{"id":1372,"date":"2015-09-24T11:34:25","date_gmt":"2015-09-24T11:34:25","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1372"},"modified":"2024-08-23T13:00:35","modified_gmt":"2024-08-23T13:00:35","slug":"exercicios-de-vestibulares-com-resolucao-comentada-sobre-energia-mecanica","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/energia-mecanica\/exercicios-de-vestibulares-com-resolucao-comentada-sobre-energia-mecanica\/","title":{"rendered":"Energia Mec\u00e2nica &#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>Energia Mec\u00e2nica<\/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>01-(UFMG-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Rita est\u00e1 esquiando numa montanha dos Andes. A energia cin\u00e9tica dela em fun\u00e7\u00e3o do tempo, durante parte do trajeto, est\u00e1 representada neste gr\u00e1fico:<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_9f41db1.jpg\" alt=\"\" width=\"179\" height=\"147\" name=\"graphics19\" align=\"BOTTOM\" border=\"0\" \/><img decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_ffc75419.jpg\" alt=\"\" width=\"242\" height=\"147\" name=\"Imagem 2\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Os pontos Q e R, indicados nesse gr\u00e1fico, correspondem a dois instantes diferentes do movimento de Rita.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Despreze todas as formas de atrito.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Com base nessas informa\u00e7\u00f5es, \u00e9 CORRETO afirmar que Rita atinge<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) velocidade m\u00e1xima em Q e altura m\u00ednima em R.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) velocidade m\u00e1xima em R e altura m\u00e1xima em Q.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) velocidade m\u00e1xima em Q e altura m\u00e1xima em R.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) velocidade m\u00e1xima em R e altura m\u00ednima em Q.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>02-(Ufsm-RS)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0A figura a seguir, representa uma barragem com a canaliza\u00e7\u00e3o que leva a \u00e1gua \u00e0 turbina.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img fetchpriority=\"high\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_93221274.jpg\" alt=\"\" width=\"423\" height=\"143\" name=\"Imagem 3\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Se n\u00e3o existe perda de energia no escoamento e se o m\u00f3dulo da velocidade da \u00e1gua em P \u00e9 v, a energia dispon\u00edvel para girar a turbina, para uma quantidade de \u00e1gua de massa m, \u00e9:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>a) (1\/2) mv<\/b><\/span><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>2<\/b><\/span><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>\u00a0+ mgh\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>b) mgh\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>c) (1\/2) mv<\/b><\/span><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>2<\/b><\/span><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>\u00a0\u2013 mgh\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) (1\/2) mv<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) (1\/2) mv<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0+ mg(20m + h)<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>03-(PUC-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Determine a massa de um avi\u00e3o viajando a 720km\/h, a uma altura de 3.000 m do solo, cuja energia mec\u00e2nica total \u00e9 de 70,0.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>6<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0J<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_b92c60b4.jpg\" alt=\"\" width=\"292\" height=\"172\" name=\"Imagem 4\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considere a energia potencial gravitacional como zero no solo.(g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_bc2e06a1.png\" alt=\"\" width=\"775\" height=\"22\" name=\"Imagem 161\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>04-(PUC-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma pedra, deixada cair de um edif\u00edcio, leva 4s para atingir o solo.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_dfc33bfe.jpg\" alt=\"\" width=\"228\" height=\"179\" name=\"graphics20\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Desprezando a resist\u00eancia do ar e considerando g = 10 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, escolha a op\u00e7\u00e3o que indica a altura do edif\u00edcio em metros.<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_11e8b782.png\" alt=\"\" width=\"775\" height=\"16\" name=\"Imagem 162\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>05-(PUC-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um ciclista desce uma rua inclinada, com forte vento contr\u00e1rio ao seu movimento, com velocidade constante.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_7be90891.jpg\" alt=\"\" width=\"142\" height=\"120\" name=\"Imagem 6\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Pode-se afirmar que:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) sua energia cin\u00e9tica est\u00e1 aumentando.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) sua energia potencial gravitacional est\u00e1 diminuindo\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) sua energia cin\u00e9tica est\u00e1 diminuindo.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) sua energia potencial gravitacional \u00e9 constante.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>06-(PUC-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Os gatos conseguem sair ilesos de muitas quedas. Suponha que a maior velocidade que ele possa atingir o solo, sem se machucar, seja de 29 km\/h. Ent\u00e3o, desprezando-se a resist\u00eancia do ar e considerando g = 10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, a altura m\u00e1xima de queda para que um gato, partindo do repouso, nada sofra \u00e9, aproximadamente, de:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_6d39b531.png\" alt=\"\" width=\"774\" height=\"18\" name=\"Imagem 163\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>07- (Uffrj-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0O salto com vara \u00e9, sem d\u00favida, uma das disciplinas mais exigentes do atletismo. Em um \u00fanico salto, o atleta executa cerca de 23 movimentos em menos de 2 segundos. Na \u00faltima Olimp\u00edada de Atenas a atleta russa, Svetlana Feofanova, bateu o recorde feminino, saltando 4,88 m.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A figura a seguir representa um atleta durante um salto com vara, em tr\u00eas instantes distintos.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_7d7b93bd.jpg\" alt=\"\" width=\"412\" height=\"150\" name=\"graphics21\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Assinale a op\u00e7\u00e3o que melhor identifica os tipos de energia envolvidos em cada uma das situa\u00e7\u00f5es I, II, e III, respectivamente.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) &#8211; cin\u00e9tica\u00a0 &#8211; cin\u00e9tica e gravitacional\u00a0 &#8211; cin\u00e9tica e gravitacional<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) &#8211; cin\u00e9tica e el\u00e1stica &#8211; cin\u00e9tica, gravitacional e el\u00e1stica &#8211; cin\u00e9tica e gravitacional<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) &#8211; cin\u00e9tica &#8211; cin\u00e9tica, gravitacional e el\u00e1stica &#8211; cin\u00e9tica e gravitacional<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) &#8211; cin\u00e9tica e el\u00e1stica &#8211; cin\u00e9tica e el\u00e1stica &#8211; gravitacional<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) &#8211; cin\u00e9tica e el\u00e1stica &#8211; cin\u00e9tica e gravitacional \u2013 gravitacional<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>08-(Ufpe)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Com base na figura a seguir, calcule a menor velocidade com que o corpo deve passar pelo ponto A para ser capaz de atingir o ponto B. Despreze o atrito e considere g = 10 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_1925cd1b.jpg\" alt=\"\" width=\"327\" height=\"130\" name=\"Imagem 8\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>09-(PUC-RS)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um bloco de 4,0 kg de massa, e velocidade de 10m\/s, movendo-se sobre um plano horizontal, choca-se contra uma mola, como mostra a figura<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_bbda5558.jpg\" alt=\"\" width=\"372\" height=\"87\" name=\"Imagem 9\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Sendo a constante el\u00e1stica da mola igual a 10000N\/m, o valor da deforma\u00e7\u00e3o m\u00e1xima que a mola poderia atingir, em cm, \u00e9<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_2eda1864.png\" alt=\"\" width=\"774\" height=\"18\" name=\"Imagem 164\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>10-(UNICAMP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um brinquedo que muito agrada \u00e0s crian\u00e7as s\u00e3o os lan\u00e7adores de objetos em uma pista. Considere que a mola da figura a seguir possui uma constante el\u00e1stica k = 8000 N\/m e massa desprez\u00edvel. Inicialmente, a mola est\u00e1 comprimida de 2,0 cm e, ao ser liberada, empurra um carrinho de massa igual a 0,20 kg. O carrinho abandona a mola quando esta atinge o seu comprimento relaxado, e percorre uma pista que termina em uma rampa. Considere que n\u00e3o h\u00e1 perda de energia mec\u00e2nica por atrito no movimento do carrinho.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_d8f8925d.jpg\" alt=\"\" width=\"410\" height=\"113\" name=\"Imagem 10\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Qual \u00e9 a velocidade do carrinho quando ele abandona a mola?<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Na subida da rampa, a que altura o carrinho tem velocidade de 2,0 m\/s?<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>11-(UFMG-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Daniel e Andr\u00e9, seu irm\u00e3o, est\u00e3o parados em um tobog\u00e3, nas posi\u00e7\u00f5es mostradas nesta figura:<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_22d7e30.jpg\" alt=\"\" width=\"429\" height=\"143\" name=\"Imagem 11\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Daniel tem o dobro do peso de Andr\u00e9 e a altura em que ele est\u00e1, em rela\u00e7\u00e3o ao solo, corresponde \u00e0 metade da altura em que est\u00e1 seu irm\u00e3o. Em um certo instante, os dois come\u00e7am a escorregar pelo tobog\u00e3. Despreze as for\u00e7as de atrito.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00c9 CORRETO afirmar que, nessa situa\u00e7\u00e3o, ao atingirem o n\u00edvel do solo, Andr\u00e9 e Daniel ter\u00e3o<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) energias cin\u00e9ticas diferentes e m\u00f3dulos de velocidade diferentes.