{"id":1313,"date":"2015-09-10T03:29:33","date_gmt":"2015-09-10T03:29:33","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1313"},"modified":"2024-08-23T13:38:32","modified_gmt":"2024-08-23T13:38:32","slug":"exercicios-de-vestibulares-com-resolucoes-comentadas-sobre-plano-inclinado-com-atrito","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/plano-inclinado-com-atrito\/exercicios-de-vestibulares-com-resolucoes-comentadas-sobre-plano-inclinado-com-atrito\/","title":{"rendered":"Plano inclinado com atrito"},"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\u00f5es comentadas sobre<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><span style=\"color: #0000cc;\"><span style=\"font-family: 'Arial Black', serif;\"><span style=\"font-size: large;\"><b>Plano inclinado com atrito<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>01-(UERJ-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um bloco de massa igual a 1,0 kg repousa em equil\u00edbrio sobre um plano inclinado. Esse plano tem comprimento igual a 50 cm e alcan\u00e7a uma altura m\u00e1xima em rela\u00e7\u00e3o ao solo igual a 30 cm. Calcule o coeficiente de atrito entre o bloco e o plano inclinado.<\/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-(UFPEL)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um caminh\u00e3o-tanque, ap\u00f3s sair do posto, segue, com velocidade constante, por uma rua plana que, num dado trecho, \u00e9 plana e inclinada.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_9e3cb365.jpg\" alt=\"\" width=\"285\" height=\"134\" name=\"Imagem 69\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O m\u00f3dulo da acelera\u00e7\u00e3o da gravidade, no local, \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>, e a massa do caminh\u00e3o, 22t, sem considerar a do combust\u00edvel.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00c9 correto afirmar que o coeficiente de atrito din\u00e2mico entre o caminh\u00e3o e a rua \u00e9<\/b><\/span><\/span><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_2bdb7390.png\" alt=\"\" width=\"775\" height=\"20\" name=\"Imagem 147\" 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>03-(Ufrrj-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um bloco se ap\u00f3ia sobre um plano inclinado, conforme representado no esquema:<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_4f7c6b6a.jpg\" alt=\"\" width=\"255\" height=\"157\" name=\"Imagem 71\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dados: sen 30\u00b0 = 0,5<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Se o bloco tem peso de 700N, a menor for\u00e7a de atrito capaz de manter o bloco em equil\u00edbrio sobre o plano \u00e9<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_d83ab5fa.png\" alt=\"\" width=\"774\" height=\"18\" name=\"Imagem 148\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>04-(UNIFESP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Conforme noticiou um site da Internet em 30.8.2006, cientistas da Universidade de Berkeley, Estados Unidos, &#8220;criaram uma malha de microfibras sint\u00e9ticas que utilizam um efeito de alt\u00edssima fric\u00e7\u00e3o para sustentar cargas em superf\u00edcies lisas&#8221;, \u00e0 semelhan\u00e7a dos &#8220;incr\u00edveis p\u00ealos das patas das lagartixas&#8221;.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(&#8220;www.inovacaotecnologica.com.br&#8221;).<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Segundo esse site, os pesquisadores demonstraram que a malha criada &#8220;consegue suportar uma moeda sobre uma superf\u00edcie de vidro inclinada a at\u00e9 80\u00b0&#8221; (veja a foto).<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_2921f57.jpg\" alt=\"\" width=\"268\" height=\"154\" name=\"Imagem 72\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dados sen 80\u00b0 = 0,98; cos 80\u00b0 = 0,17 e tg 80\u00b0 = 5,7, pode-se afirmar que, nessa situa\u00e7\u00e3o, o m\u00f3dulo da for\u00e7a de atrito est\u00e1tico m\u00e1xima entre essa malha, que reveste a face de apoio da moeda, e o vidro, em rela\u00e7\u00e3o ao m\u00f3dulo do peso da moeda, equivale a, aproximadamente,<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_89ead784.png\" alt=\"\" width=\"774\" height=\"16\" name=\"Imagem 149\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>05-(IME-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0No plano inclinado da figura, os corpos A e B, cujos pesos s\u00e3o de 200N e 400N, respectivamente, est\u00e3o ligados por um fio que passa por uma polia lisa.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_9bf86208.jpg\" alt=\"\" width=\"357\" height=\"155\" name=\"Imagem 73\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O coeficiente de atrito entre os corpos e o plano \u00e9 0,25. Determine a intensidade da for\u00e7a\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_66676f1f.jpg\" alt=\"\" width=\"12\" height=\"18\" name=\"Imagem 74\" align=\"BOTTOM\" border=\"0\" \/>\u00a0de modo que o movimento se torne iminente. 