{"id":1205,"date":"2015-08-25T23:57:13","date_gmt":"2015-08-25T23:57:13","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1205"},"modified":"2024-08-23T12:53:13","modified_gmt":"2024-08-23T12:53:13","slug":"exercicios-de-vestibulares-com-resolucao-comentada-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\/exercicios-de-vestibulares-com-resolucao-comentada-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 &#8211; Resolu\u00e7\u00e3o"},"content":{"rendered":"<p align=\"CENTER\"><span style=\"color: #c00000;\"><span style=\"font-family: 'Arial Black', serif;\"><span style=\"font-size: large;\"><b>Exerc\u00edcios de vestibulares com resolu\u00e7\u00e3o comentada sobre<\/b><\/span><\/span><\/span><b> <\/b><span style=\"color: #0000cc;\"><span style=\"font-family: 'Arial Black', serif;\"><span style=\"font-size: large;\"><b>Aplica\u00e7\u00f5es das Leis de Newton em blocos apoiados ou suspensos<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>01-(UCS-RS)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Uma for\u00e7a de intensidade 20N atua sobre os blocos A e B, de massas m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=3kg e m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=1kg, como mostra a figura.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img fetchpriority=\"high\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_b8d1c1fe.jpg\" alt=\"\" width=\"370\" height=\"155\" name=\"Imagem 8\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A superf\u00edcie sobre a qual desliza o conjunto \u00e9 horizontal e sem atrito. Considere g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e determine:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a)\u00a0 a intensidade da for\u00e7a que A aplica em B<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a intensidade da for\u00e7a que B aplica em A<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) a intensidade da for\u00e7a resultante sobre cada bloco.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>02-(UFB)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Os tr\u00eas blocos P, Q e R da figura abaixo encontram-se em repouso sobre uma superf\u00edcie plana, horizontal e perfeitamente lisa.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_7e0ffe06.jpg\" alt=\"\" width=\"397\" height=\"123\" name=\"Imagem 9\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Suas massas s\u00e3o m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>P<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=6kg, m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Q<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=4kg e m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>R<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=2kg. Uma for\u00e7a<\/b><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<img decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_9747e58d.jpg\" alt=\"\" width=\"13\" height=\"20\" name=\"graphics8\" align=\"BOTTOM\" border=\"0\" \/><\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0de intensidade F=48N \u00e9 aplicada sobre o bloco P. Considere g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/sup><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e determine a intensidade, dire\u00e7\u00e3o e sentido da for\u00e7a que o bloco R aplica no bloco Q.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>03-(FCC-BA)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Quatro blocos M, N, P e Q deslizam sobre uma superf\u00edcie horizontal, empurrados por uma for\u00e7a<\/b><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_9747e58d.jpg\" alt=\"\" width=\"13\" height=\"20\" name=\"Imagem 11\" align=\"BOTTOM\" border=\"0\" \/><\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, conforme o esquema abaixo.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_344680e4.jpg\" alt=\"\" width=\"524\" height=\"111\" name=\"graphics9\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A for\u00e7a de atrito entre os blocos e a superf\u00edcie \u00e9 desprez\u00edvel e a massa de cada bloco vale 3,0kg. Sabendo-se que a acelera\u00e7\u00e3o escalar dos blocos vale 2,0m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, a for\u00e7a do bloco M sobre o bloco N \u00e9, em newtons, igual a:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_69a0c409.png\" alt=\"\" width=\"774\" height=\"17\" name=\"Imagem 138\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>04-(FATEC-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Dois blocos A e B de massas 10 kg e 20 kg, respectivamente, unidos por um fio de massa desprez\u00edvel, est\u00e3o em repouso sobre um plano horizontal sem atrito. Uma for\u00e7a, tamb\u00e9m horizontal, de intensidade F = 60N \u00e9 aplicada no bloco B, conforme mostra a figura.<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_cea33207.jpg\" alt=\"\" width=\"197\" height=\"47\" name=\"graphics10\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O m\u00f3dulo da for\u00e7a de tra\u00e7\u00e3o no fio que une os dois blocos, em newtons, vale<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_474a7ba.png\" alt=\"\" width=\"775\" height=\"21\" name=\"Imagem 139\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>05-(F.M.Itajub\u00e1-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Tr\u00eas blocos s\u00e3o atados por fios ideais e puxados no espa\u00e7o interestelar, onde inexiste gravidade, com uma acelera\u00e7\u00e3o<\/b><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_a0fe9af7.jpg\" alt=\"\" width=\"12\" height=\"22\" name=\"Imagem 14\" align=\"BOTTOM\" border=\"0\" \/><\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0de m\u00f3dulo 10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_40d74a53.jpg\" alt=\"\" width=\"609\" height=\"110\" name=\"graphics11\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Quais as intensidades T<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, T<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e T<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sub><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>das for\u00e7as tensoras nos fios?<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>06-FUVEST-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Dois corpos A e B de massas m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=3kg e m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=1kg est\u00e3o ligados por um fio flex\u00edvel, como mostra a figura,a mover-se sob a a\u00e7\u00e3o da gravidade, sem atrito. (considere g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>).<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_e26e8b7.jpg\" alt=\"\" width=\"301\" height=\"223\" name=\"graphics12\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Determine a acelera\u00e7\u00e3o do conjunto e a intensidade da for\u00e7a de tra\u00e7\u00e3o no fio.