","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=1207"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1207\/revisions"}],"predecessor-version":[{"id":10812,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1207\/revisions\/10812"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1203"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1207"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}