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) energias cin\u00e9ticas iguais e m\u00f3dulos de velocidade iguais.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) energias cin\u00e9ticas diferentes e m\u00f3dulos de velocidade iguais.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) energias cin\u00e9ticas iguais e m\u00f3dulos de velocidade diferentes.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>12-(Ufpe-PE)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma bolinha de massa m = 200 g \u00e9 largada do repouso de uma altura h, acima de uma mola ideal, de constante el\u00e1stica k = 1240 N\/m, que est\u00e1 fixada no piso (ver figura).<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_bbabb74c.jpg\" alt=\"\" width=\"182\" height=\"157\" name=\"Imagem 12\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Ela colide com a mola comprimindo-a por\u00a0\u2206x = 10 cm. Calcule, em metros, a altura inicial h. Despreze a resist\u00eancia do ar.(g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>13-(MACKENZIE-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0A figura mostra o instante em que uma esfera de 4 kg \u00e9 abandonada do repouso, da posi\u00e7\u00e3o P, e cai sobre a mola ideal de constante el\u00e1stica 2.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0N\/m. O maior valor da velocidade atingida por essa esfera, no seu movimento descendente, \u00e9<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_dfe1e8c4.jpg\" alt=\"\" width=\"296\" height=\"166\" name=\"Imagem 13\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_6d6e129a.png\" alt=\"\" width=\"774\" height=\"18\" name=\"Imagem 165\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>14-(Ufg)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um bloco de massa igual a 0,5 kg \u00e9 abandonado, em repouso, 2 m acima de uma mola vertical de comprimento 0,8 m e constante el\u00e1stica igual a 100 N\/m, conforme o diagrama.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_abf2a944.jpg\" alt=\"\" width=\"168\" height=\"166\" name=\"Imagem 14\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Calcule o menor comprimento que a mola atingir\u00e1. Considere g = 10 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>15-(Ufpb)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um bloco de 1 kg, preso a uma mola de constante el\u00e1stica 800 N\/m e massa desprez\u00edvel, oscila sobre um plano horizontal sem atrito com amplitude A = 0,5 m. No instante em que a energia cin\u00e9tica do bloco se iguala \u00e0 energia potencial da mola, a velocidade do bloco vale:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_349629b4.png\" alt=\"\" width=\"775\" height=\"18\" name=\"Imagem 166\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>16-(Ufpe-PE)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma bolinha presa a um fio de comprimento L = 1,6 m que est\u00e1 fixado no teto, \u00e9 liberada na posi\u00e7\u00e3o indicada na figura (ponto A). Ao passar pela posi\u00e7\u00e3o vertical, o fio encontra um pino horizontal fixado a uma dist\u00e2ncia h = 1,25 m (ver figura).<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_6d4d72ce.jpg\" alt=\"\" width=\"277\" height=\"147\" name=\"Imagem 15\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>17-(UNIRIO-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um carrinho de massa m=2,0kg apresentado no desenho ao lado, desliza sobre um plano horizontal com velocidade de 10m\/s. No ponto A, a superf\u00edcie passa a ser curva, com raio de curvatura de 2,0m.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_768b0051.jpg\" alt=\"\" width=\"339\" height=\"166\" name=\"Imagem 16\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Suponha que o atrito seja desprez\u00edvel ao longo de toda a trajet\u00f3ria\u00a0 e que g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>. Determine, ent\u00e3o:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) a acelera\u00e7\u00e3o centr\u00edpeta no ponto B;<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a rea\u00e7\u00e3o da superf\u00edcie curva sobre o bloco no ponto C.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>18-(UFRS-RS)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Na figura, representamos uma pista em que o trecho final XYZD \u00e9 um arco de circunfer\u00eancia. Larga-se o carrinho de massa 0,2kg no topo da pista. Despreze os atritos, considere g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e determine:<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_5307644a.jpg\" alt=\"\" width=\"397\" height=\"203\" name=\"Imagem 17\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) a energia cin\u00e9tica no ponto X\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) o trabalho realizado para ir de X a Y\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) a velocidade m\u00ednima com que o carrinho deve ter para passar pelo ponto Z, sem perder contato com a pista.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>19-(UnB-DF)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Em uma apresenta\u00e7\u00e3o de circo em 1901, Allo Diavolo introduziu a acrobacia de bicicletas em pistas com loops,<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_70bafb31.jpg\" alt=\"\" width=\"335\" height=\"151\" name=\"Imagem 18\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>como mostra a figura. Diavolo observou que se ele partisse com velocidade zero de uma altura m\u00ednima, poderia percorrer todo o trajeto, passando inclusive pelo loop, sem cair, em um \u201cdesafio\u201d \u00e1s leis da gravidade, conforme anunciava ele. A figurta mostra o caminho do centro de massa\u00a0 do sistema acrobata-bicicleta. Nessa figura,<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0h\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00e9 a altura entre o ponto mais alto \u2013\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u2013 e o ponto mais baixo \u2013\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>C<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u2013 da trajet\u00f3ria,<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0B<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00e9 o ponto mais alto do loop e\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>R<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00e9 o raio do loop.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A partir dessas informa\u00e7\u00f5es e considerando que\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>m<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u00e9 a massa do sistema acrobata-bicicleta, que g \u00e9 a acelera\u00e7\u00e3o da gravidade, que n\u00e3o h\u00e1 for\u00e7as dissipativas, que a bicicleta n\u00e3o \u00e9 impulsionada pelo acrobata em nenhum instante da trajet\u00f3ria e que apenas o movimento do centro de massa do sistema acrobata-bicicleta \u00e9 analisado, julgue os itens abaixo.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1- No ponto<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0C<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0do caminho, mostrado na figura, a energia cin\u00e9tica \u00e9 igual a mgh.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2- A energia mec\u00e2nica total do sistema acrobata-bicicleta ser\u00e1 mgh mesmo no caso da exist\u00eancia de for\u00e7as dissipativas.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>3- Para que o sistema acrobata-bicicleta passe pelo ponto mais alto do loop sem perder contato com a pista, o sistema dever\u00e1 ter nesse ponto uma velocidade de m\u00f3dulo superior ou igual a\u00a0\u00d6Rg.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>4- A raz\u00e3o entre os m\u00f3dulos das velocidades nos pontos\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>C<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0independe da altura<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0h.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>20- (Ufam)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma bolinha de massa m \u00e9 abandonada do ponto A de um trilho, a uma altura H do solo, e descreve a trajet\u00f3ria ABCD indicada na figura abaixo.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_1a787a3f.jpg\" alt=\"\" width=\"548\" height=\"153\" name=\"Imagem 19\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A bolinha passa pelo ponto mais elevado da trajet\u00f3ria parab\u00f3lica BCD, a uma altura h do solo, com velocidade cujo m\u00f3dulo vale V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>C<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=10m\/s, e atinge o solo no ponto D com velocidade de m\u00f3dulo igual a V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>D<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=20m\/s. Podemos afirmar que as alturas referidas no texto valem: (g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) H=19m; h=14m\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) H=18m; h=10m\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) H=12m; h=4m\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) H=12m; h=15m\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) H=20m; h=15m<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>21-(UFJF-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um tren\u00f3, com um esquim\u00f3, come\u00e7a a descer por uma rampa de gelo, partindo do repouso no ponto P, \u00e0 altura de 20m. Depois de passar pelo ponto Q, atinge uma barreira de prote\u00e7\u00e3o em R, conforme a figura.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_176006b1.jpg\" alt=\"\" width=\"418\" height=\"142\" name=\"Imagem 20\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O conjunto tren\u00f3-esquim\u00f3 possui uma massa total de 90kg. O trecho QR encontra-se na horizontal. Despreze as dimens\u00f5es do conjunto, o atrito e a resist\u00eancia do ar durante o movimento.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Usando o princ\u00edpio da conserva\u00e7\u00e3o da energia mec\u00e2nica, calcule\u00a0 a velocidade com que o conjunto chega ao ponto Q na base da rampa.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Em R encontra-se uma barreira de prote\u00e7\u00e3o feita de material deform\u00e1vel usada para parar o conjunto ap\u00f3s a descida.\u00a0 Considere que, durante o choque, a\u00a0 barreira n\u00e3o se desloca e que o conjunto se choca contra esta e p\u00e1ra. \u00a0Sabendo que a barreira de prote\u00e7\u00e3o sofreu uma deforma\u00e7\u00e3o de 1,5m durante o choque, calcule a for\u00e7a m\u00e9dia exercida por ela sobre o conjunto.