{"id":1302,"date":"2015-09-02T01:35:53","date_gmt":"2015-09-02T01:35:53","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1302"},"modified":"2024-08-23T13:03:13","modified_gmt":"2024-08-23T13:03:13","slug":"exercicios-de-vestibulares-com-resolucao-comentada-sobre-forca-de-atrito","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/forca-de-atrito\/exercicios-de-vestibulares-com-resolucao-comentada-sobre-forca-de-atrito\/","title":{"rendered":"For\u00e7a de atrito &#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><\/p>\n<p align=\"CENTER\"><span style=\"color: #0000cc;\"><span style=\"font-family: 'Arial Black', serif;\"><span style=\"font-size: large;\"><b>For\u00e7a de atrito<\/b><\/span><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>01-(UFB)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Considere um bloco de massa 10kg, inicialmente em repouso sobre uma superf\u00edcie reta e horizontal com atrito e cujos coeficientes de atrito est\u00e1tico e din\u00e2mico sejam respectivamente iguais a \u03bc<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e <\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>= 0,5 e \u03bc<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d <\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>= 0,3. Aplica-se ao bloco uma for\u00e7a de intensidade crescente, a partir de zero. Analise que acontece com o bloco quando<img decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_31ec9c97.png\" alt=\"\" width=\"14\" height=\"16\" name=\"Imagem 183\" align=\"BOTTOM\" border=\"0\" \/> tiver intensidade: (g=10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) F=0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) F=20N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) F=40N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) F=50N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) F=60N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>f) F=35N, com o bloco em movimento<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>g) F=30N, com o bloco em movimento\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>h) Ele se move para a direita com velocidade de intensidade V, com F=0 e apenas o F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>at<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>agindo sobre ele.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>02-(PUC-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um bloco de borracha de massa 5,0 kg est\u00e1 em repouso sobre uma superf\u00edcie plana e horizontal. O gr\u00e1fico representa como varia a for\u00e7a de atrito sobre o bloco quando sobre ele atua uma for\u00e7a F de intensidade vari\u00e1vel paralela \u00e0 superf\u00edcie.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img fetchpriority=\"high\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_6aca5a1a.jpg\" alt=\"\" width=\"330\" height=\"208\" name=\"graphics29\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O coeficiente de atrito est\u00e1tico entre a borracha e a superf\u00edcie, e a acelera\u00e7\u00e3o adquirida pelo bloco quando a intensidade da for\u00e7a atinge 30N s\u00e3o, respectivamente, iguais a<\/b><\/span><\/span><\/span><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_f70e7332.png\" alt=\"\" width=\"775\" height=\"21\" name=\"Imagem 184\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>03-(PUC-RJ)<\/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=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Uma caixa cuja velocidade inicial \u00e9 de 10 m\/s leva 5 s deslizando sobre uma superf\u00edcie horizontal at\u00e9 parar completamente.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considerando a acelera\u00e7\u00e3o da gravidade g = 10 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, determine o coeficiente de atrito cin\u00e9tico que atua entre a superf\u00edcie e a caixa.<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_ed8adea9.png\" alt=\"\" width=\"775\" height=\"17\" name=\"Imagem 185\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>04-(FUVEST-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um autom\u00f3vel de massa 10<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>kg, movendo-se inicialmente com velocidade de 72km\/h \u00e9 freado (em movimento uniformemente desacelerado) e p\u00e1ra, ap\u00f3s percorrer 50m. Calcule a for\u00e7a, o tempo de freamento e o valor do coeficiente de atrito. (g=10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>05- (UFPB)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Dois blocos A e B de massas m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 6 kg e m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 4 kg, respectivamente, est\u00e3o apoiados sobre uma mesa horizontal e movem-se sob a a\u00e7\u00e3o de uma for\u00e7a F de m\u00f3dulo 60N, conforme representa\u00e7\u00e3o na figura a seguir.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_a39b7a72.jpg\" alt=\"\" width=\"336\" height=\"86\" name=\"graphics30\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considere que o coeficiente de atrito din\u00e2mico entre o corpo A e a mesa \u00e9 \u00a0\u03bc<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>= 0,2 e que o coeficiente entre o corpo B e a mesa \u00e9\u00a0 \u03bc<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 0,3. Com base nesses dados, o m\u00f3dulo da for\u00e7a exercida pelo bloco A sobre o bloco B \u00e9: (g=10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_69b5ea37.png\" alt=\"\" width=\"775\" height=\"18\" name=\"Imagem 186\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>06-(UNESP-SP)\u00a0<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dois blocos id\u00eanticos, A e B, se deslocam sobre uma mesa plana sob a\u00e7\u00e3o de uma for\u00e7a de 10N, aplicada em A, conforme ilustrado na figura.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_52ac3f28.jpg\" alt=\"\" width=\"422\" height=\"81\" name=\"graphics31\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Se o movimento \u00e9 uniformemente acelerado, e considerando que o coeficiente de atrito cin\u00e9tico entre os blocos e a mesa \u00e9 \u03bc = 0,5, a for\u00e7a que A exerce sobre B \u00e9:<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 20N.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 15N.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 10N.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) 5N.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) 2,5N.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>07-(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=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A figura ilustra um bloco A, de massa m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 2,0 kg, atado a um bloco B, de massa m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 1,0 kg, por um fio inextens\u00edvel de massa desprez\u00edvel. O coeficiente de atrito cin\u00e9tico entre cada bloco e a mesa \u00e9 m. Uma for\u00e7a F = 18,0 N \u00e9 aplicada ao bloco B, fazendo com que ambos se desloquem com velocidade constante.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_e1e651d9.jpg\" alt=\"\" width=\"307\" height=\"107\" name=\"graphics32\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considerando g = 10,0 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, calcule<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) o coeficiente de atrito \u03bc.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a tra\u00e7\u00e3o T no fio.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>08-(UFV-MG)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Tr\u00eas blocos id\u00eanticos, A, B e C, cada um de massa M, deslocam-se sobre uma superf\u00edcie plana com uma velocidade de m\u00f3dulo V constante. Os blocos est\u00e3o interligados pelas cordas 1 e 2 e s\u00e3o arrastados por um homem, conforme esquematizado na figura a seguir.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_5e6e0835.jpg\" alt=\"\" width=\"403\" height=\"86\" name=\"graphics33\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O coeficiente de atrito cin\u00e9tico entre os blocos e a superf\u00edcie \u00e9 \u03bc e a acelera\u00e7\u00e3o da gravidade local \u00e9 g. Calcule o que se pede em termos dos par\u00e2metros fornecidos:<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) a acelera\u00e7\u00e3o do bloco B.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a for\u00e7a de tens\u00e3o T na corda 2.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #ff6600;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>09-(UNESP-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um caixote de massa 20kg est\u00e1 em repouso sobre a carroceria de um caminh\u00e3o que percorre uma estrada plana, horizontal, com velocidade constante de 72km\/h. Os coeficientes de atrito est\u00e1tico e din\u00e2mico entre o caixote e o piso da carroceria, s\u00e3o aproximadamente iguais e valem m=0,25 (admitir g=10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_48f69832.jpg\" alt=\"\" width=\"288\" height=\"145\" name=\"graphics34\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Qual \u00e9 a intensidade da for\u00e7a de atrito que est\u00e1 agindo sobre o caixote? Justifique.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Determine o menor tempo poss\u00edvel para que esse caminh\u00e3o possa frear sem que o caixote escorregue.