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>, cos30<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=0,87 e sen30<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=0,5.<\/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-(UFB)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um corpo de massa 1kg, partindo do repouso em A, desce o plano inclinado AB e, em seguida desliza na superf\u00edcie plana e horizontal BC, parando em C, ap\u00f3s percorrer 4 m nessa superf\u00edcie. O coeficiente de atrito entre o corpo e as duas superf\u00edcies \u00e9 o mesmo e vale 0,2.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_93986ac9.jpg\" alt=\"\" width=\"393\" height=\"143\" name=\"Imagem 75\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considere cosq=0,8, senq=0,6, 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>\u00a0e determine:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) a rea\u00e7\u00e3o do plano inclinado.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a varia\u00e7\u00e3o de velocidade entre as posi\u00e7\u00f5es inicial e final.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) a intensidade da for\u00e7a de atrito nos planos AB e BC.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) a acelera\u00e7\u00e3o no plano horizontal<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) a velocidade m\u00e1xima atingida pelo m\u00f3vel<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>f) a dist\u00e2ncia percorrida no plano inclinado<\/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-(FUVEST-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um plano inclinado liso faz um \u00e2ngulo de 30<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0em rela\u00e7\u00e3o a um plano horizontal \u00e1spero.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_293d3f4c.jpg\" alt=\"\" width=\"299\" height=\"150\" name=\"Imagem 76\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um corpo de massa 10kg, abandonado no plano inclinado, demora 2s para atingir o plano horizontal. Determine: (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) a dist\u00e2ncia percorrida pelo m\u00f3vel no plano inclinado<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a dist\u00e2ncia percorrida no plano horizontal, sabendo que o coeficiente de atrito entre o corpo e o plano horizontal \u00e9 igual a 0,2.<\/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>08-(PUC-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um caixote de madeira de 4,0 kg \u00e9 empurrado por uma for\u00e7a constante\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_5dd1e5c4.jpg\" alt=\"\" width=\"15\" height=\"23\" name=\"Imagem 77\" align=\"BOTTOM\" border=\"0\" \/>e sobe com velocidade constante de 6,0 m\/s um plano inclinado de um \u00e2ngulo \u03b1, conforme 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\/plano-atrito\/i_5610828c3c8b54e6_html_2fa2e52f.jpg\" alt=\"\" width=\"454\" height=\"99\" name=\"Imagem 78\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A dire\u00e7\u00e3o da for\u00e7a\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_5dd1e5c4.jpg\" alt=\"\" width=\"15\" height=\"23\" name=\"Imagem 79\" align=\"BOTTOM\" border=\"0\" \/>\u00a0\u00e9 paralela ao plano inclinado e o coeficiente de atrito cin\u00e9tico entre as superf\u00edcies em contato \u00e9 igual a 0,5. Com base nisso, analise as seguintes afirma\u00e7\u00f5es: (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>I) O m\u00f3dulo de\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_5dd1e5c4.jpg\" alt=\"\" width=\"15\" height=\"23\" name=\"Imagem 80\" align=\"BOTTOM\" border=\"0\" \/>\u00a0\u00e9 igual a 24 N.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>II)\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_18b40349.jpg\" alt=\"\" width=\"15\" height=\"23\" name=\"Imagem 81\" align=\"BOTTOM\" border=\"0\" \/>\u00a0\u00e9 a for\u00e7a resultante do movimento na dire\u00e7\u00e3o paralela ao plano inclinado.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>III) As for\u00e7as contr\u00e1rias ao movimento de subida do caixote totalizam 40 N.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>IV) O m\u00f3dulo da for\u00e7a de atrito que atua no caixote \u00e9 igual a 16 N.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dessas afirma\u00e7\u00f5es, \u00e9 correto apenas o que se l\u00ea em<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) I e II\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) I e III\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) II e III\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) II e IV\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) III e IV<\/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-\u00a0(Mackenzie-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0A ilustra\u00e7\u00e3o refere-se a certa tarefa na qual o bloco B, dez vezes mais pesado que o bloco A, dever\u00e1 descer pelo plano inclinado com velocidade constante.