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>22-(FUVEST-SP)<\/b><\/span><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_bdf8579c.jpg\" alt=\"\" width=\"436\" height=\"168\" name=\"Imagem 21\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>No \u201dsalto com vara\u201d, um atleta corre segurando uma vara e, com per\u00edcia e treino, consegue projetar seu corpo por cima de uma barra. Para uma estimativa da altura alcan\u00e7ada nesses saltos, \u00e9 poss\u00edvel considerar que a vara sirva apenas para converter o movimento horizontal do atleta (corrida) em movimento vertical, sem perdas ou acr\u00e9scimos de energia. Na an\u00e1lise de um desses saltos, foi obtida a seq\u00fc\u00eancia de imagens reproduzida acima.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Nesse caso, \u00e9 poss\u00edvel estimar que a velocidade m\u00e1xima atingida pelo atleta, antes do salto, foi de, aproximadamente,<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>*Desconsidere os efeitos do trabalho muscular ap\u00f3s o in\u00edcio do salto.<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_b5c2e95e.png\" alt=\"\" width=\"774\" height=\"20\" name=\"Imagem 168\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>23-(UEL-PR)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma esfera de massa\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><i><b>m\u00a0<\/b><\/i><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>desliza, com atrito desprez\u00edvel, ao longo de um trilho em la\u00e7o, conforme a figura\u00a0 abaixo.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_24cc50d1.gif\" alt=\"\" width=\"284\" height=\"213\" name=\"Imagem 22\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A esfera parte do repouso no ponto\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><i><b>y\u00a0<\/b><\/i><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>= 4<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><i><b>R\u00a0<\/b><\/i><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>acima do n\u00edvel da parte mais baixa do trilho. Calcule os valore da velocidade da esfera (\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><i><b>v<\/b><\/i><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><i><b>X<\/b><\/i><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><i><b>\u00a0<\/b><\/i><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>) e da for\u00e7a normal (\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><i><b>f<\/b><\/i><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><i><b>N<\/b><\/i><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><i><b>\u00a0<\/b><\/i><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>) exercida sobre a esfera, no ponto\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><i><b>x\u00a0<\/b><\/i><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(ponto mais alto da trajet\u00f3ria circular):<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>24-(UFSCAR-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma formiga de massa\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>m<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0encontra-se no topo de uma bola de bilhar rigidamente presa ao solo.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_e6a3531d.jpg\" alt=\"\" width=\"240\" height=\"225\" name=\"Imagem 23\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A bola possui raio\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>R<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e superf\u00edcie altamente polida. Considere<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0g<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0a acelera\u00e7\u00e3o da gravidade e despreze os poss\u00edveis efeitos dissipativos. A formiga come\u00e7a a deslizar na bola com velocidade inicial nula.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a)\u00a0 Calcule o m\u00f3dulo da velocidade da formiga no ponto em que ela perde contato com a bola.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Calcule a altura do solo, em que a formiga perde contato com a bola.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>25-(ITA-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um pequeno bloco, solto com velocidade nula a uma altura h, move-se sob o efeito da gravidade e sem atrito sobre um trilho em forma de dois quartos de c\u00edrculo de raio R, que se tangenciam, como mostra a figura.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_10a062d5.jpg\" alt=\"\" width=\"297\" height=\"170\" name=\"Imagem 24\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A m\u00ednima altura inicial h que acarreta a sa\u00edda do bloco do trilho, ap\u00f3s o ponto A \u00e9:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_44515c3c.png\" alt=\"\" width=\"775\" height=\"18\" name=\"Imagem 169\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>26-(UNICAMP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um carrinho de massa 300kg percorre uma montanha russa\u00a0 cujo trecho BCD \u00e9 um arco de circunfer\u00eancia de raio R=5,4m, conforme a figura.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_39ab74fa.jpg\" alt=\"\" width=\"338\" height=\"160\" name=\"Imagem 25\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A velocidade do carrinho no ponto A \u00e9 V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=12m\/s. Considerando g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e desprezando o atrito, calcule:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) a velocidade do carrinho no ponto C,<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a acelera\u00e7\u00e3o do carrinho no ponto C,<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) a for\u00e7a feita pelos trilhos sobre o carrinho no ponto C.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>27-(FUVEST-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Para testar a elasticidade de uma bola de basquete, ela \u00e9 solta, a partir de uma altura H<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, em um equipamento no qual seu movimento \u00e9 monitorado por um sensor. Esse equipamento registra a altura do centro de massa da bola, a cada instante,<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>acompanhando seus sucessivos choques com o ch\u00e3o. A partir da an\u00e1lise dos registros, \u00e9 poss\u00edvel, ent\u00e3o, estimar a elasticidade da bola, caracterizada pelo coeficiente de restitui\u00e7\u00e3o C<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>R<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_2401e81c.jpg\" alt=\"\" width=\"138\" height=\"188\" name=\"Imagem 26\" align=\"BOTTOM\" border=\"0\" \/><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_cbc8601a.jpg\" alt=\"\" width=\"195\" height=\"187\" name=\"Imagem 27\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O gr\u00e1fico acima apresenta os registros de alturas, em fun\u00e7\u00e3o do tempo, para uma bola de massa M = 0,60kg, quando ela \u00e9 solta e inicia o movimento com seu centro de massa a uma altura H<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 1,6m, chocando-se sucessivas vezes com o ch\u00e3o. A partir dessas informa\u00e7\u00f5es:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Represente, no Gr\u00e1fico I da folha de respostas, a energia potencial da bola, E<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>P<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, em joules, em fun\u00e7\u00e3o do tempo, indicando os valores na escala.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Represente, no Gr\u00e1fico II da folha de respostas, a energia mec\u00e2nica total da bola, ET, em joules, em fun\u00e7\u00e3o do tempo, indicando os valores na escala.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) Estime o coeficiente de restitui\u00e7\u00e3o C<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>R<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0dessa bola, utilizando a defini\u00e7\u00e3o apresentada abaixo.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>NOTE E ADOTE:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>&#8211; Desconsidere a deforma\u00e7\u00e3o da bola e a resist\u00eancia do ar.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>&#8211; O coeficiente de restitui\u00e7\u00e3o, C<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>R<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>R<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\/V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>I<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, \u00e9 a raz\u00e3o entre a velocidade com que a bola \u00e9 rebatida pelo ch\u00e3o (V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>R<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>) e a velocidade com que atinge o ch\u00e3o (V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>I<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>), em cada choque. Esse coeficiente \u00e9 aproximadamente constante nas v\u00e1rias colis\u00f5es.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>28-(UFSCAR-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Id\u00e9ia para a campanha de redu\u00e7\u00e3o de acidentes: enquanto um narrador exporia fatores de riscos nas estradas, uma c\u00e2mera mostraria o trajeto de um sabonete que, a partir do repouso de um ponto sobre a borda de uma banheira, escorregaria para o interior da mesma, sofrendo um forte impacto contra a parede vertical oposta.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_b74f9432.jpg\" alt=\"\" width=\"442\" height=\"101\" name=\"Imagem 28\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Para a realiza\u00e7\u00e3o da filmagem, a equipe t\u00e9cnica, conhecendo a acelera\u00e7\u00e3o da gravidade (10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>) e desconsiderando qualquer atua\u00e7\u00e3o de for\u00e7as contr\u00e1rias ao movimento, estimou que a velocidade do sabonete, momentos antes de seu impacto contra a parede da banheira, deveria ser um valor, em m\/s, mais pr\u00f3ximo de:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_439703ee.png\" alt=\"\" width=\"774\" height=\"18\" name=\"Imagem 170\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>29-(CESGRANRIO-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma esfera de massa 0,10kg rola sobre o perfil da montanha russa mostrado na figura abaixo.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_33d50a6c.jpg\" alt=\"\" width=\"460\" height=\"159\" name=\"Imagem 29\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>No instante representado, ela se move para baixo (veja seta) com energia cin\u00e9tica igual a 0,10J. Embora o atrito seja muito pequeno, a bola acabar\u00e1 parando na posi\u00e7\u00e3o: (g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_a4ccc617.png\" alt=\"\" width=\"774\" height=\"17\" name=\"Imagem 171\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>30-(Ufla-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um parque aqu\u00e1tico tem um tobo\u00e1gua, conforme a figura.