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>10- (UFJF-MG)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um caminh\u00e3o \u00e9 carregado com duas caixas\u00a0 de madeira, de massas iguais a 500kg, conforme mostra a figura.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_b6fa8d59.jpg\" alt=\"\" width=\"308\" height=\"132\" name=\"Imagem 12\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O caminh\u00e3o \u00e9 ent\u00e3o posto em movimento numa estrada reta e plana, acelerando at\u00e9 adquirir uma velocidade de 108km\/h e depois \u00e9 freado at\u00e9 parar, conforme mostra o gr\u00e1fico. (g=10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>).<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_334f2dfd.jpg\" alt=\"\" width=\"341\" height=\"173\" name=\"graphics35\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O coeficiente de atrito est\u00e1tico entre as caixas e a carroceria do caminh\u00e3o \u00e9 m=0,1. Qual das figuras abaixo melhor representa a disposi\u00e7\u00e3o das caixas sobre a carroceria no final do movimento?<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #ff6600;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #ff6600;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #ff6600;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #ff6600;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #ff6600;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_6ec432b5.jpg\" alt=\"\" width=\"222\" height=\"89\" name=\"graphics36\" align=\"BOTTOM\" border=\"0\" \/>\u00a0\u00a0\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_91a11a61.jpg\" alt=\"\" width=\"236\" height=\"99\" name=\"graphics37\" align=\"BOTTOM\" border=\"0\" \/>\u00a0\u00a0\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_cab83c75.jpg\" alt=\"\" width=\"227\" height=\"96\" name=\"Imagem 16\" align=\"BOTTOM\" border=\"0\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_9ec59814.jpg\" alt=\"\" width=\"250\" height=\"109\" name=\"Imagem 17\" align=\"BOTTOM\" border=\"0\" \/>\u00a0\u00a0\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_f9b74bda.jpg\" alt=\"\" width=\"230\" height=\"104\" name=\"Imagem 18\" align=\"BOTTOM\" border=\"0\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>11- (UFRJ-RJ)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma for\u00e7a horizontal de m\u00f3dulo F puxa um bloco sobre uma mesa horizontal com uma acelera\u00e7\u00e3o de m\u00f3dulo a, como indica a figura 1.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_9c4888a3.jpg\" alt=\"\" width=\"443\" height=\"115\" name=\"Imagem 19\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Sabe-se que, se o m\u00f3dulo da for\u00e7a for duplicado, a acelera\u00e7\u00e3o ter\u00e1 m\u00f3dulo 3a, como indica a figura 2. Suponha que, em ambos os casos, a \u00fanica outra for\u00e7a horizontal que age sobre o bloco seja a for\u00e7a de atrito &#8211; de m\u00f3dulo invari\u00e1vel f &#8211; que a mesa exerce sobre ele.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Calcule a raz\u00e3o f\/F entre o m\u00f3dulo f da for\u00e7a de atrito e o m\u00f3dulo F da for\u00e7a horizontal que puxa o bloco.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>12- (Ufrrj-RJ)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Dois carros de corrida s\u00e3o projetados de forma a aumentar o atrito entre os pneus e a pista. Os projetos s\u00e3o id\u00eanticos, exceto que num deles os pneus s\u00e3o mais largos e no outro h\u00e1 um aerof\u00f3lio. Nessas condi\u00e7\u00f5es podemos dizer que<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) em ambos os projetos, o atrito ser\u00e1 aumentado em rela\u00e7\u00e3o ao projeto original.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) em ambos os projetos, o atrito ser\u00e1 diminu\u00eddo em rela\u00e7\u00e3o ao projeto original.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) o atrito ser\u00e1 maior no carro com aerof\u00f3lio.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) o atrito ser\u00e1 maior no carro com pneus mais largos.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) nenhum dos projetos alterar\u00e1 o atrito.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/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=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Dois blocos, A e B, com A colocado sobre B, est\u00e3o em movimento sob a\u00e7\u00e3o de uma for\u00e7a horizontal de 4,5 N aplicada sobre A, como ilustrado na figura.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_ba772bbf.jpg\" alt=\"\" width=\"268\" height=\"107\" name=\"Imagem 20\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considere que n\u00e3o h\u00e1 atrito entre o bloco B e o solo e que as massas s\u00e3o respectivamente m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 1,8 kg e m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 1,2 kg. Tomando g = 10 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, calcule<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) a acelera\u00e7\u00e3o dos blocos, se eles se locomovem juntos.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) o valor m\u00ednimo do coeficiente de atrito est\u00e1tico para que o bloco A n\u00e3o deslize sobre B.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/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=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um bloco de massa 2,0 kg repousa sobre outro de massa 3,0 kg, que pode deslizar sem atrito sobre uma superf\u00edcie plana e horizontal. Quando uma for\u00e7a de intensidade 2,0 N, agindo na dire\u00e7\u00e3o horizontal, \u00e9 aplicada ao bloco inferior, como mostra a figura, o conjunto passa a se movimentar sem que o bloco superior escorregue sobre o inferior.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_d3afc165.jpg\" alt=\"\" width=\"261\" height=\"115\" name=\"Imagem 21\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Nessas condi\u00e7\u00f5es, determine (g=10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) a acelera\u00e7\u00e3o do conjunto.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a intensidade da for\u00e7a de atrito entre os dois blocos.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>15-(PUC-RJ)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um certo bloco exige uma for\u00e7a F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0para ser posto em movimento, vencendo a for\u00e7a de atrito est\u00e1tico. Corta-se o bloco ao meio, colocando uma metade sobre a outra.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Seja agora F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0a for\u00e7a necess\u00e1ria para p\u00f4r o conjunto em movimento. Sobre a rela\u00e7\u00e3o F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0\/ F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, pode-se afirmar que:<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) ela \u00e9 igual a 2.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) ela \u00e9 igual a 1.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) ela \u00e9 igual a 1\/2.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) ela \u00e9 igual a 3\/2.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) seu valor depende da superf\u00edcie.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>16-(UNESP-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Dois corpos, A e B, atados por um cabo, com massas m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 1 kg e m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 2,5 kg, respectivamente, deslizam sem atrito no solo horizontal sob a\u00e7\u00e3o de uma for\u00e7a, tamb\u00e9m horizontal, de 12 N aplicada em B. Sobre este corpo, h\u00e1 um terceiro corpo, C, com massa m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>C<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 0,5 kg, que se desloca com B, sem deslizar sobre ele. A figura ilustra a situa\u00e7\u00e3o descrita<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_17060830.jpg\" alt=\"\" width=\"354\" height=\"122\" name=\"graphics38\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Calcule a for\u00e7a exercida sobre o corpo C.(g=10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>17-(UNIFESP-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0A figura representa uma demonstra\u00e7\u00e3o simples que costuma ser usada\u00a0 para ilustrar a Primeira Lei de Newton.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_485cedcf.jpg\" alt=\"\" width=\"256\" height=\"152\" name=\"Imagem 23\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O copo, sobre uma mesa, est\u00e1 com a boca tampada pelo cart\u00e3o\u00a0<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>C<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e, sobre este est\u00e1 a moeda\u00a0<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>. A massa da moeda \u00e9 0,010kg e o coeficiente de atrito est\u00e1tico entre a moeda e o cart\u00e3o \u00e9 0,15. O experimentador puxa o cart\u00e3o com a for\u00e7a\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_f3c4304f.jpg\" alt=\"\" width=\"13\" height=\"19\" name=\"Imagem 24\" align=\"BOTTOM\" border=\"0\" \/>, horizontal, e a moeda escorrega do cart\u00e3o e cai dentro do copo.