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_f0fdbe88.jpg\" alt=\"\" width=\"295\" height=\"187\" name=\"Imagem 82\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considerando que o fio e a polia s\u00e3o ideais, o coeficiente de atrito cin\u00e9tico entre o bloco B e o plano dever\u00e1 ser:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(Dados: sen\u03b1\u00a0=0,6\u00a0 e\u00a0 cos\u03b1\u00a0= 0,8)<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_2a011165.png\" alt=\"\" width=\"775\" height=\"18\" name=\"Imagem 150\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>10-(ITA-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um pequeno bloco de madeira de massa m = 2 kg encontra-se sobre um plano inclinado que est\u00e1 fixo no ch\u00e3o, como mostra a figura. Com que for\u00e7a F devemos pressionar o corpo sobre o plano para que o mesmo permane\u00e7a em repouso?<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_fddcc944.jpg\" alt=\"\" width=\"244\" height=\"177\" name=\"Imagem 83\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dados: Coeficiente de atrito est\u00e1tico entre o bloco e o plano inclinado,\u00a0\u03bc=0,4; comprimento do plano inclinado=1m; altura do plano inclinado=0,6m e acelera\u00e7\u00e3o da gravidade local=9,8m\/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\/plano-atrito\/i_5610828c3c8b54e6_html_1f4cb780.png\" alt=\"\" width=\"774\" height=\"16\" name=\"Imagem 151\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>11-(UNESP-SP)<\/b><\/span><\/span><\/span><span style=\"color: #c00000;\"><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>Um bloco de massa 5,0kg est\u00e1 apoiado sobre um plano inclinado de 30\u00b0 em rela\u00e7\u00e3o a um plano horizontal.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Se uma for\u00e7a constante, de intensidade F, paralela ao plano inclinado e dirigida para cima, \u00e9 aplicada ao bloco, este adquire uma acelera\u00e7\u00e3o para baixo e sua velocidade escalar \u00e9 dada por v = 2,0t (SI), (fig.1). Se uma for\u00e7a constante, de mesma intensidade F, paralela ao plano inclinado e dirigida para baixo for aplicada ao bloco, este adquire uma acelera\u00e7\u00e3o para baixo e sua velocidade escalar \u00e9 dada por v&#8217; = 3,0t (SI), (fig. 2).<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_80eed51a.jpg\" alt=\"\" width=\"494\" height=\"192\" name=\"Imagem 84\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Calcule F, adotando 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>b) Calcule o coeficiente de atrito de deslizamento entre o corpo e o plano inclinado.<\/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-(CESGRANRIO)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um corpo de massa m = 0,20kg desce um plano inclinado de 30\u00b0 em rela\u00e7\u00e3o \u00e0 horizontal. O gr\u00e1fico apresentado mostra como varia a velocidade escalar do corpo com o tempo.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_355a54ee.jpg\" alt=\"\" width=\"284\" height=\"203\" name=\"Imagem 85\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) determine o m\u00f3dulo da acelera\u00e7\u00e3o do corpo;<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) calcule a intensidade da for\u00e7a de atrito do corpo com o plano. Dados: 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>, sen 30\u00b0 = 0,50, cos 30\u00b0 = 0,87.<\/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-(PUC-PR)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Os 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> 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>, respectivamente, est\u00e3o interligados por um fio que passa pela polia, conforme a figura. A polia pode girar livremente em torno de seu eixo. A massa do fio e da polia s\u00e3o considerados desprez\u00edveis.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_3c6dc7eb.jpg\" alt=\"\" width=\"299\" height=\"163\" name=\"Imagem 86\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Se o sistema est\u00e1 em repouso \u00e9 correto afirmar:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>I. Se 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>\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>, necessariamente existe atrito entre o corpo B e o plano inclinado.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>II. Independente de existir ou n\u00e3o atrito entre o plano e o corpo B, deve-se ter 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>\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>.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>III. Se n\u00e3o existir atrito entre o corpo B e o plano inclinado, necessariamente 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>\u00a0&gt; 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><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>IV. Se n\u00e3o existir atrito entre o corpo B e o plano inclinado, necessariamente 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>\u00a0&gt; 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>.