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_d386f265.jpg\" alt=\"\" width=\"567\" height=\"232\" name=\"Imagem 30\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um indiv\u00edduo de 60kg desliza pelo tobo\u00e1gua a partir do ponto A, sendo lan\u00e7ado numa piscina de uma altura de 0,8m, ponto B, numa dire\u00e7\u00e3o que faz um \u00e2ngulo de 30<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0com a horizontal.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considerando o atrito desprez\u00edvel, g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e c\u00f3s30<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=\u00d63\/2, calcule:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) a velocidade do indiv\u00edduo ao deixar o tobo\u00e1gua no ponto B<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a energia cin\u00e9tica do indiv\u00edduo no ponto mais alto da trajet\u00f3ria, ponto C.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) a altura do ponto C, h<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>m\u00e1x<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>31- (ITA-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um aro de1 kg de massa encontra-se preso a uma mola de massa desprez\u00edvel, constante el\u00e1stica k = 10 N\/m e comprimento inicial L = 1 m quando n\u00e3o distendida, afixada no ponto O. A figura mostra o aro numa posi\u00e7\u00e3o P em uma barra horizontal fixa ao longo da qual o aro pode deslizar sem atrito.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_bc0e48c7.jpg\" alt=\"\" width=\"298\" height=\"124\" name=\"Imagem 31\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Soltando o aro do ponto P, qual deve ser sua velocidade, em m\/s, ao alcan\u00e7ar o ponto T, a 2 m de dist\u00e2ncia?<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>32- (UNESP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0No esporte conhecido como &#8220;ioi\u00f4 humano&#8221;, o praticante, preso \u00e0 extremidade de uma corda el\u00e1stica, cai da beira de uma plataforma para as \u00e1guas de um rio. Sua queda \u00e9 interrompida, a poucos metros da superf\u00edcie da \u00e1gua, pela a\u00e7\u00e3o da corda el\u00e1stica, que tem a outra extremidade firmemente presa \u00e0 beira da plataforma.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><a name=\"0.1_graphic02\"><\/a><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_2523c519.jpg\" alt=\"\" width=\"352\" height=\"120\" name=\"Imagem 32\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Suponha que, nas condi\u00e7\u00f5es citadas acima, a distens\u00e3o m\u00e1xima sofrida pela corda, quando usado por um atleta de peso 750 N, \u00e9 de 10 metros, e que seu comprimento, quando n\u00e3o distendida, \u00e9 de 30 metros. Nestas condi\u00e7\u00f5es:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) A que dist\u00e2ncia da plataforma est\u00e1 o atleta, quando chega ao ponto mais pr\u00f3ximo da \u00e1gua?<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Qual o valor da constante el\u00e1stica da corda?<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(Despreze o atrito com o ar e a massa da corda, e considere igual a zero o valor da velocidade do atleta no in\u00edcio da queda.)<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p align=\"CENTER\"><span style=\"color: #0000cc;\"><span style=\"font-family: 'Arial Black', serif;\"><span style=\"font-size: large;\"><b>Sistemas dissipativos<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>33-(UFSCAR-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0O trabalho realizado por uma for\u00e7a conservativa independe da trajet\u00f3ria, o que n\u00e3o acontece com as for\u00e7as dissipativas, cujo trabalho realizado depende da trajet\u00f3ria. S\u00e3o bons exemplos de for\u00e7as conservativas e dissipativas, respectivamente,<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) peso e massa.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) peso e resist\u00eancia do ar.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) for\u00e7a de contato e for\u00e7a normal.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) for\u00e7a el\u00e1stica e for\u00e7a centr\u00edpeta.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) for\u00e7a centr\u00edpeta e for\u00e7a centr\u00edfuga.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>34-(UNIFESP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Na figura est\u00e3o representadas duas situa\u00e7\u00f5es f\u00edsicas cujo objetivo \u00e9 ilustrar o conceito de trabalho de for\u00e7as conservativas e dissipativas.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_a5740cda.jpg\" alt=\"\" width=\"486\" height=\"86\" name=\"Imagem 33\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Em I, o bloco \u00e9 arrastado pela for\u00e7a \u00f9 sobre o plano horizontal; por causa do atrito, quando a for\u00e7a\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_ab28f08e.jpg\" alt=\"\" width=\"12\" height=\"20\" name=\"Imagem 34\" align=\"BOTTOM\" border=\"0\" \/>\u00a0cessa o bloco p\u00e1ra. Em II, o bloco, preso \u00e0 mola e em repouso no ponto O, \u00e9 puxado pela for\u00e7a\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_ab28f08e.jpg\" alt=\"\" width=\"12\" height=\"20\" name=\"Imagem 35\" align=\"BOTTOM\" border=\"0\" \/>\u00a0sobre o plano horizontal, sem que sobre ele atue nenhuma for\u00e7a de resist\u00eancia; depois de um pequeno deslocamento, a for\u00e7a cessa e o bloco volta, puxado pela mola, e passa a oscilar em torno do ponto O.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Essas figuras ilustram:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) I: exemplo de trabalho de for\u00e7a dissipativa (for\u00e7a de atrito), para o qual a energia mec\u00e2nica n\u00e3o se conserva;<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>II: exemplo de trabalho de for\u00e7a conservativa (for\u00e7a el\u00e1stica), para o qual a energia mec\u00e2nica se conserva.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) I: exemplo de trabalho de for\u00e7a dissipativa (for\u00e7a de atrito), para o qual a energia mec\u00e2nica se conserva;<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>II: exemplo de trabalho de for\u00e7a conservativa (for\u00e7a el\u00e1stica), para o qual a energia mec\u00e2nica n\u00e3o se conserva.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) I: exemplo de trabalho de for\u00e7a conservativa (for\u00e7a de atrito), para o qual a energia mec\u00e2nica n\u00e3o se conserva;<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>II: exemplo de trabalho de for\u00e7a dissipativa (for\u00e7a el\u00e1stica), para o qual a energia mec\u00e2nica se conserva.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) I: exemplo de trabalho de for\u00e7a conservativa (for\u00e7a de atrito), para o qual a energia mec\u00e2nica se conserva;<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>II: exemplo de trabalho de for\u00e7a dissipativa (for\u00e7a el\u00e1stica), para o qual a energia mec\u00e2nica n\u00e3o se conserva.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) I: exemplo de trabalho de for\u00e7a dissipativa (for\u00e7a de atrito);<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>II: exemplo de trabalho de for\u00e7a conservativa (for\u00e7a el\u00e1stica), mas em ambos a energia mec\u00e2nica se conserva.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>35-(FGV-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Devido a for\u00e7as dissipativas, parte da energia mec\u00e2nica (E) de um sistema foi convertida em calor, circunst\u00e2ncia caracterizada pelo gr\u00e1fico apresentado<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_81c59c83.jpg\" alt=\"\" width=\"378\" height=\"173\" name=\"Imagem 36\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Sabendo-se que a varia\u00e7\u00e3o da energia potencial desse sistema foi nula, o trabalho realizado sobre o sistema nos primeiros 4 segundos, em J, foi, em m\u00f3dulo,<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_9f07bba9.png\" alt=\"\" width=\"774\" height=\"18\" name=\"Imagem 172\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>36-(FGV-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Ao passar pelo ponto A, a uma altura de 3,5m do n\u00edvel de refer\u00eancia B, uma esfera de massa 2kg, que havia sido abandonada de um ponto mais alto que A, possui velocidade de 2m\/s. A esfera passa por B e, em C, a 3,0m do mesmo n\u00edvel de refer\u00eancia, sua velocidade torna-se zero.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_4c682bf3.jpg\" alt=\"\" width=\"313\" height=\"158\" name=\"Imagem 37\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A parcela de energia dissipada por a\u00e7\u00f5es resistentes sobre a esfera \u00e9, em J.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dados: g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_1b08e90e.png\" alt=\"\" width=\"774\" height=\"18\" name=\"Imagem 173\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>37-(Uece)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Na presen\u00e7a da atmosfera terrestre, um proj\u00e9til, lan\u00e7ado verticalmente para cima, perde parte de sua energia devido a for\u00e7as viscosas com o ar. Tal perda pode ser minimizada tornando o proj\u00e9til mais aerodin\u00e2mico. Caso fosse poss\u00edvel eliminar uma perda de 40 kJ neste processo, devido a essas melhorias aerodin\u00e2micas, de quanto aumentaria, aproximadamente, a altura m\u00e1xima atingida por um proj\u00e9til de 10 kg lan\u00e7ado verticalmente para cima?<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Admita que a acelera\u00e7\u00e3o da gravidade n\u00e3o varie e que seja igual a 10 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>.<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_45737b2e.png\" alt=\"\" width=\"775\" height=\"19\" name=\"Imagem 174\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>38-(UFMG-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Observe o perfil de uma montanha russa representado nesta figura:<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_651c9ccd.jpg\" alt=\"\" width=\"404\" height=\"114\" name=\"Imagem 38\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um carrinho \u00e9 solto do ponto M, passa pelos pontos N e P e s\u00f3 consegue chegar at\u00e9 o ponto Q. Suponha que a superf\u00edcie dos trilhos apresenta as mesmas caracter\u00edsticas em toda a sua extens\u00e3o. Sejam E(cn) e E(cp) as energias cin\u00e9ticas do carrinho, respectivamente, nos pontos N e P e E(tp) e E(tq) as energias mec\u00e2nicas totais do carrinho, tamb\u00e9m respectivamente, nos pontos P e Q. Considerando-se essas informa\u00e7\u00f5es, \u00e9 CORRETO afirmar que<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) E(cn) = E(cp) e E(tp) = E(tq).