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Supondo que num certo instante, ap\u00f3s iniciado o movimento, o fio de liga\u00e7\u00e3o se rompa, o que acontecer\u00e1 com os movimentos dos corpos A e B<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>07-(UFMG-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Na montagem abaixo, sabendo-se que F=40N, m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=1,0kg e que g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, qual \u00e9 o valor de T? Despreze qualquer atrito.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_cda32377.jpg\" alt=\"\" width=\"305\" height=\"183\" name=\"Imagem 17\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>08-(UFB)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Na figura abaixo os blocos 1, 2 e 3 tem massas m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=40kg, m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=20kg e m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=60kg. Considere os fios A e B e a polia ideais, despreze todos os atritos e calcule:<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_9f395d1f.jpg\" alt=\"\" width=\"310\" height=\"221\" name=\"Imagem 18\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) a acelera\u00e7\u00e3o do sistema\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a intensidade da for\u00e7a de tra\u00e7\u00e3o no fio B<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>09-(ITA-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> O arranjo experimental esquematizado na figura consiste de uma roldana por onde passa um fio perfeitamente flex\u00edvel e sem peso. Este fio sustenta em uma de suas extremidades a massa de 10kg e na outra, um dinam\u00f4metro no qual est\u00e1 pendurada uma massa de 6kg. A roldana pode girar sem atrito e sua massa, bem como a do dinam\u00f4metro, \u00e9 desprez\u00edvel em rela\u00e7\u00e3o \u00e0quela do sistema.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_540cfc61.jpg\" alt=\"\" width=\"265\" height=\"249\" name=\"Imagem 19\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O sistema, a partir do repouso, vai se movimentar pela a\u00e7\u00e3o da gravidade. Sendo g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, determine:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) o m\u00f3dulo da acelera\u00e7\u00e3o de cada bloco\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a intensidade da for\u00e7a, em newtons, indicada pelo dinam\u00f4metro.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>10- (UNIFESP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Na representa\u00e7\u00e3o da figura, o bloco A desce verticalmente e traciona o bloco B, que se movimenta em um plano horizontal por meio de um fio inextens\u00edvel. Considere desprez\u00edveis as massas do fio e da roldana e todas as for\u00e7as de resist\u00eancia ao movimento.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_8eef309c.jpg\" alt=\"\" width=\"290\" height=\"171\" name=\"Imagem 20\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Suponha que, no instante representado na figura, o fio se quebre. Pode-se afirmar que, a partir desse instante,<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) o bloco A adquire acelera\u00e7\u00e3o igual \u00e0 da gravidade; o bloco B p\u00e1ra.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) o bloco A adquire acelera\u00e7\u00e3o igual \u00e0 da gravidade; o bloco B passa a se mover com velocidade constante.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) o bloco A adquire acelera\u00e7\u00e3o igual \u00e0 da gravidade; o bloco B reduz sua velocidade e tende a parar.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) os dois blocos passam a se mover com velocidade constante.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) os dois blocos passam a se mover com a mesma acelera\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>11- (FGV-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Dois carrinhos de supermercado podem ser acoplados um ao outro por meio de uma pequena corrente, de modo que uma \u00fanica pessoa, ao inv\u00e9s de empurrar dois carrinhos separadamente, possa puxar o conjunto pelo interior do supermercado. Um cliente aplica uma for\u00e7a horizontal de intensidade F, sobre o carrinho da frente, dando ao conjunto uma acelera\u00e7\u00e3o de intensidade 0,5 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_bdc9728a.jpg\" alt=\"\" width=\"388\" height=\"143\" name=\"Imagem 21\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Sendo o piso plano e as for\u00e7as de atrito desprez\u00edveis, o m\u00f3dulo da for\u00e7a F e o da for\u00e7a de tra\u00e7\u00e3o na corrente s\u00e3o, em N, respectivamente:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 70 e 20.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 70 e 40.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 70 e 50.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) 60 e 20.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) 60 e 50.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>12-(UFRJ-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Um bloco de massa m \u00e9 abaixado e levantado por meio de um fio ideal. Inicialmente, o bloco \u00e9 abaixado com acelera\u00e7\u00e3o constante vertical, para baixo, de m\u00f3dulo a (por hip\u00f3tese, menor do que o m\u00f3dulo g da acelera\u00e7\u00e3o da gravidade), como mostra a figura 1.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Em seguida, o bloco \u00e9 levantado com acelera\u00e7\u00e3o constante vertical, para cima, tamb\u00e9m de m\u00f3dulo a, como mostra a figura 2. Sejam T a tens\u00e3o do fio na descida e T&#8217; a tens\u00e3o do fio na subida.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_4c08c7b1.jpg\" alt=\"\" width=\"296\" height=\"181\" name=\"Imagem 22\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Determine a raz\u00e3o T&#8217;\/T em fun\u00e7\u00e3o de a e g.