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Est\u00e1 correta ou est\u00e3o corretas:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Somente I.\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) Somente II .\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) I e III.\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) I e IV.\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) Somente III.<\/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>14-(UFPel-RS)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma empresa de transportes faz a entrega de produtos para um supermercado. Um desses produtos \u00e9 de dimens\u00f5es consider\u00e1veis e peso elevado, o que requer o uso de uma m\u00e1quina simples (plano inclinado) para facilitar a descarga.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_2dea7c82.jpg\" alt=\"\" width=\"309\" height=\"132\" name=\"Imagem 87\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Suponha que a inclina\u00e7\u00e3o do plano de apoio, em rela\u00e7\u00e3o \u00e0 horizontal, n\u00e3o seja suficiente para provocar o deslizamento da caixa rampa abaixo. Resolva, para a situa\u00e7\u00e3o proposta, as quest\u00f5es que se seguem:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Represente graficamente as for\u00e7as que atuam sobre a caixa.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Qual \u00e9 a intensidade da for\u00e7a resultante na dire\u00e7\u00e3o do plano de apoio? Justifique.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) qual \u00e9 o valor do coeficiente de atrito entre a caixa e o plano, considerando, para esse caso, que a inclina\u00e7\u00e3o do plano de apoio, igual a 30<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, \u00e9 m\u00e1xima, sem que a caixa deslize? Dados: sen30<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=0,5; cos30<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=0,87).<\/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-(Olimp\u00edada Brasileira de F\u00edsica)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um bloco desliza sobre um plano inclinado com atrito (ver figura).<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_1f04a403.jpg\" alt=\"\" width=\"305\" height=\"216\" name=\"Imagem 88\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>No ponto A, a velocidade \u00e9 V<\/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>=2m\/s, e no ponto B, distante 1m do ponto A ao longo do plano, V<\/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>=3m\/s. Considere sen60<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=\u221a3\/2; cos60<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=1\/2 e 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>) e obtenha o valor do coeficiente de atrito cin\u00e9tico entre o bloco e o plano.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a)\u00a0\u221a3\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)\u00a0\u221a3\/2\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)\u00a0\u221a3\/2 + 1\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)\u00a0\u221a3 + \u00bd\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)\u00a0\u221a3 \u2013 \u00bd<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>16-(UNESP-SP)<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Ao modificar o estilo se uma casa para o colonial, deseja-se fazer a troca do modelo de telhas existentes. Com o intuito de preservar o jardim, foi montada uma rampa de 10m de comprimento, apoiada na beirada do madeiramento do telhado, a 6m de altura.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_35269ef.jpg\" alt=\"\" width=\"316\" height=\"170\" name=\"Imagem 89\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>No momento em que uma telha &#8212; que tem massa de 2,5kg &#8212; \u00e9 colocada sobre a rampa, ela desce acelerada, sofrendo, no entanto, a a\u00e7\u00e3o do atrito. Nessas condi\u00e7\u00f5es, determine o valor da acelera\u00e7\u00e3o desenvolvida pela telha. Dado: coeficiente de atrito\u00a0\u03bc=0,2; 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>17-(MACKENZIE-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Os corpos A e B da figura s\u00e3o id\u00eanticos e est\u00e3o ligados por meio de um fio suposto ideal. A polia possui in\u00e9rcia desprez\u00edvel, a superf\u00edcie I \u00e9 altamente polida e o coeficiente de atrito cin\u00e9tico entre a superf\u00edcie II e o corpo B \u00e9\u00a0\u03bc=0,2. 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\/plano-atrito\/i_5610828c3c8b54e6_html_c0581b86.jpg\" alt=\"\" width=\"315\" height=\"195\" name=\"Imagem 90\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Em determinado instante, o corpo A est\u00e1 descendo com velocidade escalar de 3m\/s. Calcule sua velocidade escalar ap\u00f3s 2s.