\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) E(cn) = E(cp) e E(tp) &gt; E(tq).\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) E(cn) &gt; E(cp) e E(tp) = E(tq).\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) E(cn) &gt; E(cp) e E(tp) &gt; E(tq).<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>39-(UEA-AM)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Na situa\u00e7\u00e3o descrita a seguir, uma esfera de massa 4,0kg \u00e9 abandonada do repouso da altura de 8,0m.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_f23487d3.jpg\" alt=\"\" width=\"230\" height=\"185\" name=\"Imagem 39\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Ela percorre a rampa passando pelo ponto horizontal com velocidade de 10m\/s. (g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>) Qual a porcentagem da energia dissipada por atrito entre os pontos A e C?<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_147ca20a.png\" alt=\"\" width=\"774\" height=\"17\" name=\"Imagem 175\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>40-(PUC-SP<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0O carrinho da figura tem massa 100g e encontra-se encostado em uma mola de constante el\u00e1stica 100N\/m, comprimida de 10cm.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_9db7d97b.jpg\" alt=\"\" width=\"392\" height=\"106\" name=\"Imagem 40\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Ao ser liberado, o carrinho sobe a rampa at\u00e9 uma altura m\u00e1xima de 30cm. O m\u00f3dulo da quantidade de energia mec\u00e2nica dissipada no processo, em joules, \u00e9:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_a897b4f0.png\" alt=\"\" width=\"774\" height=\"18\" name=\"Imagem 176\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>41-(UNESP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Um carrinho de 2,0kg, que disp\u00f5e de um gancho, movimenta-se sobre um plano horizontal, com velocidade de 1,0m\/s, em dire\u00e7\u00e3o \u00e0 argola presa na extremidade do fio mostrado na figura. A outra extremidade do fio est\u00e1 presa a um bloco, de peso 5,0N, que se encontra em repouso sobre uma prateleira.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Enganchando-se na argola, o carrinho puxa o fio e eleva o bloco, parando momentaneamente quando o bloco atinge a altura m\u00e1xima h acima da prateleira.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_bdee67f7.jpg\" alt=\"\" width=\"271\" height=\"187\" name=\"graphics22\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Nessas condi\u00e7\u00f5es, determine: (g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>).<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) a energia cin\u00e9tica inicial do carrinho<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a altura h, supondo que ocorra perda de 20% da energia cin\u00e9tica inicial do carrinho quando o gancho se prende na argola.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Observa\u00e7\u00f5es: Despreze quaisquer atritos e as massas das polias.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>42-(PUC-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma crian\u00e7a de massa 25 kg, inicialmente no ponto A, distante 2,4 m do solo, percorre, a partir do repouso, o escorregador esquematizado na figura. O escorregador pode ser considerado um plano inclinado cujo \u00e2ngulo com a horizontal \u00e9 de 37\u00b0. Supondo o coeficiente de atrito cin\u00e9tico entre a roupa da crian\u00e7a e o escorregador igual a 0,5, a velocidade com que a crian\u00e7a chega \u00e0 base do escorregador (ponto B) \u00e9, em m\/s, (g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>)<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_36f6799f.jpg\" alt=\"\" width=\"321\" height=\"138\" name=\"Imagem 42\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dados: sen 37\u00b0 = 0,6; cos 37\u00b0 = 0,8; tg 37\u00b0 = 0,75<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_9042ee97.png\" alt=\"\" width=\"775\" height=\"18\" name=\"Imagem 177\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>43-(FUVEST-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma pista \u00e9 formada por duas rampas inclinadas, A e B, e por uma regi\u00e3o horizontal de comprimento L. Soltando-se, na rampa A, de uma altura H\u00c1, um bloco de massa m, verifica-se que ele atinge uma altura H<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0(figura), em experimento realizado na Terra.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_5f2ebe55.jpg\" alt=\"\" width=\"485\" height=\"137\" name=\"graphics23\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O coeficiente de atrito cin\u00e9tico entre o bloco e a pista \u00e9 nulo nas rampas e igual a\u00a0m\u00a0na regi\u00e3o horizontal. Suponha que esse mesmo experimento seja realizado em Marte, onde a acelera\u00e7\u00e3o da gravidade \u00e9 g<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>M<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=g\/3, e considere que o bloco seja solto na mesma rampa A e da mesma altura H<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>. Determine:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) a raz\u00e3o R<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>xTerra<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\/V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>xMarte<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0entre as velocidades do bloco no final da rampa A (ponto x), em cada uma das experi\u00eancias (Terra e Marte)<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a raz\u00e3o R<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=W<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Terra<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\/W<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Marte<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, entre as energias mec\u00e2nicas dissipadas pela for\u00e7a de atrito na regi\u00e3o horizontal, em cada uma das experi\u00eancias (Terra e Marte).<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>44-(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=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A partir do repouso, um carrinho de montanha russa desliza de uma altura H=20\u00d63m sobre uma rampa de 60<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0de inclina\u00e7\u00e3o\u00a0 e corre 2om num trecho horizontal antes de chegar em um loop circular, de pista sem atrito.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_724c6e01.jpg\" alt=\"\" width=\"387\" height=\"162\" name=\"graphics24\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Sabendo que o coeficiente de atrito da rampa e do plano horizontal \u00e9 0,5, assinale o valor do raio m\u00e1ximo que pode ter esse loop para que o carrinho fa\u00e7a todo o percurso sem perder contato com a pista.\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) R=8\u221a3m\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) R=4(\u221a3 \u2013 1)m\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) R=8(\u221a3 \u2013 1)m\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) R=4(2\u221a3 \u2013 1)m\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) R=40(\u221a3 \u2013 1)\/3m<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>45-(UF-RR)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0Uma bola de borracha, de massa igual a 1,0kg, cai de uma altura de 2,0m, em rela\u00e7\u00e3o ao solo, com velocidade inicial nula. Ao tocar o solo, a bola transfere para este 12J, na forma de calor e, e volta a subir verticalmente.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considere a acelera\u00e7\u00e3o da gravidade g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>. A altura, em cm, atingida pela bola na subida \u00e9 de:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_56799076.png\" alt=\"\" width=\"774\" height=\"18\" name=\"Imagem 178\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>46-(UFOP-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Um jogador de basquete treina com uma bola cuja massa \u00e9 de 2 kg. A bola \u00e9<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_d4cf9e7b.jpg\" alt=\"\" width=\"153\" height=\"193\" name=\"graphics25\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>abandonada a 1 m de altura e,ao chocar-se com o solo, perde 50 % de sua energia. Usando g= 10 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, calcule:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) a energia cin\u00e9tica da bola imediatamente ap\u00f3s o primeiro choque;<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a velocidade da bola ao atingir o solo pela segunda vez;<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) depois de qual choque a bola ir\u00e1 adquirir a energia aproximada de 0,08 J.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>47-(UFF-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Dois brinquedos id\u00eanticos, que lan\u00e7am dardos usando molas, s\u00e3o disparados<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_2491931d.jpg\" alt=\"\" width=\"309\" height=\"143\" name=\"graphics26\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>simultaneamente na vertical para baixo. As molas com os respectivos dardos foram inicialmente comprimidas at\u00e9 a posi\u00e7\u00e3o 1 e, ent\u00e3o, liberadas. A \u00fanica diferen\u00e7a entre os dardos I e II, conforme mostra a figura, \u00e9 que I tem um peda\u00e7o de chumbo grudado nele, o que n\u00e3o existe em II.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Escolha o gr\u00e1fico que representa as velocidades dos dardos I e II, como fun\u00e7\u00e3o do tempo, a partir do instante em que eles saem dos canos dos brinquedos.