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>13- (UNESP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Dois blocos, A e B, de massas m e 2m, respectivamente, ligados por um fio inextens\u00edvel e de massa desprez\u00edvel, est\u00e3o inicialmente em repouso sobre um plano horizontal sem atrito. Quando o conjunto \u00e9 puxado para a direita pela for\u00e7a horizontal<\/b><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_c1d3502c.jpg\" alt=\"\" width=\"15\" height=\"22\" name=\"Imagem 23\" align=\"BOTTOM\" border=\"0\" \/><\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0aplicada em B, como mostra a figura, o fio fica sujeito \u00e0 tra\u00e7\u00e3o T<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>. Quando puxado para a esquerda por uma for\u00e7a de mesma intensidade que a anterior, mas agindo em sentido contr\u00e1rio, o fio fica sujeito \u00e0 tra\u00e7\u00e3o T<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_5a2f17f4.jpg\" alt=\"\" width=\"364\" height=\"175\" name=\"Imagem 24\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Nessas condi\u00e7\u00f5es, pode-se afirmar que T<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u201a \u00e9 igual a<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_5d9d98fc.png\" alt=\"\" width=\"774\" height=\"24\" name=\"Imagem 140\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>14-(UNESP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Uma barra AC homog\u00eanea de massa m e comprimento L, colocada numa mesa lisa e horizontal, desliza sem girar sob a\u00e7\u00e3o de uma for\u00e7a<\/b><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_c1d3502c.jpg\" alt=\"\" width=\"15\" height=\"22\" name=\"Imagem 25\" align=\"BOTTOM\" border=\"0\" \/><\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, tamb\u00e9m horizontal, aplicada na sua extremidade esquerda.\u00a0<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_eb3182fe.jpg\" alt=\"\" width=\"365\" height=\"104\" name=\"Imagem 26\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Mostre que a for\u00e7a<\/b><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_c1d3502c.jpg\" alt=\"\" width=\"15\" height=\"22\" name=\"Imagem 27\" align=\"BOTTOM\" border=\"0\" \/><\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0com que a fra\u00e7\u00e3o BC de comprimento 2L\/3, atua sobre a fra\u00e7\u00e3o AB \u00e9 igual a &#8211; 2<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_c1d3502c.jpg\" alt=\"\" width=\"15\" height=\"22\" name=\"Imagem 28\" align=\"BOTTOM\" border=\"0\" \/>\/3.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>15- (UFRJ-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> O sistema representado na figura \u00e9 abandonado sem velocidade inicial. Os tr\u00eas blocos t\u00eam massas iguais. Os fios e a roldana s\u00e3o ideais e s\u00e3o desprez\u00edveis os atritos no eixo da roldana. S\u00e3o tamb\u00e9m desprez\u00edveis os atritos entre os blocos (2) e (3) e a superf\u00edcie horizontal na qual est\u00e3o apoiados.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_c85ebad2.jpg\" alt=\"\" width=\"328\" height=\"155\" name=\"Imagem 29\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O sistema parte do repouso e o bloco (1) adquire uma acelera\u00e7\u00e3o de m\u00f3dulo igual a a. Ap\u00f3s alguns instantes, rompe-se o fio que liga os blocos (2) e (3). A partir de ent\u00e3o, a acelera\u00e7\u00e3o do bloco (1) passa a ter um m\u00f3dulo igual a a&#8217;.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Calcule a raz\u00e3o a&#8217; \/ a.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>16-(UFRJ-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Analise as figuras a seguir e leia com aten\u00e7\u00e3o o texto.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_e374849d.jpg\" alt=\"\" width=\"496\" height=\"110\" name=\"Imagem 30\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dois blocos de massas m e M, sendo M&gt;m est\u00e3o em repouso e em contato um ao lado do outro, sobre uma superf\u00edcie plana. Se empurrarmos um dos blocos com uma for\u00e7a F, paralela \u00e0 superf\u00edcie, o conjunto ir\u00e1 mover-se com uma dada acelera\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Determine se faria diferen\u00e7a para as magnitudes da acelera\u00e7\u00e3o do conjunto e das for\u00e7as de contato entre os blocos, se tiv\u00e9ssemos empurrado o outro bloco.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>17- (Ufrrj)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Em uma obra, realizada na cobertura de um pr\u00e9dio, h\u00e1 um sistema para subir e descer material entre o t\u00e9rreo e o \u00faltimo andar atrav\u00e9s de baldes e cordas. Um dos oper\u00e1rios, interessado em F\u00edsica, colocou um dinam\u00f4metro na extremidade de uma corda. Durante o transporte de um dos baldes, ele percebeu que o dinam\u00f4metro marcava 100 N com o balde em repouso e 120 N quando o balde passava por um ponto A no meio do trajeto.(considere g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_f7a49b32.jpg\" alt=\"\" width=\"320\" height=\"206\" name=\"Imagem 31\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Determine a acelera\u00e7\u00e3o do balde nesse instante em que ele passa pelo ponto A.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) \u00c9 poss\u00edvel concluir se, nesse instante, o balde est\u00e1 subindo ou descendo? Justifique.\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>18- (Ufpb)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Uma locomotiva desenvolvendo uma acelera\u00e7\u00e3o de 2m\/<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, puxa tr\u00eas vag\u00f5es ao longo de uma ferrovia retil\u00ednea, conforme a figura. (g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>).<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_30fc7de5.jpg\" alt=\"\" width=\"491\" height=\"90\" name=\"Imagem 32\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Se o vag\u00e3o 3 pesa 2 \u00d7 10<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sup><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>N, determine a intensidade da for\u00e7a a for\u00e7a exercida sobre ele pelo vag\u00e3o 2.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>19-(UFRJ-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Um sistema \u00e9 constitu\u00eddo por um barco de 100 kg, uma pessoa de 58 kg e um pacote de 2,0 kg que ela carrega consigo. O barco \u00e9 puxado por uma corda de modo que a for\u00e7a resultante sobre o sistema seja constante, horizontal e de m\u00f3dulo 240 newtons.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_f257e3e9.jpg\" alt=\"\" width=\"316\" height=\"125\" name=\"Imagem 33\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Supondo que n\u00e3o haja movimento relativo entre as partes do sistema, calcule o m\u00f3dulo da for\u00e7a horizontal que a pessoa exerce sobre o pacote.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>20-(UERJ-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Os corpos A e B, ligados ao dinam\u00f4metro D por fios inextens\u00edveis, deslocam-se em movimento uniformemente acelerado. Observe a representa\u00e7\u00e3o desse sistema, posicionado sobre a bancada de um laborat\u00f3rio.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_ec8b9a9b.jpg\" alt=\"\" width=\"270\" height=\"199\" name=\"Imagem 34\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A massa de A \u00e9 igual a 10 kg e a indica\u00e7\u00e3o no dinam\u00f4metro \u00e9 igual a 40 N.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Desprezando qualquer atrito e as massas das roldanas e dos fios, estime a massa de B.