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Represente todas as for\u00e7as que atuam sobre a moeda\u00a0<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00a0quando ela est\u00e1 escorregando sobre o cart\u00e3o puxado pela for\u00e7a\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_f3c4304f.jpg\" alt=\"\" width=\"13\" height=\"19\" name=\"graphics39\" align=\"BOTTOM\" border=\"0\" \/>. Nomeie cada uma das for\u00e7as representadas.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Costuma-se explicar o que ocorre com a afirma\u00e7\u00e3o de que, devido \u00e0 in\u00e9rcia, a moeda escorrega e cai dentro do copo. Isso \u00e9 sempre verdade ou \u00e9 necess\u00e1rio que o m\u00f3dulo de\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_4009c6aa.jpg\" alt=\"\" width=\"13\" height=\"19\" name=\"graphics40\" align=\"BOTTOM\" border=\"0\" \/>\u00a0tenha uma intensidade m\u00ednima para que a moeda escorregue sobre o cart\u00e3o? Se for necess\u00e1ria essa for\u00e7a m\u00ednima, qual \u00e9, nesse caso, o seu valor? (Despreze a massa do cart\u00e3o, o atrito entre o cart\u00e3o e o copo e admita g=10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>).<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>18-(UFAL)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00a0<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A figura ilustra um pequeno bloco A, de massa 1 kg, sobre um grande bloco B, de massa 4 kg.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_98ad983f.jpg\" alt=\"\" width=\"289\" height=\"123\" name=\"Imagem 27\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>N\u00e3o h\u00e1 atrito entre os blocos. As for\u00e7as horizontais paralelas possuem m\u00f3dulos constantes F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 24 N e F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 12 N. Considerando a acelera\u00e7\u00e3o da gravidade g = 10 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e o coeficiente de atrito cin\u00e9tico entre o bloco B e a superf\u00edcie horizontal igual a 0,2,\u00a0 determine o m\u00f3dulo da acelera\u00e7\u00e3o relativa entre os blocos, enquanto um bloco estiver sobre o outro.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>19-(PUC-MG)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um bloco de massa 3,0 kg \u00e9 pressionado contra uma parede vertical por uma for\u00e7a de intensidade F conforme ilustra\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_2631b0d0.jpg\" alt=\"\" width=\"316\" height=\"134\" name=\"graphics41\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considere a gravidade como 10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, o coeficiente de atrito est\u00e1tico entre o bloco e a parede como 0,20 e o coeficiente de atrito cin\u00e9tico como 0,15.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O valor m\u00e1ximo da for\u00e7a F para que o bloco des\u00e7a em equil\u00edbrio din\u00e2mico \u00e9 de:<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_b40f97b7.png\" alt=\"\" width=\"775\" height=\"19\" name=\"Imagem 187\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>20-(UFAL)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0O bloco da figura possui peso P e se encontra na imin\u00eancia de movimento sob a a\u00e7\u00e3o de uma for\u00e7a de m\u00f3dulo constante F e dire\u00e7\u00e3o perpendicular \u00e0 parede vertical. Se o coeficiente de atrito est\u00e1tico entre a parede e o bloco \u00e9 menor que 1, assinale a rela\u00e7\u00e3o correta entre P e F.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_194da551.jpg\" alt=\"\" width=\"201\" height=\"145\" name=\"graphics42\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>A) 0 &lt; P &lt; F\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>B) F &lt; P &lt; 2F\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>C) 0 &lt; F &lt; P\/2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>D) P\/2 &lt; F &lt; P\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>E) 0 &lt; F &lt; P<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>21- (UERJ-RJ)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma pessoa de massa igual a 80 kg encontra-se em repouso, em p\u00e9 sobre o solo, pressionando perpendicularmenteuma parede com uma for\u00e7a de magnitude igual a 120 N, como mostra a ilustra\u00e7\u00e3o a seguir.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_6c3f4ef1.jpg\" alt=\"\" width=\"199\" height=\"186\" name=\"Imagem 30\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A melhor representa\u00e7\u00e3o gr\u00e1fica para as distintas for\u00e7as externas que atuam sobre a pessoa est\u00e1 indicada em:<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_c52293cb.jpg\" alt=\"\" width=\"734\" height=\"102\" name=\"Imagem 31\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>22-(UERJ-RJ)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Com rela\u00e7\u00e3o ao exerc\u00edcio anterior, considerando a acelera\u00e7\u00e3o da gravidade igual a 10 m . s\u00a0<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>-2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, o coeficiente de atrito entre a superf\u00edcie do solo e a sola do cal\u00e7ado da pessoa \u00e9 da ordem de:<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_8a56853b.png\" alt=\"\" width=\"775\" height=\"17\" name=\"Imagem 188\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>23-(UNIFESP-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0A figura representa um bloco B de massa m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0apoiado sobre um plano horizontal e um bloco A de massa m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0a ele pendurado. O conjunto n\u00e3o se movimenta por causa do atrito entre o bloco B e o plano, cujo coeficiente de atrito est\u00e1tico \u00e9 \u03bc<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_f832487b.jpg\" alt=\"\" width=\"229\" height=\"181\" name=\"Imagem 32\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>N\u00e3o leve em conta a massa do fio, considerado inextens\u00edvel, nem o atrito no eixo da roldana. Sendo g o m\u00f3dulo da acelera\u00e7\u00e3o da gravidade local, pode-se afirmar que o m\u00f3dulo da for\u00e7a de atrito est\u00e1tico entre o bloco B e o plano<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) \u00e9 igual ao m\u00f3dulo do peso do bloco A.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) n\u00e3o tem rela\u00e7\u00e3o alguma com o m\u00f3dulo do peso do bloco A.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) \u00e9 igual ao produto m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0. g . m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, mesmo que esse valor seja maior que o m\u00f3dulo do peso de A.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) \u00e9 igual ao produto m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0. g . m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, desde que esse valor seja menor que o m\u00f3dulo do peso de A.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) \u00e9 igual ao m\u00f3dulo do peso do bloco B.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>24-(Unisanta-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0No sistema representado a seguir, os corpos A, B e C tem massas respectivamente iguais a 3kg, 2kg e 7kg.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_98afd381.jpg\" alt=\"\" width=\"261\" height=\"187\" name=\"Imagem 33\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Os blocos A e C s\u00e3o ligados por um fio leve e flex\u00edvel. A polia \u00e9 ideal e o coeficiente de atrito dos blocos A e B com a superf\u00edcie \u00e9 igual a \u03bc = 0,2. A acelera\u00e7\u00e3o dos blocos e a for\u00e7a de contato entre os blocos A e B valem, respectivamente (g=10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>):<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 5m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e 35N\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 5m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e 14N\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 6m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2\u00a0<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e 14N\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) 8m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e 35N\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) 6m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e 35N.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>25-(PUC-MG)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Determine massa m\u00ednima que deve ser colocada sobre o bloco de 10 kg para mant\u00ea-lo<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_11453a3a.jpg\" alt=\"\" width=\"276\" height=\"181\" name=\"Imagem 34\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>em equil\u00edbrio, sabendo-se que o coeficiente de atrito est\u00e1tico entre ele e a mesa \u00e9 0,20. (g=10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2.<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>).<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>26-(UnB-DF)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0O coeficiente de atrito est\u00e1tico entre os blocos A e B, montados como mostra a figura abaixo, \u00e9 de 0,9.