<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_e423e6c5.jpg\" alt=\"\" width=\"768\" height=\"139\" name=\"graphics27\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(<\/b><\/span><\/span><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>UFU-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> TEXTO PARA AS PR\u00d3XIMAS 3 QUEST\u00d5ES:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>O tiro com arco \u00e9 um esporte ol\u00edmpico desde a realiza\u00e7\u00e3o da segunda olimp\u00edada em Paris, no ano de 1900. O arco \u00e9 um dispositivo que converte energia potencial el\u00e1stica, armazenada quando a corda do arco \u00e9 tensionada, em energia cin\u00e9tica, que \u00e9<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_8838bf64.jpg\" alt=\"\" width=\"370\" height=\"170\" name=\"graphics28\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>transferida para a flecha.Num experimento, medimos a for\u00e7a F necess\u00e1ria para tensionar o arco at\u00e9 uma certa dist\u00e2ncia x, obtendo os seguintes valores:<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_2482d9e1.jpg\" alt=\"\" width=\"348\" height=\"68\" name=\"Imagem 49\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>48-(UFU-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 O valor e unidades da constante el\u00e1stica, k, do arco s\u00e3o:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_aeadebdd.png\" alt=\"\" width=\"775\" height=\"15\" name=\"Imagem 179\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>49-(UFU-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Ao tensionar o arco, armazena-se energia potencial el\u00e1stica no sistema. Sendo assim, a express\u00e3o para a energia potencial armazenada \u00e9:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_65229428.png\" alt=\"\" width=\"775\" height=\"18\" name=\"Imagem 180\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>50-(UFU-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Se a massa da flecha \u00e9 de 10 gramas, a altura h=1,40 m e a dist\u00e2ncia x=1m, a velocidade com que ela \u00e9 disparada \u00e9:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_c19c9fb2.png\" alt=\"\" width=\"775\" height=\"19\" name=\"Imagem 181\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>51-(UERJ-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Os esquemas a seguir mostram quatro rampas AB, de mesma altura AC e perfis distintos, fixadas em mesas id\u00eanticas, nas quais uma pequena pedra \u00e9 abandonada, do ponto A, a partir do repouso.<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_177225e5.jpg\" alt=\"\" width=\"768\" height=\"130\" name=\"Imagem 50\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Ap\u00f3s deslizar sem atrito pelas rampas I, II, III e IV, a pedra toca o solo, pela primeira vez, a uma dist\u00e2ncia do ponto B respectivamente igual a dI, dII, dIII e dIV.A rela\u00e7\u00e3o entre essas dist\u00e2ncias est\u00e1 indicada na seguinte alternativa:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) dI &gt; dII = dIII &gt; dIV\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) dIII &gt; dII &gt; dIV &gt; dI\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) dII &gt; dIV = dI &gt; dIII\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) dI = dII = dIII = dIV\u00a0<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>52-(FATEC-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Um skatista brinca numa rampa de skate conhecida por \u201chalf pipe\u201d. Essa pista tem como corte transversal uma semicircunfer\u00eancia de raio 3 metros, conforme mostra a figura. O atleta, saindo do extremo A da pista com velocidade de 4<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_fbd44840.jpg\" alt=\"\" width=\"330\" height=\"171\" name=\"Imagem 51\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>m\/s, atinge um ponto B de altura m\u00e1xima h. Desconsiderando a a\u00e7\u00e3o de for\u00e7as dissipativas e adotando a acelera\u00e7\u00e3o da gravidade g = 10 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, o valor de h, em metros, \u00e9 de<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_c338a576.png\" alt=\"\" width=\"774\" height=\"19\" name=\"Imagem 182\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>53-(UNESP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> O Skycoaster \u00e9 uma atra\u00e7\u00e3o existente em grandes parques de divers\u00e3o, representado nas figuras a seguir. Considere que em um desses brinquedos, tr\u00eas aventureiros s\u00e3o presos a cabos de a\u00e7o e i\u00e7ados a grande altura. Os jovens, que\u00a0se<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_8f3c8382.jpg\" alt=\"\" width=\"552\" height=\"175\" name=\"Imagem 52\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>movem juntos no brinquedo, t\u00eam massas iguais a 50 kg cada um. Depois de solto um dos cabos, passam a oscilar tal como um p\u00eandulo simples, atingindo uma altura m\u00e1xima de 60 metros e chegando a uma altura m\u00ednima do ch\u00e3o de apenas 2 metros. Nessas\u00a0condi\u00e7\u00f5es e desprezando a a\u00e7\u00e3o de for\u00e7as de resist\u00eancias, qual \u00e9, aproximadamente, a m\u00e1xima velocidade, em m\/s, dos participantes durante essa oscila\u00e7\u00e3o e qual o valor da maior energia cin\u00e9tica, em kJ, a que eles ficam submetidos?<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>54-(UECE-CE)\u00a0 A figura a seguir mostra quatro trajet\u00f3rias de uma bola de futebol lan\u00e7ada no espa\u00e7o.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_106c8b5f.jpg\" alt=\"\" width=\"477\" height=\"131\" name=\"Imagem 53\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Desconsiderando o atrito viscoso com o ar, assinale o correto.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) A trajet\u00f3ria que exigiu a maior energia foi a I.\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) A trajet\u00f3ria que exigiu a maior energia foi a II.\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) A trajet\u00f3ria que exigiu a maior energia foi a III.\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) A energia exigida \u00e9 a mesma para todas as trajet\u00f3rias.\u00a0<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>55-(UECE-CE)\u00a0 Um carrinho de montanha russa tem velocidade igual a zero na posi\u00e7\u00e3o 1, indicada na<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_a2e995ea.jpg\" alt=\"\" width=\"476\" height=\"182\" name=\"Imagem 54\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>figura a seguir, e desliza no trilho sem atrito, completando o c\u00edrculo at\u00e9 a posi\u00e7\u00e3o 3. A menor altura h, em metros, para o carro iniciar o movimento sem que venha a sair do trilho na posi\u00e7\u00e3o 2 \u00e9<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_a68083e9.png\" alt=\"\" width=\"774\" height=\"21\" name=\"Imagem 183\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>56-(FGV-RJ)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 O gr\u00e1fico abaixo representa a energia potencial\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><i><b>E<\/b><\/i><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><i><b>P<\/b><\/i><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, em fun\u00e7\u00e3o do tempo, de uma<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_74218b5.jpg\" alt=\"\" width=\"391\" height=\"137\" name=\"Imagem 55\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>pequena esfera em movimento oscilat\u00f3rio, presa na extremidade de uma mola. Dentre os gr\u00e1ficos I, II, III e IV, aqueles que representam a energia\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span style=\"text-decoration: underline;\"><b>cin\u00e9tica<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e a energia\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span style=\"text-decoration: underline;\"><b>total<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0do sistema, quando n\u00e3o h\u00e1 efeitos dissipativos, s\u00e3o,\u00a0<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span style=\"text-decoration: underline;\"><b>respectivamente<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>,<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_ab99a7fa.jpg\" alt=\"\" width=\"767\" height=\"125\" name=\"Imagem 56\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) I e II. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) I e III.\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) II e III.\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) II e IV.\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) III e I.\u00a0<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>57-(FUVEST-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Um esqueitista treina em uma pista cujo perfil est\u00e1 representado na figura abaixo. O trecho horizontal AB est\u00e1 a uma altura h = 2,4 m em rela\u00e7\u00e3o ao trecho, tamb\u00e9m horizontal, CD. O esqueitista percorre a pista no sentido de A para D.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_d5985289.jpg\" alt=\"\" width=\"360\" height=\"173\" name=\"Imagem 57\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>\u00a0<span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>No trecho AB, ele est\u00e1 com velocidade constante, de m\u00f3dulo v = 4 m\/s; em seguida, desce a rampa BC, percorre o trecho CD, o mais baixo da pista, e sobe a outra rampa at\u00e9 atingir uma altura m\u00e1xima H, em rela\u00e7\u00e3o a CD. A velocidade do esqueitista no trecho CD e a altura m\u00e1xima H s\u00e3o, respectivamente, iguais a<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_2503e35.jpg\" alt=\"\" width=\"371\" height=\"107\" name=\"Imagem 58\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) 5 m\/s e 2,4 m.\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) 7 m\/s e 2,4 m.\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) 7 m\/s e 3,2 m.\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) 8 m\/s e 2,4 m.\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) 8 m\/s e 3,2 m.\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>58-(UNIFESP-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Um dos brinquedos prediletos de crian\u00e7as no ver\u00e3o \u00e9 o tobo\u00e1gua. A emo\u00e7\u00e3o do brinquedo est\u00e1 associada \u00e0 grande velocidade atingida durante a descida, uma vez que o atrito pode ser desprezado devido \u00e0 presen\u00e7a da \u00e1gua em todo o percurso do brinquedo, bem como \u00e0 exist\u00eancia das curvas fechadas na horizontal, de forma que a crian\u00e7a percorra esses trechos<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_8c19cc3.jpg\" alt=\"\" width=\"315\" height=\"172\" name=\"Imagem 59\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>encostada na parede lateral (vertical) do tobo\u00e1gua. Sabendo que a crian\u00e7a de 36 kg parte do repouso, de uma altura de 6,0 m acima da base do tobo\u00e1gua, colocado \u00e0 beira de uma piscina, calcule: Dado: g = 10,0 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) A for\u00e7a normal, na horizontal, exercida sobre a crian\u00e7a pela parede lateral do tobo\u00e1gua, no ponto indicado na figura (curva do tobo\u00e1gua situada a 2,0 m da sua base) onde o raio de curvatura \u00e9 igual a 80 cm.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) A for\u00e7a dissipativa m\u00e9dia exercida pela \u00e1gua da piscina, necess\u00e1ria para fazer a crian\u00e7a parar ao atingir 1,5 m de profundidade, considerando que a crian\u00e7a entra na \u00e1gua da piscina com velocidade, na vertical, aproximadamente igual a 10,9 m\/s, desprezando-se, neste c\u00e1lculo, a perda de energia mec\u00e2nica no impacto da crian\u00e7a com a \u00e1gua da piscina.