(g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>).<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>21-(UNESP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Um bloco de massa m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/sub><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>deslisa no solo horizontal, sem atrito, sob a\u00e7\u00e3o de uma for\u00e7a constante, quando um bloco de massa m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00e9 depositado sobre ele.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Ap\u00f3s a uni\u00e3o, a for\u00e7a aplicada continua sendo a mesma, por\u00e9m a acelera\u00e7\u00e3o dos dois blocos fica reduzida \u00e0 quarta parte da acelera\u00e7\u00e3o que o bloco A possu\u00eda. Pode-se afirmar que a raz\u00e3o entre as massas, m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\/m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, \u00e9<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_1f49df56.png\" alt=\"\" width=\"774\" height=\"21\" name=\"Imagem 141\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>22-(MACKENZIE-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> O conjunto abaixo, constitu\u00eddo de fios e polias ideais, \u00e9 abandonado do repouso no instante t=0 e a velocidade do corpo A varia\u00a0 em fun\u00e7\u00e3o do tempo segundo o diagrama dado.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_25e44cce.jpg\" alt=\"\" width=\"456\" height=\"207\" name=\"Imagem 35\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Desprezando o atrito e admitindo g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, calcule<\/b><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/sup><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a rela\u00e7\u00e3o entre as massas de A (m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>) e de B (m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>).<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>23-(PUC-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Uma caminhonete de 2.000kg tenta resgatar um caixote a partir de um precip\u00edcio, usando um cabo inextens\u00edvel que liga o ve\u00edculo ao objeto, de massa 80kg. Considere a polia ideal. Se o caixote sobe com acelera\u00e7\u00e3o de 1m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, responda: (g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_46022099.jpg\" alt=\"\" width=\"319\" height=\"206\" name=\"Imagem 36\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Qual a for\u00e7a que movimenta a caminhonete?<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) O cabo suporta no m\u00e1ximo uma tra\u00e7\u00e3o de 2.000N. Ser\u00e1 poss\u00edvel o resgate com essa acelera\u00e7\u00e3o sem que ele arrebente?<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>24-(FUVEST-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Uma esfera de massa m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sub><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>est\u00e1 pendurada por um fio, ligado em sua outra extremidade a um caixote, de massa M=3 m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, sobre uma mesa horizontal. Quando o fio entre eles permanece n\u00e3o esticado e a esfera \u00e9 largada, ap\u00f3s percorrer uma dist\u00e2ncia H<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, ela atingir\u00e1 uma velocidade V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, sem que o caixote se mova. Na situa\u00e7\u00e3o em que o fio entre eles estiver esticado, a esfera, puxando o caixote, ap\u00f3s percorrer a mesma dist\u00e2ncia H<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, atingir\u00e1 uma velocidade V. Determine V em fun\u00e7\u00e3o de V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_7febb620.jpg\" alt=\"\" width=\"366\" height=\"210\" name=\"Imagem 37\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>25-(UNESP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Dois blocos est\u00e3o suspensos em um campo gravitacional de acelera\u00e7\u00e3o g, por duas cordas A e B de massas desprez\u00edveis, como indica a figura.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_569f26b1.jpg\" alt=\"\" width=\"166\" height=\"254\" name=\"Imagem 38\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Determine as tens\u00f5es em cada corda nos seguintes casos:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) os corpos s\u00e3o mantidos suspensos em repouso pela for\u00e7a<\/b><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_6e42674a.jpg\" alt=\"\" width=\"15\" height=\"21\" name=\"Imagem 39\" align=\"BOTTOM\" border=\"0\" \/><\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) os corpos s\u00e3o submetidos a uma for\u00e7a<\/b><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_cadee834.jpg\" alt=\"\" width=\"16\" height=\"23\" name=\"Imagem 40\" align=\"BOTTOM\" border=\"0\" \/><\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>tal que os acelera a 2,0m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, para cima.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>26-(FUVEST-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Um carrinho A de 20kg de massa \u00e9 unido a um bloco B de 5kg por meio de um fio leve e inextens\u00edvel, conforme a figura.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_32cad08c.jpg\" alt=\"\" width=\"366\" height=\"157\" name=\"Imagem 41\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Inicialmente o sistema est\u00e1 em repouso, devido \u00e0 presen\u00e7a do anteparo C que bloqueia o carrinho A. (g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>).<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Qual o valor da for\u00e7a que o anteparo C exerce sobre o carrinho A<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Retirando C, com que acelera\u00e7\u00e3o o carrinho A se movimenta?