<\/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-(UFG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um catador de recicl\u00e1veis de massa m sobe uma ladeira puxando seu carrinho. O coeficiente de atrito est\u00e1tico entre o piso e os seus sapatos \u00e9\u00a0\u03bc<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e o \u00e2ngulo que a ladeira forma com a horizontal \u00e9 \u03b8. O carrinho, por estar sobre rodas, pode ser considerado livre de atrito. A maior massa do carrinho com os recicl\u00e1veis que ele pode suportar, sem escorregar, e de:<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_5c9a084e.jpg\" alt=\"\" width=\"210\" height=\"132\" name=\"Imagem 91\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_f78321d.jpg\" alt=\"\" width=\"768\" height=\"50\" name=\"Imagem 92\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>19-(ITA-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0A partir do n\u00edvel P, com velocidade inicial de 5 m\/s, um corpo sobe a superf\u00edcie de um plano inclinado PQ de 0,8 m de comprimento. Sabe-se que o coeficiente de atrito cin\u00e9tico entre o plano e o corpo \u00e9 igual a 1\/3.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considere a acelera\u00e7\u00e3o da gravidade 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>, sen\u00a0\u03b8\u00a0= 0,8, cos\u00a0\u03b8\u00a0= 0,6 e que o ar n\u00e3o oferece resist\u00eancia. O tempo m\u00ednimo de percurso do corpo para que se torne nulo o componente vertical de sua velocidade \u00e9<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_934480f2.jpg\" alt=\"\" width=\"392\" height=\"116\" name=\"Imagem 93\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_eed23172.png\" alt=\"\" width=\"775\" height=\"19\" name=\"Imagem 152\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>20-(Ufrrj-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um homem puxa uma caixa de massa 2 kg para cima de um plano inclinado de um \u00e2ngulo 30\u00b0 em rela\u00e7\u00e3o \u00e0 horizontal, por meio de um fio ideal, que faz um \u00e2ngulo tamb\u00e9m de 30\u00b0 com o<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_5b50c327.jpg\" alt=\"\" width=\"236\" height=\"161\" name=\"Imagem 94\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>plano, conforme mostra a figura. O coeficiente de atrito entre a caixa e o plano \u00e9\u00a0\u03bc\u00a0= 0,2. (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=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>21-(FUVEST)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0&#8211; Um bloco de massa m, montado sobre rodas (para tornar o atrito desprez\u00edvel), parte do repouso em A e leva um tempo t para atingir B. A massa das rodas \u00e9 desprez\u00edvel. Retirando-se as rodas, verifica-se que o bloco, partindo do repouso em A, leva um tempo 2t para atingir B. A alturas entre A e o solo \u00e9 h.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Determinar o valor de t.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Determinar o valor do coeficiente de atrito entre o plano e o bloco sem rodas, em fun\u00e7\u00e3o de\u00a0a.<\/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>22-(MACKENZIE-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Certo corpo come\u00e7a a deslizar, em linha reta, por um plano inclinado, a partir do repouso na posi\u00e7\u00e3o x<\/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>= 0. Sabendo-se que ap\u00f3s 1,00 s de movimento, ele passa pela posi\u00e7\u00e3o x<\/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>\u00a0= 1,00 m e que, com mais 3,00 s, ele chega \u00e0 posi\u00e7\u00e3o x<\/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>, o coeficiente de atrito cin\u00e9tico entre as superf\u00edcies em contato (\u03bc<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>) e a posi\u00e7\u00e3o x<\/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>\u00a0s\u00e3o, respectivamente, iguais a:<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_b82dc3ce.jpg\" alt=\"\" width=\"334\" height=\"145\" name=\"Imagem 95\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dados\u00a0 &#8212;\u00a0 sen \u03b2 = 0,6\u00a0 &#8212;\u00a0 cos \u03b2 = 0,8\u00a0 &#8212;\u00a0 g = 10m\/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><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a)\u00a0 0,25 e 16,00 m\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)\u00a0 0,50 e 8,00 m\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)\u00a0 0,25 e 8,00 m\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)\u00a0 0,50 e 16,00 m\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)\u00a0 0,20 e 16,00 m\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>23-(PUC-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Um bloco escorrega a partir do repouso por um plano inclinado que faz um \u00e2ngulo de 45\u00ba com a horizontal. Sabendo que durante a queda a acelera\u00e7\u00e3o do bloco \u00e9 de 5,0 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>\u00a0e considerando 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>, podemos dizer que o coeficiente de atrito cin\u00e9tico entre o bloco e o plano \u00e9<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 0,1\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 0,2\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 0,3\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) 0,4\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) 0,5\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>24-(UERJ-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> A figura abaixo representa o plano inclinado ABFE, inserido em um paralelep\u00edpedo ret\u00e2ngulo ABCDEFGH de base horizontal, com 6 m de altura CF, 8 m de comprimento BC e 15 m de largura AB, em repouso, apoiado no solo.