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>59-(UFG-GO)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Uma mola ideal \u00e9 usada para fornecer energia a um bloco de massa m, inicialmente em repouso, o qual mover-se sem atrito em toda a superf\u00edcie, exceto entre os pontos A e B. Ao liberar o sistema massa-mola, o bloco passa pelo ponto P com energia cin\u00e9tica de 1\/20 da energia potencial gravitacional.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_90f7f6a4.jpg\" alt=\"\" width=\"472\" height=\"142\" name=\"Imagem 60\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considerando o exposto, com h = 0,15H e d = 3H, calcule:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) o valor num\u00e9rico do coeficiente de atrito para que o bloco pare no ponto B;<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) a porcentagem da energia total dissipada pela for\u00e7a de atrito.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>60-(UFLA-MG)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0 Um esquilo \u201cvoador\u201d consegue planar do alto de uma \u00e1rvore, a uma altura de 10 m at\u00e9 o ch\u00e3o, com<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_19a5380.jpg\" alt=\"\" width=\"311\" height=\"154\" name=\"Imagem 61\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>velocidade constante de 5 m\/s. Considerando a acelera\u00e7\u00e3o da gravidade g = 10 m\/s2 e a massa do esquilo 2 kg, \u00e9 CORRETO afirmar que o trabalho da for\u00e7a de sustenta\u00e7\u00e3o que atua sobre o esquilo ao longo desse deslocamento \u00e9 de<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_6f9c92a0.png\" alt=\"\" width=\"774\" height=\"16\" 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>61-(FUVEST-SP)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Usando um sistema formado por uma corda e uma roldana, um homem levanta uma caixa de massa m, aplicando na corda uma for\u00e7a F que forma um \u00e2ngulo \u03b8 com a dire\u00e7\u00e3o vertical, como mostra a figura. O trabalho realizado pela resultante das for\u00e7as que atuam na caixa &#8211; peso e for\u00e7a da corda -, quando o centro de massa da caixa \u00e9 elevado, com velocidade constante v, desde a altura y<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0at\u00e9 a altura y<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, \u00e9:<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_d7143276.jpg\" alt=\"\" width=\"392\" height=\"180\" name=\"Imagem 62\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) nulo.\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>b) F (y<\/b><\/span><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>b<\/b><\/span><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>\u00a0\u2013 y<\/b><\/span><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>a<\/b><\/span><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>).\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=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>c) mg (y<\/b><\/span><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>b<\/b><\/span><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>\u00a0\u2013 y<\/b><\/span><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>a<\/b><\/span><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>).\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=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>d) F cos <\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u03b8<\/b><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b> (yb \u2013 y<\/b><\/span><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>a<\/b><\/span><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>).\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) mg (y<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u2013 y<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>) + mv<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\/2.\u00a0<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>62-(UFSM-RS<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>) N\u00e3o se percebe a exist\u00eancia do ar num dia sem vento; contudo, isso n\u00e3o significa que ele n\u00e3o existe.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_c727140e.jpg\" alt=\"\" width=\"186\" height=\"210\" name=\"Imagem 63\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um corpo com massa de 2kg \u00e9 abandonado de uma altura de 10m, caindo verticalmente num referencial fixo no solo. Por<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>efeito da resist\u00eancia do ar, 4J da energia mec\u00e2nica do sistema corpo-Terra se transformam em energia interna do ar e do<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>corpo. Considerando o m\u00f3dulo de acelera\u00e7\u00e3o da gravidade como g= 10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>, o corpo atinge o solo com velocidade de m\u00f3dulo,<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>em m\/s, de<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_9ed5b855.png\" alt=\"\" width=\"774\" height=\"16\" 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>63-(UFJF-MG)<\/b><\/span><\/span><\/span><b> <\/b><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A figura ao lado mostra um sistema composto por dois blocos de massas id\u00eanticas m m kg A B = = 3,0 e uma mola de constante el\u00e1stica k N m = 4,0 \/ . O bloco A est\u00e1 preso a um fio de<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_e07b62dc.jpg\" alt=\"\" width=\"386\" height=\"179\" name=\"Imagem 64\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>massa desprez\u00edvel e suspenso de uma altura h m = 0,8 em rela\u00e7\u00e3o \u00e0 superf\u00edcie S , onde est\u00e1 posicionado o bloco B . Sabendo que a dist\u00e2ncia entre o bloco B e a mola \u00e9 d m = 3,0 e que a colis\u00e3o entre os blocos A e B \u00e9 el\u00e1stica, fa\u00e7a o que se pede nos itens seguintes<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Usando a lei de conserva\u00e7\u00e3o da quantidade de movimento (momento linear), calcule a velocidade do bloco B imediatamente ap\u00f3s a colis\u00e3o do bloco A .<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Calcule o deslocamento m\u00e1ximo sofrido pela mola se o atrito entre o bloco B e o solo for desprez\u00edvel.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) Calcule a dist\u00e2ncia deslocada pelo bloco B em dire\u00e7\u00e3o \u00e0 mola, se o atrito cin\u00e9tico entre o bloco B e o solo for igual\u00a0 a \u03bc=0,4. Nesse caso, a mola ser\u00e1 comprimida pelo bloco B ? Justifique.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>64-(UFBA-BA)<\/b><\/span><\/span><\/span><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b> Uma esfera r\u00edgida de massa m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 0,5kg, presa por um fio de comprimento L = 45,0cm e massa desprez\u00edvel, \u00e9 suspensa em uma posi\u00e7\u00e3o tal que, como mostra a figura, o fio suporte faz um \u00e2ngulo de 90<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>o<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0com a dire\u00e7\u00e3o vertical.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_6852b150.jpg\" alt=\"\" width=\"404\" height=\"206\" name=\"Imagem 65\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Em um dado momento, a esfera \u00e9 solta, indo se chocar com outra esfera de massa m<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0= 0,5kg, posicionada em repouso no solo.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Considerando o di\u00e2metro das esferas desprez\u00edvel e o choque entre elas perfeitamente el\u00e1stico, determine a velocidade das esferas ap\u00f3s o choque, supondo todas as for\u00e7as dissipativas desprez\u00edveis, o m\u00f3dulo da acelera\u00e7\u00e3o da gravidade local igual a 10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e o<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>coeficiente de restitui\u00e7\u00e3o \u03b5=(V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u2019 \u2013 V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u2019)\/(V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0\u2013 V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>) em que V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u2019 e V<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u2019 s\u00e3o as velocidades finais das esferas e v<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>1<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e v<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0as velocidades iniciais.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>65-(UFF-RJ)\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_86396424.jpg\" alt=\"\" width=\"681\" height=\"134\" name=\"Imagem 66\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dois objetos feitos do mesmo material repousam sobre um trecho sem atrito de uma superf\u00edcie horizontal, enquanto comprimem uma mola de massa desprez\u00edvel.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_22856917.jpg\" alt=\"\" width=\"450\" height=\"131\" name=\"Imagem 67\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Quando abandonados, um deles, de massa 2,0 kg, alcan\u00e7a a velocidade de 1,0 m\/s ao perder o contato com a mola. Em seguida, alcan\u00e7a um trecho rugoso da superf\u00edcie, passa a sofrer o efeito do atrito cin\u00e9tico e percorre 0,5 m nesse trecho at\u00e9 parar.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Qual o coeficiente de atrito cin\u00e9tico entre esse bloco e o trecho rugoso da superf\u00edcie horizontal?<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) Qual \u00e9 a velocidade alcan\u00e7ada pelo 2\u00ba bloco, de massa 1,0 kg, ao perder o contato com a mola?<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) Sabendo-se que a constante el\u00e1stica da mola \u00e9 6,0.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>4<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0N\/m, de quanto a mola estava comprimida inicialmente?<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>66-(FGV-SP)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_972a5f5b.jpg\" alt=\"\" width=\"354\" height=\"116\" name=\"Imagem 68\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Em algumas esta\u00e7\u00f5es de trem, h\u00e1 r\u00edgidas molas no fim dos trilhos com a finalidade de amortecer eventual colis\u00e3o de um trem, cujo maquinista n\u00e3o consiga par\u00e1-lo corretamente junto \u00e0 plataforma. Certa composi\u00e7\u00e3o, de massa total 2 m, parada bem pr\u00f3xima \u00e0 mola de constante k, relaxada, recebe um impacto de outra composi\u00e7\u00e3o, de massa m, vindo a uma velocidade v, que acaba engatando na primeira. Ambas v\u00e3o comprimir a mola, causando-lhe uma deforma\u00e7\u00e3o m\u00e1xima x ao pararem instantaneamente, como mostram os esquemas.