<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>27-(MACKENZIE-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> O sistema abaixo \u00e9 constitu\u00eddo por fios e polias ideais, num local onde g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_92db041b.jpg\" alt=\"\" width=\"426\" height=\"237\" name=\"Imagem 42\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Desprezando-se qualquer tipo de resist\u00eancia e abandonando-se o conjunto quando o corpo A se encontra na posi\u00e7\u00e3o X, a sua velocidade, ao passar por Y, \u00e9, em m\/s:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_3331d781.png\" alt=\"\" width=\"774\" height=\"17\" name=\"Imagem 142\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>28-(MACKENZIE-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> No sistema abaixo, o corpo 1, de massa 6,0kg, est\u00e1 preso na posi\u00e7\u00e3o A. O corpo 2, tem massa de 4kg. Despreze os atritos e adote g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_f45ed4cc.jpg\" alt=\"\" width=\"283\" height=\"223\" name=\"Imagem 43\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Abandonando o corpo 1, a sua velocidade, em m\/s, ao passar pela posi\u00e7\u00e3o B ser\u00e1 de:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_670df3ce.png\" alt=\"\" width=\"774\" height=\"21\" name=\"Imagem 143\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>29-(Ceub-DF)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Na figura a seguir temos dois blocos, A e B, de massas respectivamente iguais a m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=4,0kg e m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=6,0kg, que deslizam, sem atrito, em uma superf\u00edcie plana e horizontal, sob a\u00e7\u00e3o de uma for\u00e7a horizontal e constante e de intensidade F. Os blocos est\u00e3o ligados por fios ideais a um dinam\u00f4metro tamb\u00e9m ideal (massa desprez\u00edvel), calibrado em newtons.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_de02a1cb.jpg\" alt=\"\" width=\"591\" height=\"130\" name=\"Imagem 44\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>N\u00e3o considere o efeito do ar e admita que os blocos tem uma acelera\u00e7\u00e3o horizontal, para a direita, constante e de m\u00f3dulo igual a 2,0m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Julgue os itens a seguir.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(1) a for\u00e7a tensora no fio (1) tem intensidade igual a 12N.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(2) O valor de F \u00e9 20N.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(3) como o dinam\u00f4metro tem massa desprez\u00edvel, as for\u00e7as que tracionam os fios (1) e (2) tem intensidades iguais.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(4) o dinam\u00f4metro indica 12N.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>30-(FUVEST-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Um sistema mec\u00e2nico \u00e9 formado por duas polias ideais que suportam tr\u00eas corpos A, B e C de mesma massa m, suspensos por fios ideais como representado na figura.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_5257832c.jpg\" alt=\"\" width=\"254\" height=\"238\" name=\"Imagem 45\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O corpo B est\u00e1 suspenso simultaneamente por dois fios, um ligado a A e outro a C. Podemos afirmar que a acelera\u00e7\u00e3o do corpo B ser\u00e1:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) zero\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) (g\/3) para baixo\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) (g\/3) para cima\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) (2g\/3) para baixo\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) (2g\/3) para cima<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>31-(Aman-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> No sistema apresentado na figura, n\u00e3o h\u00e1 for\u00e7as de atrito e o fio tem massa desprez\u00edvel. (g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>).<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_cd62c0d7.jpg\" alt=\"\" width=\"335\" height=\"214\" name=\"Imagem 46\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>S\u00e3o dados F=500N; m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=15kg e m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=10kg.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Determine a intensidade da for\u00e7a de tra\u00e7\u00e3o no fio e a acelera\u00e7\u00e3o do sistema.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>32-(UFMG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> A figura mostra uma corrente formada por tr\u00eas elos. A massa de cada elo \u00e9 de 100g e uma for\u00e7a vertical<\/b><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_5f832b29.jpg\" alt=\"\" width=\"14\" height=\"16\" name=\"Imagem 47\" align=\"BOTTOM\" border=\"0\" \/><\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0puxa essa corrente para cima. A corrente sobe com uma acelera\u00e7\u00e3o de 3,0m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_f7eacedb.jpg\" alt=\"\" width=\"70\" height=\"200\" name=\"Imagem 48\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considerando essas informa\u00e7\u00f5es calcule:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) o m\u00f3dulo da for\u00e7a<\/b><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_5f832b29.jpg\" alt=\"\" width=\"14\" height=\"16\" name=\"Imagem 49\" align=\"BOTTOM\" border=\"0\" \/><\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>que puxa a corrente.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) o m\u00f3dulo da for\u00e7a resultante que atua sobre o elo do meio.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) o m\u00f3dulo da for\u00e7a que o elo do meio faz sobre o elo de baixo.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>33-(UFRJ) O sistema ilustrado na figura abaixo \u00e9 uma m\u00e1quina de Atwood. A roldana tem massa desprez\u00edvel e gira livremente em torno de um eixo fixo perpendicular ao plano da figura, passando pelo centro geom\u00e9trico da roldana. Uma das massas vale m e a outra 2m. O sistema encontra-se inicialmente na situa\u00e7\u00e3o ilustrada pela figura a, isto \u00e9, com as duas massas no mesmo n\u00edvel. O sistema \u00e9 ent\u00e3o abandonado a partir do repouso e, ap\u00f3s um certo intervalo de tempo, a dist\u00e2ncia vertical entre as massas \u00e9 h (figura b).<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_8b532152.jpg\" alt=\"\" width=\"277\" height=\"262\" name=\"Imagem 50\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Calcule o m\u00f3dulo da velocidade de cada uma das massas na situa\u00e7\u00e3o mostrada na figura (b).<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>34-(UFRN)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Uma corrente constitu\u00edda de sete an\u00e9is, cada um com massa de 200g, est\u00e1 sendo puxada verticalmente para cima, com acelera\u00e7\u00e3o constante de 2,0m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>. A for\u00e7a para cima no anel do meio \u00e9: (g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>).<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_ff97e0e1.png\" alt=\"\" width=\"774\" height=\"16\" name=\"Imagem 144\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>35-(UNESP-SP) <\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Em uma circular t\u00e9cnica da Embrapa, depois da figura,<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_d5c3f85f.jpg\" alt=\"\" width=\"476\" height=\"158\" name=\"Imagem 51\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Encontramos uma recomenda\u00e7\u00e3o que, em resumo, diz:<\/b><\/span><\/span><\/p>\n<p>\u201c<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>No caso do arraste com a carga junto ao solo (se por algum motivo n\u00e3o pode ou n\u00e3o deve e ser erguida . . .) o ideal \u00e9 arrast\u00e1-la. . . reduzindo a porca necess\u00e1ria para moviment\u00e1-la, causando menos dano ao solo . . . e facilitando as manobras. Mas neste caso o peso da tora aumenta. (www.cpafac.embrapa.br\/pdficirtec39.pdf.Modificado.)