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_9b09635f.jpg\" alt=\"\" width=\"396\" height=\"160\" name=\"Imagem 96\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Admita um\u00a0 corpo de massa igual a 20 kg que desliza com atrito, em movimento retil\u00edneo, do ponto F ao ponto B, com velocidade constante. A for\u00e7a de atrito, em newtons, entre a superf\u00edcie deste corpo e o plano inclinado \u00e9 cerca de:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_3bc8e9e6.png\" alt=\"\" width=\"774\" height=\"20\" name=\"Imagem 153\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>25-(UPE-PE)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Um bloco de a\u00e7o \u00e9 colocado sobre uma t\u00e1bua de apoio que vai se inclinando aos poucos. Quando o bloco fica na imin\u00eancia de escorregar, a t\u00e1bua forma com a horizontal um \u00e2ngulo \u03b2 de acordo com 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\/plano-atrito\/i_5610828c3c8b54e6_html_77598d34.jpg\" alt=\"\" width=\"356\" height=\"180\" name=\"Imagem 97\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Sabendo-se que o coeficiente de atrito est\u00e1tico entre o bloco e a t\u00e1bua vale \u03bc<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=0,40 \u00e9 correto afirmar que a dist\u00e2ncia x indicada na figura, em cent\u00edmetros, vale:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 25\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 12\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) 20\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) 4<\/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-(UFRJ-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Um bloco de massa 2,0 kg est\u00e1 sobre a superf\u00edcie de um plano inclinado, que est\u00e1 em movimento retil\u00edneo para a direita, com acelera\u00e7\u00e3o de 2,0 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>, tamb\u00e9m para a direita, como indica a figura a seguir. A inclina\u00e7\u00e3o do plano \u00e9 de 30<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0em<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_b3abe339.jpg\" alt=\"\" width=\"438\" height=\"136\" name=\"Imagem 98\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>rela\u00e7\u00e3o \u00e0 horizontal. Suponha que o bloco n\u00e3o deslize sobre o plano inclinado e que a acelera\u00e7\u00e3o da gravidade seja 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>.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Usando a aproxima\u00e7\u00e3o \u221a3\u22481,7, calcule o m\u00f3dulo e indique a dire\u00e7\u00e3o e o sentido da for\u00e7a de atrito exercida pelo plano inclinado sobre o 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>27-(MACKENZIE-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Com rela\u00e7\u00e3o \u00e0 rampa de apoio, os corpos C<\/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>\u00a0e C<\/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>\u00a0est\u00e3o em repouso e na imin\u00eancia de movimento. Ao abandonar-se o conjunto, o corpo C<\/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>\u00a0sobe a rampa, com a qual existe atrito cin\u00e9tico de coeficiente \u03bc = 0,2. Considerando-se os<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>dados da tabela abaixo e fios e polias ideais, o ganho de energia cin\u00e9tica do<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_c8771b6.png\" alt=\"\" width=\"729\" height=\"198\" name=\"Imagem 155\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>corpo C<\/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>, durante o deslocamento do corpo C<\/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>, do ponto A ao ponto B, \u00e9 de<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_fadcd412.png\" alt=\"\" width=\"775\" height=\"16\" name=\"Imagem 156\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>28-(USS-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Uma pequena esfera de massa m \u00e9 abandonada em repouso no ponto A de um plano inclinado sem atrito, AC.