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_210ab095.jpg\" alt=\"\" width=\"298\" height=\"161\" name=\"Imagem 35\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considerando que as massas dos blocos A e B sejam, respectivamente, iguais a 5,0kg e 0,4kg e que g=10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, calcule em newtons, o menor valor do m\u00f3dulo da for\u00e7a\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_4d91da32.jpg\" alt=\"\" width=\"14\" height=\"19\" name=\"Imagem 36\" align=\"BOTTOM\" border=\"0\" \/>\u00a0para que o bloco B n\u00e3o caia.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>27-(UNIFESP-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0A figura representa um caixote transportado por uma esteira horizontal. Ambos tem velocidade de m\u00f3dulo V, constante,, suficientemente pequeno para que a resist\u00eancia do ar sobre o caixote possa ser considerada desprez\u00edvel.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_eccec57d.jpg\" alt=\"\" width=\"282\" height=\"111\" name=\"Imagem 37\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Pode-se afirmar que sobre esse caixote, na situa\u00e7\u00e3o da figura.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) atuam quatro for\u00e7as: o seu peso, a rea\u00e7\u00e3o normal da esteira, a for\u00e7a de atrito entre a esteira e o caixote e a for\u00e7a motora que a esteira exerce sobre o caixote.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) ) atuam tr\u00eas for\u00e7as: o seu peso, a rea\u00e7\u00e3o normal da esteira, a for\u00e7a de atrito entre a esteira e o caixote, no sentido oposto ao do movimento.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) ) atuam tr\u00eas for\u00e7as: o seu peso, a rea\u00e7\u00e3o normal da esteira, a for\u00e7a de atrito entre a esteira e o caixote, no sentido do movimento..<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) ) atuam duas for\u00e7as: o seu peso, a rea\u00e7\u00e3o normal da esteira.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) n\u00e3o atua for\u00e7a nenhuma, pois ele tem movimento retil\u00edneo uniforme.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>28-(UFRJ-RJ)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Considere um caminh\u00e3o de frutas trafegando em movimento retil\u00edneo numa estrada horizontal, com velocidade uniforme de V=20m\/s. O caminh\u00e3o transporta, na ca\u00e7amba, uma caixa de ma\u00e7\u00e3s de massa m=30kg.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_de99aa7f.jpg\" alt=\"\" width=\"334\" height=\"124\" name=\"Imagem 38\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Ao avistar um sinal de tr\u00e2nsito a 100m, o motorista come\u00e7a a frear uniformemente, de modo a parar junto a ele.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Fa\u00e7a um esquema das for\u00e7as que atuam sobre a caixa durante a frenagem.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Calcule o m\u00f3dulo da componente horizontal da for\u00e7a que o ch\u00e3o da ca\u00e7amba do caminh\u00e3o exerce sobre a caixa durante a frenagem.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>29-(UFPE)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma caixa de massa 10kg \u00e9 ligada a um bloco de massa 5kg, por meio de um fio fino e inextens\u00edvel que passa por uma pequena polia sem atrito, como mostra a figura.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_cfa4524a.jpg\" alt=\"\" width=\"323\" height=\"167\" name=\"Imagem 39\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Determine o valor da for\u00e7a horizontal\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_5db0a7f6.jpg\" alt=\"\" width=\"15\" height=\"24\" name=\"Imagem 40\" align=\"BOTTOM\" border=\"0\" \/>, em N,\u00a0 que deve ser aplicada a caixa de modo que o bloco suba, com acelera\u00e7\u00e3o a=2,0m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>. O coeficiente de atrito din\u00e2mico entre a caixa e o piso \u00e9 \u03bc=0,10. Considere g=10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>30-(UFPE)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma vassoura, de massa 0,4 kg, est\u00e1 posicionada sobre um piso horizontal como indicado na figura. Uma for\u00e7a, de m\u00f3dulo F(cabo), \u00e9 aplicada para baixo ao longo do cabo da vassoura.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_6fc58c93.jpg\" alt=\"\" width=\"277\" height=\"141\" name=\"Imagem 41\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Sabendo-se que o coeficiente de atrito est\u00e1tico entre o piso e a base da vassoura \u00e9 \u03bc = 1\/8, calcule F(cabo), em newtons, para que a vassoura fique na imin\u00eancia de se deslocar. Considere desprez\u00edvel a massa do cabo, quando comparada com a base da vassoura<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>31-(UERJ-RJ)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma caixa est\u00e1 sendo puxada por um trabalhador, conforme mostra a Figura 1. \u00a0Para diminuir a for\u00e7a de atrito entre a caixa e o ch\u00e3o, aplica-se, no ponto X, uma for\u00e7a f. O segmento orientado que pode representar esta for\u00e7a est\u00e1 indicado na alternativa:<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_f1d68452.jpg\" alt=\"\" width=\"779\" height=\"138\" name=\"Imagem 42\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>32-(UFSCAR-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um menino deseja deslocar um bloco de madeira sobre o ch\u00e3o horizontal puxando uma corda amarrada ao bloco.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Sabendo-se que o coeficiente de atrito est\u00e1tico entre a madeira e o ch\u00e3o vale 0,4, que a massa do bloco \u00e9 42 kg e que a acelera\u00e7\u00e3o da gravidade \u00e9 igual a 10 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, e considerando \u221a3 = 1,7, qual a menor intensidade da for\u00e7a que o menino deve puxar a corda para deslocar o bloco, se a dire\u00e7\u00e3o da corda forma com o ch\u00e3o um \u00e2ngulo de 60\u00b0?<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 100 N.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 200 N.\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 220 N.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) 250 N.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) 300 N.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>33-(UNESP-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Na figura est\u00e1 representada esquematicamente a for\u00e7a\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_f78ec0e9.jpg\" alt=\"\" width=\"14\" height=\"20\" name=\"Imagem 43\" align=\"BOTTOM\" border=\"0\" \/>\u00a0arrastando o bloco de massa 2,0kg com acelera\u00e7\u00e3o constante de 0,1m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0sobre o plano horizontal.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_cf17e6b3.jpg\" alt=\"\" width=\"228\" height=\"124\" name=\"Imagem 44\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(Dados: cos37<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=0,8; sen37<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=0,6 e g=10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>). Sendo F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>at<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=0,6N a for\u00e7a de atrito entre o bloco e o plano, pode-se afirmar que o m\u00f3dulo de\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_f78ec0e9.jpg\" alt=\"\" width=\"14\" height=\"20\" name=\"Imagem 45\" align=\"BOTTOM\" border=\"0\" \/>, em N, \u00e9:<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_a7bb6212.png\" alt=\"\" width=\"775\" height=\"18\" name=\"Imagem 189\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>34-(UFES)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Dois corpos de massas m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0(m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>&gt;m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>) est\u00e3o ligados por um fio inextens\u00edvel e de massa desprez\u00edvel conforme a figura abaixo.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_1bbf4010.jpg\" alt=\"\" width=\"308\" height=\"163\" name=\"Imagem 46\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dois mancais exercem, cada um, uma for\u00e7a horizontal de intensidade N sobre o corpo A. O coeficiente de atrito din\u00e2mico entre os mancais \u00e9 m, e a acelera\u00e7\u00e3o da gravidade\u00a0 g \u00e9 conhecida. Considere que o fio desliza livremente sobre as duas polias e que estas possuem massas desprez\u00edveis. Estando os corpos em movimento, determine:<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) a acelera\u00e7\u00e3o com que os corpos A e B se deslocam.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a intensidade da for\u00e7a N que cada um dos mancais deve exercer sobre o corpo A, para que os corpos A e B se desloquem com velocidade constante.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>35-(UNICAMP-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Considere, na figura ao lado, dois blocos A e B, de massas conhecidas, ambos em repouso.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_244ac255.jpg\" alt=\"\" width=\"276\" height=\"155\" name=\"Imagem 47\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Uma for\u00e7a de 5,0N \u00e9 aplicada no bloco A, que permanece em repouso. H\u00e1 atrito entre o bloco A e a mesa, e entre os blocos A e B.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) O que acontece com o bloco B?<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Reproduza a figura, indicando as for\u00e7as horizontais (sentido, m\u00f3dulo e onde est\u00e3o aplicadas) que atuam sobre os blocos A e B.