<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Pode-se afirmar que a frase destacada \u00e9 conceitualmente<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A) inadequada, pois o peso da tora diminui, j\u00e1 que se distribui sobre uma \u00e1rea maior.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B) inadequada, pois o peso da tora \u00e9 sempre o mesmo, mas \u00e9 correto afirmar que em II a for\u00e7a exercida pela tora sobre o solo aumenta;<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>C) inadequada: o peso da tora \u00e9 sempre o mesmo e, al\u00e9m disso, a for\u00e7a a for\u00e7a exercida pela tora sobre o solo em II diminui, pois se distribui por uma \u00e1rea maior.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>D) adequada, pois nessa situa\u00e7\u00e3o a tora est\u00e1 integralmente apoiada sobre o solo.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>E) adequada, pois nessa situa\u00e7\u00e3o a \u00e1rea sobre a qual a tora est\u00e1 apoiada sobre o solo tamb\u00e9m aumenta<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>36-(UNESP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Um rebocador puxa duas barca\u00e7as pela \u00e1guas de um lago tranq\u00fcilo. A primeira delas tem massa de 30 toneladas e a segunda, 20 toneladas. Por uma quest\u00e3o de economia, o cabo de a\u00e7o I que conecta o rebocador \u00e0 primeira barca\u00e7a suporta, no m\u00e1ximo, 6.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>5<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>N, e o cabo II, 8.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>4<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>N.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_5a8e83ea.jpg\" alt=\"\" width=\"551\" height=\"99\" name=\"Imagem 52\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Desprezando o efeito de for\u00e7as resistivas, calcule a acelera\u00e7\u00e3o m\u00e1xima do conjunto, a fim de evitar o rompimento de um dos cabos.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>37-(UFSC-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Em repouso, o sistema de vasos comunicantes apresentado est\u00e1 em equil\u00edbrio, de acordo com a figura.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_68fe2594.jpg\" alt=\"\" width=\"271\" height=\"122\" name=\"Imagem 53\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Quando o sistema \u00e9 submetido a um movimento uniformemente variado devido \u00e0 a\u00e7\u00e3o de uma for\u00e7a horizontal voltada para direita, o l\u00edquido dever\u00e1 permanecer em uma posi\u00e7\u00e3o tal qual o esquematizado em<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_2b2ff8dc.jpg\" alt=\"\" width=\"775\" height=\"75\" name=\"Imagem 54\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>38-(UEL-PR)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Considere o sistema constitu\u00eddo por tr\u00eas blocos de massas m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, apoiados um sobre o outro, em repouso sobre uma superf\u00edcie horizontal, como mostra a figura a seguir.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_56b8982a.jpg\" alt=\"\" width=\"303\" height=\"148\" name=\"Imagem 55\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Observe que uma for\u00e7a F \u00e9 aplicada ao bloco de massa m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, conforme a representa\u00e7\u00e3o. Entretanto, esta for\u00e7a \u00e9 incapaz de vencer as for\u00e7as de f<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>ij<\/b><\/span><\/span><\/sub><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>entre os blocos m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>i<\/b><\/span><\/span><\/sub><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>j<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, onde i e j variam de 1 a 3.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Desprezando a resist\u00eancia do ar, assinale a alternativa que representa todas as for\u00e7as que atuam no bloco de massa m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, onde os N<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>i<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, representam as normais que atuam nos blocos e P<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>i<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, correspondem aos pesos dos respectivos blocos com i variando de 1 a 3.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_57b4ea55.jpg\" alt=\"\" width=\"767\" height=\"133\" name=\"Imagem 56\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>39-(UFCG-PB)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Durante uma viagem, Lucinha observou as enormes curvas que os cabos das linhas de transmiss\u00e3o de energia<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_437a1e04.jpg\" alt=\"\" width=\"371\" height=\"150\" name=\"Imagem 57\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>el\u00e9trica apresentavam (figura). Ao comentar a observa\u00e7\u00e3o, disse que os engenheiros poderiam economizar o material dos cabos se os esticassem entre as torres de sustenta\u00e7\u00e3o at\u00e9 que estivessem dispostos horizontalmente.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Proponha um modelo, fundamentado nas Leis de Newton, para a situa\u00e7\u00e3o observada e discuta o coment\u00e1rio feito por Lucinha.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>40-(PUC-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Alberto (A) desafiou seu colega Cabral (C) para uma competi\u00e7\u00e3o de cabo de guerra, de uma maneira especial, mostrada na figura. Alberto segurou no peda\u00e7o de corda que passava ao redor da polia enquanto que Cabral segurou no peda\u00e7o atado ao centro da polia. Apesar de mais forte, Cabral n\u00e3o conseguiu puxar Alberto, que lentamente foi arrastando o seu advers\u00e1rio at\u00e9 ganhar o jogo. Sabendo que a for\u00e7a com que Alberto puxa a corda \u00e9 de 200 N e que a polia n\u00e3o tem massa nem<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_ce02f799.jpg\" alt=\"\" width=\"345\" height=\"141\" name=\"Imagem 58\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>atritos:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) especifique a tens\u00e3o na corda que Alberto est\u00e1 segurando;<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) desenhe as for\u00e7as que agem sobre a polia, fazendo um diagrama de corpo livre;<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) calcule a for\u00e7a exercida pelo Cabral sobre a corda que ele puxava;<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) considerando que Cabral foi puxado por 2,0 m para frente, indique quanto Alberto andou para tr\u00e1s.