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>altura do plano \u00e9 AB=12 metros. A partir do ponto C, a esfera passa a se mover sobre a superf\u00edcie<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_4ec29fa4.jpg\" alt=\"\" width=\"523\" height=\"127\" name=\"Imagem 101\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>horizontal com atrito, e p\u00e1ra em D, tal que CD=20 metros. O coeficiente de atrito din\u00e2mico entre a esfera e a superf\u00edcie horizontal vale:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_87df944d.png\" alt=\"\" width=\"774\" height=\"19\" name=\"Imagem 157\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>29-(UECE-CE)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_93ff0ffb.jpg\" alt=\"\" width=\"450\" height=\"127\" name=\"Imagem 102\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um bloco, sob a\u00e7\u00e3o da gravidade, desce um plano inclinado com acelera\u00e7\u00e3o de 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=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>. Considere o<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_d157666e.jpg\" alt=\"\" width=\"283\" height=\"186\" name=\"Imagem 103\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>m\u00f3dulo da acelera\u00e7\u00e3o da gravidade 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>. Sabendo-se que o \u00e2ngulo de inclina\u00e7\u00e3o do plano \u00e9 45<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0com a horizontal, o coeficiente de atrito cin\u00e9tico entre o bloco e o plano \u00e9, aproximadamente,<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A) 0,7. \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) 0,3. \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) 0,5. \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) 0,9.<\/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-(UFPE-PE)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_32babe13.jpg\" alt=\"\" width=\"451\" height=\"147\" name=\"Imagem 104\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um bloco de massa m = 4,0 kg \u00e9 impulsionado sobre um plano inclinado com velocidade inicial 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>\u00a0=<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_4f83a703.jpg\" alt=\"\" width=\"431\" height=\"179\" name=\"Imagem 105\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>mostra a figura. Ele desliza em um movimento descendente por uma dist\u00e2ncia L = 5,0 m, at\u00e9 parar. Calcule o m\u00f3dulo da for\u00e7a resultante que atua no bloco, ao longo da descida, em newtons.<\/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-(UEPA-PA)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_f6ec07a2.jpg\" alt=\"\" width=\"513\" height=\"134\" name=\"Imagem 106\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Num parque de divers\u00f5es h\u00e1 um escorregador infantil, conforme indica a figura abaixo.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_d9553499.jpg\" alt=\"\" width=\"378\" height=\"178\" name=\"Imagem 107\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Neste brinquedo, as crian\u00e7as, inicialmente em repouso, partem do ponto A e atingem o ponto B. Suponha que o coeficiente de atrito entre as superf\u00edcies de contato seja igual a 0,5.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considerando que, quando uma crian\u00e7a escorrega, a dissipa\u00e7\u00e3o de energia ocorra apenas pela a\u00e7\u00e3o da for\u00e7a de atrito, e sabendo que\u00a0 a\u00a0 ingest\u00e3o de um sorvete fornece 112.000 J, o n\u00famero de vezes que uma crian\u00e7a de 20 kg dever\u00e1 escorregar pelo brinquedo para perder a energia correspondente \u00e0 ingest\u00e3o de um sorvete \u00e9:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dados: g = 10 m\/s2; sen 45\u00b0 = cos 45\u00b0 = 0,7<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/plano-atrito\/i_5610828c3c8b54e6_html_41e09bbf.png\" alt=\"\" width=\"774\" height=\"16\" name=\"Imagem 158\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<h3><span style=\"color: #000080;\"><a style=\"color: #000080;\" title=\"Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Plano inclinado com atrito\" href=\"http:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/plano-inclinado-com-atrito\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-plano-inclinado-com-atrito\/\">Confira a resolu\u00e7\u00e3o comentada<\/a><\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"<p>Exerc\u00edcios de vestibulares com resolu\u00e7\u00f5es comentadas sobre Plano inclinado com atrito 01-(UERJ-RJ)\u00a0Um bloco de massa igual a 1,0 kg repousa em equil\u00edbrio sobre um plano inclinado. Esse plano tem comprimento igual a 50 cm e alcan\u00e7a uma altura m\u00e1xima em rela\u00e7\u00e3o ao solo igual a 30 cm. Calcule o coeficiente de atrito entre o bloco e o plano inclinado. &nbsp;<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1311,"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-1313","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1313","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=1313"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1313\/revisions"}],"predecessor-version":[{"id":10848,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1313\/revisions\/10848"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1311"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1313"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}