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>36-(UFBA)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um corpo A pesa 100N e est\u00e1 em repouso sobre o corpo B, que pesa 200N. O corpo A est\u00e1 ligado por uma corda ao anteparo C, enquanto o corpo B est\u00e1 sendo solicitado por uma for\u00e7a horizontal F, de 125N. O coeficiente de atrito de escorregamento entre os corpos A e B \u00e9 0,25.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_a4813e11.jpg\" alt=\"\" width=\"249\" height=\"126\" name=\"Imagem 48\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Determine o coeficiente de atrito entre o corpo B e a superf\u00edcie de apoio e a tra\u00e7\u00e3o na corda, considerando o corpo B em movimento iminente.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>37-(ITA-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Os blocos A e B da figura tem massa m. O coeficiente de atrito entre todas as superf\u00edcies \u00e9 m. A for\u00e7a\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_4fe5f75b.jpg\" alt=\"\" width=\"16\" height=\"27\" name=\"Imagem 49\" align=\"BOTTOM\" border=\"0\" \/>\u00a0imprime ao bloco B da figura (I) velocidade uniforme.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_c5bbf35a.jpg\" alt=\"\" width=\"572\" height=\"121\" name=\"Imagem 50\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Calcule as rela\u00e7\u00f5es F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\/F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\/F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, nas quais F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00e9 a for\u00e7a indicada na figura (II) e F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>3<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00e9 indicada na figura (III). Para que o bloco B nessas figuras tenha velocidade constante.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>38-(UFF-RJ)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um pano de prato retangular, com 60cm de comprimento e constitui\u00e7\u00e3o homog\u00eanea, est\u00e1 em repouso sobre uma mesa, parte sobre sua superf\u00edcie, horizontal e fina, e parte pendente, como mostra a figura.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_768349fd.jpg\" alt=\"\" width=\"256\" height=\"131\" name=\"Imagem 51\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Sabendo-se que o coeficiente de atrito est\u00e1tico entre a superf\u00edcie da mesa\u00a0 e o pano \u00e9 igual a 0,5 e que o pano est\u00e1 na imin\u00eancia de deslizar, pode-se afirmar que o comprimento L da parte sobre a mesa \u00e9:<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_c37ae9d6.png\" alt=\"\" width=\"775\" height=\"19\" name=\"Imagem 190\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>39-(FUVEST-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0O sistema indicado na figura a seguir, onde as polias s\u00e3o ideais, permanece em repouso gra\u00e7as \u00e0 for\u00e7a de atrito entre o corpo de 10kg e a superf\u00edcie de apoio.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_100b89c3.jpg\" alt=\"\" width=\"357\" height=\"181\" name=\"Imagem 52\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Determine o valor da for\u00e7a de atrito.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>40-(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=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um trator\u00a0 se desloca em uma estrada, da esquerda para a direita, com movimento acelerado. O sentido das for\u00e7as de atrito que a estrada faz sobre as rodas do carro \u00e9 indicado na figura a seguir:<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_4bd57bea.jpg\" alt=\"\" width=\"263\" height=\"156\" name=\"Imagem 53\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0\u00a0 <span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00c9 correto afirmar que:<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) o trator tem tra\u00e7\u00e3o nas quatro rodas;\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) o trator tem tra\u00e7\u00e3o traseira;\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) o trator tem tra\u00e7\u00e3o dianteira<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) o trator est\u00e1 com o motor desligado;\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) a situa\u00e7\u00e3o apresentada \u00e9 imposs\u00edvel de acontecer.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>41-(UERJ-RJ)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considere um carro de tra\u00e7\u00e3o dianteira que acelera no sentido indicado na figura em destaque. O motor \u00e9 capaz de impor \u00e0s rodas de tra\u00e7\u00e3o um determinado sentido de rota\u00e7\u00e3o. S\u00f3 h\u00e1 movimento quando h\u00e1 atrito est\u00e1tico, pois, na sua aus\u00eancia, as rodas de tra\u00e7\u00e3o patinam sobre o solo, como acontece em um terreno enlameado. O diagrama que representa corretamente as for\u00e7as de atrito est\u00e1tico que o solo exerce sobre as rodas \u00e9:<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_b8f25a5d.jpg\" alt=\"\" width=\"767\" height=\"126\" name=\"Imagem 54\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>42-(ITA-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=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um autom\u00f3vel desloca-se sobre uma estrada, da esquerda para a direita, conforme as figuras de\u00a0<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0a\u00a0<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>. As setas nas rodas indicam os sentidos das for\u00e7as de atrito (sem rela\u00e7\u00e3o com os m\u00f3dulos) exercidas sobre elas pelo ch\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_1ed99e6a.jpg\" alt=\"\" width=\"180\" height=\"80\" name=\"Imagem 55\" align=\"BOTTOM\" border=\"0\" \/>\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_1d95a4f5.jpg\" alt=\"\" width=\"180\" height=\"82\" name=\"Imagem 56\" align=\"BOTTOM\" border=\"0\" \/>\u00a0\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_63eb0c1b.jpg\" alt=\"\" width=\"166\" height=\"79\" name=\"Imagem 57\" align=\"BOTTOM\" border=\"0\" \/>\u00a0\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_7bf2cfe3.jpg\" alt=\"\" width=\"170\" height=\"79\" name=\"Imagem 58\" align=\"BOTTOM\" border=\"0\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Associe as alternativas apresentadas com os algarismos romanos de I a IV.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>I \u2013 Tra\u00e7\u00e3o somente nas rodas dianteiras\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>II \u2013 Tra\u00e7\u00e3o nas quatro rodas\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>III \u2013 Motor desligado (desacoplado)<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>IV \u2013 Tra\u00e7\u00e3o somente nas rodas traseiras<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>43-(UNICAMP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um caminh\u00e3o transporta um bloco de ferro de 3,0t, trafegando horizontalmente e em linha reta, com\u00a0velocidade constante. O motorista v\u00ea o sinal (sem\u00e1foro) ficar vermelho e aciona os freios, aplicando\u00a0uma\u00a0desacelera\u00e7\u00e3o constante de valor 3,0 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>. O bloco n\u00e3o escorrega.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_cb071227.jpg\" alt=\"\" width=\"302\" height=\"117\" name=\"graphics43\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O coeficiente de atrito est\u00e1tico entre o\u00a0bloco e a carroceria \u00e9 0,40. Adote g = 10 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Qual a intensidade da for\u00e7a de atrito que a carroceria aplica sobre o bloco, durante a desacelera\u00e7\u00e3o?<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Qual \u00e9 a m\u00e1xima desacelera\u00e7\u00e3o que o caminh\u00e3o pode ter para o bloco n\u00e3o escorregar?<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>44- (UNIFESP-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma bonequinha est\u00e1 presa por um \u00edm\u00e3 a ela colado, \u00e0 porta vertical de uma geladeira.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Desenhe esquematicamente essa bonequinha no caderno de respostas, representando e nomeando as for\u00e7as que atuam sobre ela.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Sendo m = 20g a massa total da bonequinha com o \u00edm\u00e3 e \u03bc = 0,50 o coeficiente de atrito est\u00e1tico entre o \u00edm\u00e3 e a porta da geladeira, qual deve ser o menor valor da for\u00e7a magn\u00e9tica entre o \u00edm\u00e3 e a geladeira para que a bonequinha n\u00e3o caia? Dado: g = 10 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>45-(UFRJ-RJ)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma for\u00e7a horizontal e de intensidade 30 N \u00e9 aplicada num corpo A de massa 4,0 kg, preso a um corpo B de massa 2,0 kg que, por sua vez, se prende a um corpo C.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_824ab8df.jpg\" alt=\"\" width=\"380\" height=\"86\" name=\"Imagem 60\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O coeficiente de atrito entre cada corpo e a superf\u00edcie horizontal de apoio \u00e9 0,10 e verifica-se que a acelera\u00e7\u00e3o do sistema \u00e9, nessas condi\u00e7\u00f5es, 2,0 m\/s2. Adote g = 10 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e analise as afirma\u00e7\u00f5es.