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>41-(UFT-TO)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Uma pequena esfera de chumbo com massa igual a 50 g \u00e9 amarrada por um fio, de comprimento igual a 10 cm e massa desprez\u00edvel, e fixada no interior de um autom\u00f3vel conforme figura. O carro se move horizontalmente com acelera\u00e7\u00e3o constante. Considerando-se hipoteticamente o \u00e2ngulo que o fio faz com a vertical igual a 45 graus, qual seria o melhor valor para representar o m\u00f3dulo da acelera\u00e7\u00e3o do carro?<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Desconsidere o atrito com o ar, e considere o m\u00f3dulo da acelera\u00e7\u00e3o da gravidade igual a 9,8 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_975fdf28.jpg\" alt=\"\" width=\"303\" height=\"179\" name=\"Imagem 59\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 5,3 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\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><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 8,2 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\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><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 9,8 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\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><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) 7,4 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\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><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) 6,8 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/p>\n<p>&nbsp;<\/p>\n<p align=\"CENTER\"><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>42-(UFLA-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Um corpo, ao se deslocar em um meio fluido (l\u00edquido ou gasoso) fica sujeito a uma <img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_e91c6c67.jpg\" alt=\"\" width=\"244\" height=\"136\" name=\"Imagem 60\" align=\"BOTTOM\" border=\"0\" \/><\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>for\u00e7a de resist\u00eancia que \u00e9 expressa por: F<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>R<\/b><\/span><\/span><\/sub><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>= kv<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, em que k \u00e9 uma constante de proporcionalidade e v a velocidade do corpo no meio. Considerando o Sistema Internacional de Unidades (SI), \u00e9 CORRETO afirmar que a constante k \u00e9 dada pelas unidades:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_7487101a.png\" alt=\"\" width=\"775\" height=\"20\" name=\"Imagem 145\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>43-(UFRN-RN)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> \u00c9 muito comum observarmos nas fachadas de edif\u00edcios em constru\u00e7\u00e3o andaimes constitu\u00eddos por uma t\u00e1bua horizontal sustentada por cordas que passam por roldanas presas no topo da edifica\u00e7\u00e3o. O fato de um dos oper\u00e1rios se deslocar sobre o andaime em dire\u00e7\u00e3o ao outro, por exemplo, quando vai entregar alguma ferramenta ao companheiro, afeta a distribui\u00e7\u00e3o de for\u00e7as sobre as cordas. Nesse sentido, considere a situa\u00e7\u00e3o mostrada na Figura abaixo. Nela, um dos oper\u00e1rios se encontra na<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_bc0823b3.jpg\" alt=\"\" width=\"356\" height=\"169\" name=\"Imagem 61\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>extremidade esquerda do andaime, enquanto o outro, ap\u00f3s ter caminhado em dire\u00e7\u00e3o a ele, conduzindo uma marreta, encontra-se parado no meio do andaime.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considerando a situa\u00e7\u00e3o mostrada na Figura, pode-se afirmar que a<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A) for\u00e7a resultante sobre o andaime \u00e9 diferente de zero e a tens\u00e3o na corda Y \u00e9 maior que na corda X.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B) for\u00e7a resultante sobre o andaime \u00e9 igual a zero e a tens\u00e3o na corda Y \u00e9 maior que na corda X.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>C) for\u00e7a resultante sobre o andaime \u00e9 diferente de zero e a tens\u00e3o na corda X \u00e9 maior que na corda Y.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>D) for\u00e7a resultante sobre o andaime \u00e9 igual a zero e a tens\u00e3o na corda X \u00e9 maior que na corda<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>44-(UFV-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Nas extremidades de um fio inextens\u00edvel e de massa desprez\u00edvel, que passa por uma polia, est\u00e3o pendurados dois blocos maci\u00e7os A e B, feitos de um mesmo material de densidade de massa \u03c1. O bloco B se encontra suspenso no ar, enquanto que o bloco A esta com a metade de seu volume imerso em um liquido, conforme a figura.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_aae31c07.jpg\" alt=\"\" width=\"202\" height=\"215\" name=\"Imagem 62\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Sabe-se que o volume do bloco A \u00e9 tr\u00eas vezes maior que o do bloco B. Desprezando qualquer tipo de atrito e qualquer influencia do ar sobre os blocos, e CORRETO afirmar que a densidade de massa do liquido \u00e9:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_fbddf19d.jpg\" alt=\"\" width=\"766\" height=\"35\" name=\"Imagem 63\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>45-(UEPG-PR)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_e18e05a0.jpg\" alt=\"\" width=\"329\" height=\"126\" name=\"Imagem 64\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O corpo abaixo ilustrado est\u00e1 sob a a\u00e7\u00e3o de um sistema de for\u00e7as, onde a for\u00e7a F poder\u00e1 desequilibrar esse sistema.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_c1c85ca0.jpg\" alt=\"\" width=\"397\" height=\"179\" name=\"Imagem 65\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Analise as assertivas e assinale a alternativa correta.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>I \u2013 Para que o corpo se desloque \u00e9 necess\u00e1rio que a for\u00e7a F seja maior que (Fsen\u04e8).<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>II \u2013 O corpo estar\u00e1 pronto a romper o repouso quando a for\u00e7a F<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>atrito<\/b><\/span><\/span><\/sub><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>for igual ao valor Fcos\u04e8.