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_3784c994.jpg\" alt=\"\" width=\"436\" height=\"169\" name=\"Imagem 69\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Desprezando a a\u00e7\u00e3o de agentes externos e dissipativos, a express\u00e3o de x, em fun\u00e7\u00e3o de k, m e v, ser\u00e1<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>(A) x = (m \u00b7 v) \u2215 (3 \u00b7 k).\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>(B) x = (m \u00b7 v2) \u2215 (3 \u00b7 k).\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><span lang=\"en-US\"><b>(C) x = (v \/ 3) \u00b7 \u221a(m\/ k)\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(D) x = v \u00b7 \u221a(3\u00b7m) \/ k .\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(E) x = v \u00b7\u221a m\/ (3k).<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>67-(UEPG-PR)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_6e164488.jpg\" alt=\"\" width=\"369\" height=\"108\" name=\"Imagem 70\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um corpo est\u00e1 suspenso por um fio inextens\u00edvel. Conforme \u00e9 mostrado abaixo, o corpo \u00e9 afastado<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_de6c3940.jpg\" alt=\"\" width=\"273\" height=\"229\" name=\"Imagem 71\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>da sua posi\u00e7\u00e3o inicial A e deslocado at\u00e9 B, onde \u00e9 solto e come\u00e7a a oscilar. Com\u00a0 rela\u00e7\u00e3o \u00e0 energia mec\u00e2nica do sistema e desprezando as for\u00e7as externas que poder\u00e3o atuar sobre ele, assinale a alternativa correta.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>a) Quando o corpo passa pela posi\u00e7\u00e3o A, a energia mec\u00e2nica do sistema \u00e9 nula.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>b) A energia total do sistema independe do afastamento do corpo da posi\u00e7\u00e3o inicial.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>c) Em qualquer ponto do sistema o somat\u00f3rio das energias potencial e cin\u00e9tica \u00e9 igual \u00e0 energia mec\u00e2nica do sistema.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>d) A energia mec\u00e2nica total do sistema depende do comprimento do fio (L).<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>e) Em B, a energia cin\u00e9tica do sistema \u00e9 m\u00e1xima e a potencial nula.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>68-(PUC-RJ)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_dd815d65.jpg\" alt=\"\" width=\"405\" height=\"124\" name=\"Imagem 72\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um ciclista tentando bater um recorde de velocidade em uma bicicleta desce, a partir do repouso, a dist\u00e2ncia de 1440 m em uma<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_b525caa7.jpg\" alt=\"\" width=\"266\" height=\"158\" name=\"Imagem 73\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>montanha cuja inclina\u00e7\u00e3o \u00e9 de 30\u00b0.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Calcule a velocidade atingida pelo ciclista ao chegar \u00e0 base da montanha.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dados: N\u00e3o h\u00e1 atrito\u00a0 e\u00a0 g = 10 m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(A)\u00a0\u00a0 84 m\/s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(B) 120 m\/s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(C) 144 m\/s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(D) 157 m\/s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(E) 169 m\/s<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>69&#8211;COL\u00c9GIO NAVAL)<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_6cb0bd1f.jpg\" alt=\"\" width=\"410\" height=\"133\" name=\"Imagem 74\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Analise\u00a0 a\u00a0 figura\u00a0 a\u00a0 seguir.<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_5632c4ad.jpg\" alt=\"\" width=\"579\" height=\"180\" name=\"Imagem 75\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Dados: g=10m\/s<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>2<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>\u00a0e M<\/b><\/span><\/span><sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>tren\u00f3<\/b><\/span><\/span><\/sub><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>=800kg.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Numa\u00a0 determinada\u00a0 montanha\u00a0 russa\u00a0 um\u00a0 tren\u00f3, sob\u00a0 a\u00a0 a\u00e7\u00e3o\u00a0 de\u00a0 uma for\u00e7a\u00a0 resultante\u00a0 constante, que\u00a0 atua\u00a0 de\u00a0 A\u00a0 at\u00e9\u00a0 B, parte\u00a0 do repouso\u00a0 do\u00a0 ponto\u00a0 A\u00a0 e, ap\u00f3s\u00a0 2\u00a0 segundos, atinge\u00a0 a\u00a0 velocidade de\u00a0 180\u00a0 km\/h\u00a0 no\u00a0 ponto\u00a0 B, iniciando\u00a0 uma\u00a0 subida\u00a0 que\u00a0 o\u00a0 leva\u00a0 at\u00e9 o\u00a0 ponto\u00a0 C, onde\u00a0 passa\u00a0 com\u00a0 velocidade\u00a0 de\u00a0 18\u00a0 km\/h.\u00a0 Sabendo\u00a0 que a\u00a0 energia\u00a0 perdida\u00a0 pelos\u00a0 atritos\u00a0 entre\u00a0 os\u00a0 pontos\u00a0 B\u00a0 e\u00a0 C\u00a0 foi\u00a0 de 19.10<\/b><\/span><\/span><sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>4<\/b><\/span><\/span><\/sup><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>J, \u00e9\u00a0 correto\u00a0 afirmar\u00a0 que\u00a0 a\u00a0 for\u00e7a\u00a0 resultante\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_2f3f059f.jpg\" alt=\"\" width=\"21\" height=\"21\" name=\"Imagem 76\" align=\"BOTTOM\" border=\"0\" \/>\u00a0que\u00a0 atuou sobre\u00a0 o\u00a0 tren\u00f3\u00a0 entre\u00a0 os\u00a0 pontos\u00a0 A\u00a0 e\u00a0 B\u00a0 e\u00a0 a\u00a0 altura\u00a0 atingida\u00a0 por ele\u00a0 no\u00a0 ponto\u00a0 C\u00a0 s\u00e3o, respectivamente:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(A)\u00a0 10000\u00a0 N\u00a0 e\u00a0 h = 80m\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(B)\u00a0 20000\u00a0 N\u00a0 e\u00a0 h = 80m\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(C)\u00a0 20000\u00a0 N\u00a0 e\u00a0 h = 100m\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(D)\u00a0 40000\u00a0 N\u00a0 e\u00a0 h = 100m<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>(E)\u00a0 80000\u00a0 N\u00a0 e\u00a0 h = 120m<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>70-(EsPCEx)<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_837dbc07.jpg\" alt=\"\" width=\"421\" height=\"149\" name=\"Imagem 77\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Um corpo de massa 4 kg est\u00e1 em queda livre no campo gravitacional da Terra e n\u00e3o h\u00e1 nenhuma for\u00e7a dissipativa<\/b><\/span><\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_dc0893db.jpg\" alt=\"\" width=\"166\" height=\"283\" name=\"Imagem 78\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>atuando. Em determinado ponto, ele possui uma energia potencial, em rela\u00e7\u00e3o ao solo, de 9 J, e sua energia cin\u00e9tica vale 9 J. A velocidade do corpo, ao atingir o solo, \u00e9 de:<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>[A] 5 m\/s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>[B] 4 m\/s \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>[C] 3 m\/s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>[D] 2 m\/s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>[E] 1 m\/s<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #c00000;\"><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>71-(ENEM-MEC)<\/b><\/span><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_f4af4e5e.jpg\" alt=\"\" width=\"269\" height=\"104\" name=\"Imagem 79\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Uma das modalidades presentes nas olimp\u00edadas \u00e9 o salto com vara. As etapas de um dos saltos de um atleta est\u00e3o representadas na figura:<\/b><\/span><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/fisicaevestibular.com.br\/novo\/wp-content\/uploads\/migracao\/energiamecanica\/i_8fb1d79c9d4ebcbf_html_88004fcf.jpg\" alt=\"\" width=\"742\" height=\"144\" name=\"Imagem 80\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>Desprezando-se as for\u00e7as dissipativas (resist\u00eancia do ar e atrito), para que o salto atinja a maior altura poss\u00edvel, ou seja, o m\u00e1ximo de energia seja conservada, \u00e9 necess\u00e1rio que<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>A. a energia cin\u00e9tica, representada na etapa I, seja totalmente convertida em energia potencial el\u00e1stica, representada na etapa IV.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>B. a energia cin\u00e9tica, representada na etapa II,\u00a0 seja totalmente convertida em energia potencial gravitacional,<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>representada na etapa IV.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>C. a energia cin\u00e9tica, representada na etapa I, seja totalmente convertida em energia potencial el\u00e1stica representada na etapa III.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>D. a energia potencial gravitacional, representada na etapa II, seja totalmente convertida em energia potencial el\u00e1stica, representada na etapa IV.<\/b><\/span><\/span><\/p>\n<p><span style=\"font-family: Arial, serif;\"><span style=\"font-size: medium;\"><b>E. a energia potencial gravitacional, representada na etapa I, seja totalmente convertida em energia potencial el\u00e1stica, representada na etapa III.<\/b><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<h3><a title=\"Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Energia Mec\u00e2nica\" href=\"http:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/energia-mecanica\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-energia-mecanica\/\"><span style=\"color: #000080;\">Confira o gabarito e resolu\u00e7\u00e3o comentada!<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"<p>Exerc\u00edcios de vestibulares com resolu\u00e7\u00e3o comentada sobre Energia Mec\u00e2nica \u00a0 01-(UFMG-MG)\u00a0Rita est\u00e1 esquiando numa montanha dos Andes. A energia cin\u00e9tica dela em fun\u00e7\u00e3o do tempo, durante parte do trajeto, est\u00e1 representada neste gr\u00e1fico: Os pontos Q e R, indicados nesse gr\u00e1fico, correspondem a dois instantes diferentes do movimento de Rita. Despreze todas as formas de atrito. Com base nessas informa\u00e7\u00f5es,<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1370,"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-1372","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1372","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=1372"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1372\/revisions"}],"predecessor-version":[{"id":10820,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1372\/revisions\/10820"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1370"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1372"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}