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1) A massa do corpo \u00e9 de 5,0kg\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2) A tra\u00e7\u00e3o no fio que une A a B tem m\u00f3dulo 18N<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>3) A for\u00e7a de atrito sofrida pelo corpo A vale 4,0N\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>4) a tra\u00e7\u00e3o no fio que une B a C tem intensidade 8,0N<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>5) A for\u00e7a resultante no corpo B tem m\u00f3dulo 4,0N<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>46-(UNICAMP-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Ao se usar um saca-rolha, a for\u00e7a m\u00ednima que deve ser aplicada para que a rolha de uma garrafa comece a sair \u00e9 igual a 360N.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_5e46912f.jpg\" alt=\"\" width=\"147\" height=\"163\" name=\"graphics44\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Sendo \u03bc<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 0,2 o coeficiente de atrito est\u00e1tico entre a rolha e o bocal da garrafa, encontre a for\u00e7a normal que a rolha exerce no bocal da garrafa. Despreze o peso da rolha.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>47-(UFPE)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Os blocos A, B e C da figura possuem a mesma massa m = 7,0 kg. O coeficiente de atrito cin\u00e9tico entre todas as superf\u00edcies \u00e9 0,3. Calcule o m\u00f3dulo da for\u00e7a F, em N, que imprime uma<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_ab938146.jpg\" alt=\"\" width=\"432\" height=\"145\" name=\"graphics45\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>velocidade constante ao bloco B, levando-o desde a situa\u00e7\u00e3o (1) at\u00e9 a situa\u00e7\u00e3o (2). (g=10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>).<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>48-(FUVEST-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Procedimento de seguran\u00e7a, em auto-estradas, recomenda que o motorista mantenha uma \u201cdist\u00e2ncia\u201d de dois segundos do carro que est\u00e1 \u00e0 sua frente, para que, se necess\u00e1rio, tenha espa\u00e7o para frear (\u201cregra dos dois segundos\u201d). Por essa regra, a dist\u00e2ncia D que o carro percorre, em dois segundos, com velocidade constante V<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, deve ser igual \u00e0 dist\u00e2ncia necess\u00e1ria para que o carro pare completamente ap\u00f3s frear. Tal procedimento, por\u00e9m, depende da velocidade V<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0com que o carro trafega e da desacelera\u00e7\u00e3o m\u00e1xima a, fornecida pelos freios.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Determine o intervalo de tempo T<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o,<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0em segundos, necess\u00e1rio para que o carro pare completamente , percorrendo a dist\u00e2ncia D referida.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Represente, no sistema de eixos abaixo, a varia\u00e7\u00e3o da desacelera\u00e7\u00e3o a, em fun\u00e7\u00e3o da velocidade V<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, para situa\u00e7\u00f5es em que o carro p\u00e1ra completamente em um intervalo T<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(determinado no item anterior).<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_3a21ceaa.jpg\" alt=\"\" width=\"191\" height=\"174\" name=\"graphics46\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) Considerando que a desacelera\u00e7\u00e3o a depende principalmente do coeficiente de atrito m entre os pneus e o asfalto, sendo 0,6 o valor de m, determine, a partir do gr\u00e1fico, o valor m\u00e1ximo de velocidade V<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>M<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, em m\/s, para o qual a regra dos dois segundos permanece v\u00e1lida.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>49-(MACKENZIE-SP) <\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um bloco A, de massa 6 kg, est\u00e1 preso a outro B, de massa 4 kg, por meio de uma mola ideal de constante el\u00e1stica 800 N\/m. Os blocos est\u00e3o apoiados sobre uma superf\u00edcie horizontal e se movimentam devido \u00e0 a\u00e7\u00e3o da for\u00e7a\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_a9eabf2f.gif\" alt=\"\" width=\"12\" height=\"18\" name=\"Imagem 64\" align=\"BOTTOM\" border=\"0\" \/>\u00a0horizontal, de intensidade 60 N. Sendo o coeficiente de atrito cin\u00e9tico entre as superf\u00edcies em contato igual a 0,4, a distens\u00e3o da<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_8f6df558.jpg\" alt=\"\" width=\"462\" height=\"101\" name=\"Imagem 65\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>mola \u00e9 de: Dado: g = 10m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 3 cm\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 4 cm\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 5 cm\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) 6 cm\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) 7 cm<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>50-(MACKENZIE-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Um corpo de peso 30 N repousa sobre uma superf\u00edcie horizontal de coeficiente de atrito est\u00e1tico 0,4. Por meio de uma mola de massa desprez\u00edvel, de comprimento natural 20 cm e constante el\u00e1stica 20N\/m, prende-se esse corpo em uma parede como mostra a figura. A m\u00e1xima dist\u00e2ncia a que podemos manter esse corpo da parede e em equil\u00edbrio ser\u00e1 de<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_f5defa5a.jpg\" alt=\"\" width=\"341\" height=\"111\" name=\"Imagem 66\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 26 cm\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 40 cm\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 80 cm\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) 90 cm\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) 100 cm\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>51-(CPS)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Para evitar que seus pais, que j\u00e1 s\u00e3o idosos, n\u00e3o sofram acidentes no piso escorregadio do<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_2c441d2f.jpg\" alt=\"\" width=\"197\" height=\"163\" name=\"Imagem 67\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>quintal da casa, \u00a0Sandra contratou uma pessoa para fazer ranhuras na superf\u00edcie desse piso \u2013 atitude ecopr\u00e1tica que n\u00e3o gera entulho, pois torna<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>desnecess\u00e1ria a troca do piso. O fato de o piso com ranhuras evitar que pessoas escorreguem est\u00e1 ligado ao conceito f\u00edsico de<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) atrito.\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) empuxo.\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) press\u00e3o.\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) viscosidade.\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) condutibilidade.\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>52-(CFT-MG)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Em uma superf\u00edcie horizontal, uma caixa \u00e9 arrastada para a direita, sob a a\u00e7\u00e3o de uma for\u00e7a constante F e de uma for\u00e7a de atrito F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>at<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0conforme a figura.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_72b89910.jpg\" alt=\"\" width=\"392\" height=\"119\" name=\"graphics47\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considerando essa situa\u00e7\u00e3o, a alternativa correta \u00e9<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Se a caixa est\u00e1 em movimento retil\u00edneo, temos as seguintes hip\u00f3teses:<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_5a091d2d.jpg\" alt=\"\" width=\"472\" height=\"166\" name=\"graphics48\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>53-(UFG-GO)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 A for\u00e7a muscular origina-se nas fibras musculares, conforme figura (a), como resultado das intera\u00e7\u00f5es entre certas prote\u00ednas que experimentam mudan\u00e7as de configura\u00e7\u00e3o e proporcionam a contra\u00e7\u00e3o r\u00e1pida e volunt\u00e1ria do m\u00fasculo.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A for\u00e7a m\u00e1xima que um m\u00fasculo pode exercer depende da sua \u00e1rea da se\u00e7\u00e3o reta e vale cerca de 30 N\/cm<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>. Considere um oper\u00e1rio que movimenta com uma velocidade constante uma caixa de 120 kg sobre uma superf\u00edcie rugosa, de coeficiente de atrito 0,8, usando os dois bra\u00e7os, conforme ilustrado na figura (b).<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_b8aae4de.jpg\" alt=\"\" width=\"501\" height=\"171\" name=\"Imagem 70\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dessa forma, a menor se\u00e7\u00e3o reta dos m\u00fasculos de um dos bra\u00e7os do oper\u00e1rio, em cm2, e uma das prote\u00ednas respons\u00e1veis pela contra\u00e7\u00e3o das miofibrilas s\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dados: g =10,0 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 16 e actina. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 16 e mielina. \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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 20 e miosina. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) 32 e actina. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) 32 e miosina. \u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>54- (UFLA-AL)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Um trator utiliza uma for\u00e7a motriz de 2000 N e arrasta, com velocidade constante, um<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_99231aef.jpg\" alt=\"\" width=\"358\" height=\"130\" name=\"Imagem 71\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>tronco de massa 200Kg ao longo de um terreno horizontal e irregular. Considerando g = 10 m\/s2, \u00e9 correto afirmar que o coeficiente de atrito cin\u00e9tico \u03bc<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0entre o tronco e o terreno \u00e9:<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 1,0 \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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 0,25 \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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) zero \u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>55-(UNESP-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 As figuras 1 e 2 representam dois esquemas experimentais utilizados para a determina\u00e7\u00e3o do coeficiente de atrito est\u00e1tico entre um bloco B e uma t\u00e1bua plana, horizontal.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_39f566dc.jpg\" alt=\"\" width=\"492\" height=\"181\" name=\"Imagem 72\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>No esquema da figura 1, um aluno exerceu uma for\u00e7a horizontal\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_f2729fa4.gif\" alt=\"\" width=\"12\" height=\"18\" name=\"Imagem 73\" align=\"BOTTOM\" border=\"0\" \/>\u00a0no fio A e mediu o valor 2,0 cm para a deforma\u00e7\u00e3o da mola, quando a for\u00e7a\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_f2729fa4.gif\" alt=\"\" width=\"12\" height=\"18\" name=\"Imagem 74\" align=\"BOTTOM\" border=\"0\" \/>\u00a0atingiu seu m\u00e1ximo valor poss\u00edvel, imediatamente antes que o bloco B se movesse. Para determinar a massa do bloco B, este foi suspenso verticalmente, com o fio A fixo no teto, conforme indicado na figura 2, e o aluno mediu a deforma\u00e7\u00e3o da mola igual a 10,0 cm, quando o sistema estava em equil\u00edbrio. Nas condi\u00e7\u00f5es descritas, desprezando a resist\u00eancia do ar, o coeficiente de atrito entre o bloco e a t\u00e1bua vale<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><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<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><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\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><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\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><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\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) 0,5. \u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>56-(UNICAMP-SP)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Acidentes de tr\u00e2nsito causam milhares de mortes todos os anos nas estradas do pa\u00eds. Pneus desgastados (\u201ccarecas\u201d), freios em p\u00e9ssimas condi\u00e7\u00f5es e excesso de velocidade s\u00e3o fatores que contribuem para elevar o n\u00famero de acidentes de tr\u00e2nsito.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O sistema de freios ABS (do alem\u00e3o \u201cAntiblockier-Bremssystem\u201d) impede o travamento das rodas do ve\u00edculo, de forma que elas<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_61992152.jpg\" alt=\"\" width=\"525\" height=\"163\" name=\"Imagem 75\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>n\u00e3o deslizem no ch\u00e3o, o que leva a um menor desgaste do pneu. N\u00e3o havendo deslizamento, a dist\u00e2ncia percorrida pelo ve\u00edculo at\u00e9 a parada completa \u00e9 reduzida, pois a for\u00e7a de atrito aplicada pelo ch\u00e3o nas rodas \u00e9 est\u00e1tica, e seu valor m\u00e1ximo \u00e9 sempre maior que a for\u00e7a de atrito cin\u00e9tico. O coeficiente de atrito est\u00e1tico entre os pneus e a pista \u00e9 \u03bc<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>= 0,80 e o cin\u00e9tico vale \u03bc<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 0,60. Sendo g = 10 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e a massa do carro m = 1200 kg, o m\u00f3dulo da for\u00e7a de atrito est\u00e1tico m\u00e1xima e a da for\u00e7a de atrito cin\u00e9tico s\u00e3o, respectivamente, iguais a<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 1200 N e 12000 N. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 12000 N e 120 N. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 20000 N e 15000 N. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) 9600 N e 7200 N. \u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>57-(UFAL-AL)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Uma crian\u00e7a tenta puxar a sua caixa de brinquedos, de peso P, exercendo uma for\u00e7a de tens\u00e3o numa corda ideal, de m\u00f3dulo F e dire\u00e7\u00e3o perfazendo um \u00e2ngulo \u03b8 com a horizontal (ver figura)<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_f78490f2.jpg\" alt=\"\" width=\"330\" height=\"177\" name=\"Imagem 76\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O coeficiente de atrito est\u00e1tico entre a caixa e o solo horizontal \u00e9 denotado por \u03bc<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>. Assinale a express\u00e3o para o m\u00e1ximo valor de<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>F de modo que a caixa ainda permane\u00e7a em repouso. (Para efeito de c\u00e1lculo, considere a caixa como uma part\u00edcula material.)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_c67078ba.jpg\" alt=\"\" width=\"767\" height=\"39\" name=\"Imagem 77\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>58-(UPE-PE)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Sejam os blocos P e Q de massas m e M, respectivamente, ilustrados na figura a seguir. O coeficiente de atrito est\u00e1tico entre os blocos \u00e9 \u03bc, entretanto n\u00e3o existe atrito entre o bloco Q e a superf\u00edcie A. Considere g a acelera\u00e7\u00e3o da gravidade.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A express\u00e3o que representa o menor valor do m\u00f3dulo da for\u00e7a horizontal F para que o bloco P n\u00e3o caia, \u00e9<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_5fda81da.jpg\" alt=\"\" width=\"356\" height=\"134\" name=\"Imagem 78\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_5ae1855.jpg\" alt=\"\" width=\"775\" height=\"67\" name=\"Imagem 79\" align=\"BOTTOM\" border=\"0\" \/><span style=\"color: #000000;\">\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>59-(UFPR-PR)<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Um esporte muito popular em paises do Hemisf\u00e9rio Norte \u00e9 o \u201ccurling\u201d, em que pedras<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_9af4808b.jpg\" alt=\"\" width=\"301\" height=\"134\" name=\"Imagem 80\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>de granito polido s\u00e3o lan\u00e7adas sobre uma pista horizontal de gelo. Esse esporte lembra o nosso popular jogo de bocha. Considere que um jogador tenha arremessado uma dessas pedras de modo que ela percorreu 45 m em linha reta antes de parar, sem a interven\u00e7\u00e3o de nenhum jogador. Considerando que a massa da pedra \u00e9 igual a 20 kg e o coeficiente de atrito entre o gelo e o granito \u00e9 de 0,02, assinale a alternativa que d\u00e1 a estimativa correta para o tempo que a pedra leva para parar.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Menos de 18 s.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Entre 18 s e 19 s.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) Entre 20 s e 22 s.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) Entre 23 s e 30 s.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) Mais de 30 s.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>60-(UEM-PR)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_88bf26e3.jpg\" alt=\"\" width=\"441\" height=\"135\" name=\"Imagem 81\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Supondo que um bloco de massa m kg esteja sobre uma superf\u00edcie plana e horizontal e que para<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_526dafef.jpg\" alt=\"\" width=\"314\" height=\"136\" name=\"Imagem 82\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>mover esse bloco uma for\u00e7a ligeiramente maior que\u00a0 X N \u00e9 necess\u00e1ria, assinale o que for correto.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>01) A for\u00e7a de atrito est\u00e1tico m\u00e1xima \u00e9 igual a X N.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>02) O coeficiente de atrito est\u00e1tico entre a superf\u00edcie e o bloco \u00e9 igual a X\/(mg), em que\u00a0 g \u00e9 a acelera\u00e7\u00e3o da gravidade, dada em metros por segundo ao quadrado.