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>III \u2013 Se a for\u00e7a F for maior que P quando \u04e8 for igual a 90<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o corpo, primeiramente, rotacionar\u00e1 at\u00e9 que a linha de a\u00e7\u00e3o de F coincida com a linha de a\u00e7\u00e3o da for\u00e7a de P, para ent\u00e3o se elevar.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>IV \u2013 Se a\u00a0 for\u00e7a N for maior que o peso ocorrer\u00e1 a eleva\u00e7\u00e3o do corpo.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) As assertivas I e II s\u00e3o corretas.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) As assertivas I e III s\u00e3o corretas.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) As assertivas II e IV s\u00e3o corretas.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) As assertivas II e III s\u00e3o corretas.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) As assertivas II, III e IV s\u00e3o corretas.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>46-(UNESP-SP)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_5a1bb010.jpg\" alt=\"\" width=\"729\" height=\"126\" name=\"Imagem 66\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Em uma obra, para permitir o transporte de objetos para cima, foi montada uma m\u00e1quina constitu\u00edda por uma polia, fios e \u00a0duas plataformas A e B horizontais, todos de massas desprez\u00edveis, como mostra a figura.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um objeto de massa m = 225 kg, colocado na plataforma A, inicialmente em repouso no solo, deve ser levado verticalmente para cima e atingir um ponto a 4,5 m de altura, em movimento uniformemente acelerado, num intervalo de tempo de 3 s.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A partir da\u00ed, um sistema de freios passa a atuar, fazendo a plataforma A parar na posi\u00e7\u00e3o onde o objeto ser\u00e1 descarregado.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_39e219bf.jpg\" alt=\"\" width=\"435\" height=\"243\" name=\"Imagem 67\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considerando g = 10 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, desprezando os efeitos do ar sobre o sistema e os atritos durante o movimento acelerado, a massa M,<\/b><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/sup><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>em kg, do corpo que deve ser colocado na plataforma B para acelerar para cima a massa m no intervalo de 3 s \u00e9 igual a<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_6dff3b4b.png\" alt=\"\" width=\"774\" height=\"19\" name=\"Imagem 146\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>47-(UNICAMP-SP)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_f774185f.jpg\" alt=\"\" width=\"732\" height=\"117\" name=\"Imagem 68\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O tempo de viagem de qualquer entrada da Unicamp at\u00e9 a regi\u00e3o central do campus \u00e9 de apenas<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_1727ffc.jpg\" alt=\"\" width=\"402\" height=\"122\" name=\"Imagem 69\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>alguns minutos. Assim,\u00a0 a<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>economia de tempo obtida, desrespeitando-se o limite de velocidade, \u00e9\u00a0 muito pequena, enquanto o risco de acidentes aumenta significativamente.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a)\u00a0 Considere que um \u00f4nibus de massa M = 9000 kg , viajando a 80 km\/h,\u00a0 colide\u00a0 na traseira de um carro de massa\u00a0 m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=1000 kg<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_7e013daa.jpg\" alt=\"\" width=\"521\" height=\"117\" name=\"Imagem 70\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>que se encontrava parado. A colis\u00e3o \u00e9\u00a0 inel\u00e1stica, ou seja, carro e \u00f4nibus seguem grudados ap\u00f3s a batida.\u00a0 Calcule a\u00a0 velocidade do<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>conjunto logo ap\u00f3s a colis\u00e3o.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b)\u00a0 Al\u00e9m do excesso de velocidade, a falta de manuten\u00e7\u00e3o do ve\u00edculo pode causar acidentes. Por exemplo, o\u00a0 desalinhamento das rodas faz com que o carro sofra a a\u00e7\u00e3o de uma for\u00e7a lateral. Considere um carro com um pneu dianteiro desalinhado de 3\u00b0, conforme\u00a0 a figura abaixo, gerando uma componente lateral da for\u00e7a de atrito<\/b><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_31628bf6.jpg\" alt=\"\" width=\"12\" height=\"21\" name=\"Imagem 71\" align=\"BOTTOM\" border=\"0\" \/><\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>em uma das rodas.\u00a0<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/newton-muv\/i_73941f68cedf69d4_html_96dbeee3.jpg\" alt=\"\" width=\"381\" height=\"226\" name=\"Imagem 72\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Para um carro de massa\u00a0 m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=1600 kg, calcule o m\u00f3dulo da acelera\u00e7\u00e3o lateral do carro, sabendo que o m\u00f3dulo da for\u00e7a de atrito em cada roda vale F<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>at<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>= 8000 N . Dados:<\/b><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/sub><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>sen 3\u00b0 = 0,05 e cos 3\u00b0 = 0,99.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<h3><a title=\"Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Aplica\u00e7\u00f5es das Leis de Newton em blocos apoiados ou suspensos\" href=\"http:\/\/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\/\"><span style=\"color: #000080;\">Confira a resolu\u00e7\u00e3o comentada<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"<p>Exerc\u00edcios de vestibulares com resolu\u00e7\u00e3o comentada sobre Aplica\u00e7\u00f5es das Leis de Newton em blocos apoiados ou suspensos 01-(UCS-RS) Uma for\u00e7a de intensidade 20N atua sobre os blocos A e B, de massas mA=3kg e mB=1kg, como mostra a figura. \u00a0A superf\u00edcie sobre a qual desliza o conjunto \u00e9 horizontal e sem atrito. Considere g=10m\/s2\u00a0e determine: a)\u00a0 a intensidade da for\u00e7a<\/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-1205","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1205","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=1205"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1205\/revisions"}],"predecessor-version":[{"id":10811,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1205\/revisions\/10811"}],"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=1205"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}