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>04) O coeficiente de atrito cin\u00e9tico entre a superf\u00edcie e o bloco \u00e9 maior que X\/(mg), em que g \u00e9 a acelera\u00e7\u00e3o da gravidade, dada em metros por segundo ao quadrado.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>08) No S.I., tanto os coeficientes de atrito cin\u00e9tico e est\u00e1tico s\u00e3o dados em newtons.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>16) A for\u00e7a de atrito est\u00e1tico \u00e9 sempre maior que a for\u00e7a de atrito cin\u00e9tico.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>61-(PUC-RS)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_9d1822e2.jpg\" alt=\"\" width=\"467\" height=\"131\" name=\"Imagem 83\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Freios com sistema antibloqueio (ABS) s\u00e3o eficientes em frenagens bruscas porque evitam que as<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_603c9c4f.jpg\" alt=\"\" width=\"276\" height=\"162\" name=\"Imagem 84\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>rodas sejam bloqueadas e que os pneus deslizem no pavimento. Essa efici\u00eancia decorre do fato de que a for\u00e7a de atrito que o pavimento exerce sobre as rodas \u00e9 m\u00e1xima quando<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A) os pneus est\u00e3o deslizando, porque o atrito cin\u00e9tico \u00e9 maior que o est\u00e1tico m\u00e1ximo.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B) os pneus est\u00e3o na imin\u00eancia de deslizar, porque o atrito est\u00e1tico m\u00e1ximo \u00e9 maior que o cin\u00e9tico.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>C) o carro est\u00e1 parado, porque o atrito est\u00e1tico \u00e9 sempre m\u00e1ximo nessa situa\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>D) a velocidade do carro \u00e9 constante, porque o atrito cin\u00e9tico \u00e9 constante.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>E) a velocidade do carro come\u00e7a a diminuir, porque nessa situa\u00e7\u00e3o o atrito cin\u00e9tico est\u00e1 aumentando.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>62-(UEPA-PA)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_21edb226.jpg\" alt=\"\" width=\"477\" height=\"117\" name=\"Imagem 85\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A faixa de pedestres \u00e9 uma conquista do cidad\u00e3o, a qual vem se consolidando na constru\u00e7\u00e3o de novas avenidas nas grandes cidades<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_daba3e76.jpg\" alt=\"\" width=\"254\" height=\"157\" name=\"Imagem 86\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>brasileiras. Um motorista trafegando em uma avenida a 54 km\/h observa um pedestre atravessando a faixa e aciona os freios, aplicando uma desacelera\u00e7\u00e3o constante no ve\u00edculo, o qual p\u00e1ra\u00a0 depois\u00a0 de\u00a0 5 s. Sabendo-se que o motorista conseguiu respeitar a faixa, afirma-se que o coeficiente de atrito entre os pneus e a estrada vale: (Dado: g = 10 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_3203fb70.png\" alt=\"\" width=\"775\" height=\"19\" name=\"Imagem 191\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>63-(MACKENZIE-SP)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_18bc96c4.jpg\" alt=\"\" width=\"454\" height=\"124\" name=\"Imagem 87\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um corpo de 5 kg est\u00e1 em movimento devido \u00e0 a\u00e7\u00e3o da for\u00e7a\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_6770f237.jpg\" alt=\"\" width=\"11\" height=\"18\" name=\"Imagem 88\" align=\"BOTTOM\" border=\"0\" \/>, de intensidade 50 N, como mostra a figura. \u00a0<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_6a47cc30.jpg\" alt=\"\" width=\"329\" height=\"185\" name=\"Imagem 89\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O coeficiente de atrito cin\u00e9tico entre a superf\u00edcie de apoio horizontal e o bloco \u00e9 0,6 e a acelera\u00e7\u00e3o da gravidade no local tem m\u00f3dulo igual a 10 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2.<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/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=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A acelera\u00e7\u00e3o com a qual o corpo est\u00e1 se deslocando tem intensidade<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 2,4 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2\u00a0<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><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<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 3,6 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><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\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 4,2m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) 5,6m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) 6,2m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>64-(UENP-PR)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_95fddd85.jpg\" alt=\"\" width=\"492\" height=\"129\" name=\"Imagem 90\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\">\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um bloco de massa M<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 10 kg, inicialmente em repouso em um plano horizontal, est\u00e1 ligado por um cabo ao bloco M<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 5 kg.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_73d8f487.jpg\" alt=\"\" width=\"259\" height=\"200\" name=\"Imagem 91\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Admitindo que o sistema esteja em equil\u00edbrio est\u00e1tico, assinale o valor do coeficiente de atrito entre a superf\u00edcie e o bloco. (g = 10 m\/s<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 0,18\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 0,27\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 0,50\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) 0,60\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><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) 0,754<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0<\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>65-(UFF-RJ)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_bfb3c6ff.jpg\" alt=\"\" width=\"536\" height=\"108\" name=\"Imagem 92\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00cdm\u00e3s s\u00e3o frequentemente utilizados para prender pequenos objetos\u00a0 em superf\u00edcies met\u00e1licas planas e verticais, como quadros de avisos e portas de geladeiras.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considere que um \u00edm\u00e3, colado a um grampo, esteja em contato com a porta de uma geladeira . Suponha que a for\u00e7a magn\u00e9tica que o \u00edm\u00e3 faz sobre a superf\u00edcie da geladeira \u00e9 perpendicular a ela e tem m\u00f3dulo F<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>M<\/b><\/span><\/span><\/sub><\/span><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>. O conjunto im\u00e3\/grampo tem massa m<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sub><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.O coeficiente de atrito est\u00e1tico entre a superf\u00edcie da geladeira e a do \u00edm\u00e3 \u00e9\u00a0 \u03bc<\/b><\/span><\/span><\/span><span style=\"color: #000000;\"><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e<\/b><\/span><\/span><\/sub><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.Uma massa M est\u00e1 pendurada no grampo por um fio de massa desprez\u00edvel, como mostra a figura.<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/fat\/i_d9c168372436545f_html_811254fe.jpg\" alt=\"\" width=\"214\" height=\"224\" name=\"Imagem 93\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Desenhe no diagrama as for\u00e7as que agem sobre o conjunto \u00edm\u00e3\/grampo (representado pelo ponto preto no<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>cruzamento dos eixos x e y na figura), identificando cada uma dessas for\u00e7as.<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Qual o maior valor da massa\u00a0 M\u00a0 que pode ser pendurada no grampo sem que o conjunto caia?<\/b><\/span><\/span><\/span><\/p>\n<h3><a title=\"Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre For\u00e7a de atrito\" href=\"http:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/forca-de-atrito\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-forca-de-atrito\/\"><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 For\u00e7a de atrito &nbsp; \u00a001-(UFB)\u00a0Considere um bloco de massa 10kg, inicialmente em repouso sobre uma superf\u00edcie reta e horizontal com atrito e cujos coeficientes de atrito est\u00e1tico e din\u00e2mico sejam respectivamente iguais a \u03bce = 0,5 e \u03bcd = 0,3. Aplica-se ao bloco uma for\u00e7a de intensidade crescente, a partir de zero. Analise<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1300,"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-1302","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1302","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=1302"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1302\/revisions"}],"predecessor-version":[{"id":10823,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1302\/revisions\/10823"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1300"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1302"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}