{"id":3425,"date":"2021-09-08T09:00:39","date_gmt":"2021-09-08T12:00:39","guid":{"rendered":"https:\/\/professorpinguim.com.br\/blog\/?p=3425"},"modified":"2023-08-06T21:16:58","modified_gmt":"2023-08-07T00:16:58","slug":"energia-mecanica-exercicios-resolvidos","status":"publish","type":"post","link":"https:\/\/professorpinguim.com.br\/blog\/energia-mecanica-exercicios-resolvidos\/","title":{"rendered":"Energia mec\u00e2nica: exerc\u00edcios resolvidos"},"content":{"rendered":"\n<p>Fala, pessoal, tudo belezinha?<\/p>\n\n\n\n<p>Sem enrola\u00e7\u00e3o, vamos direto para a resolu\u00e7\u00e3o de alguns exerc\u00edcios de Energia Mec\u00e2nica. Bons estudos!<\/p>\n\n\n\n<!--more-->\n\n\n\n<h2 class=\"wp-block-heading\">Exerc\u00edcios resolvidos de Energia Mec\u00e2nica<\/h2>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 1<\/h3>\n\n\n\n<p><strong>(Puccamp-SP) A massa m de um p\u00eandulo simples, cujo fio tem comprimento L = 0,90 m, \u00e9 abandonada a partir do repouso quando o fio forma \u00e2ngulo de 60\u00b0 com a vertical, como mostra a figura.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/pRyTIu9I_JW4P7_EDYUoIpsr4GZhvECfCMuv7VRZk6ZwgY4vlYbOo7hK9Si9RsuNPrNp_V5QylE14tiLsvyoBuPzsGoDngnOI4q-3Q4zYuB1ONbeAd_nz9ew3z1ZEMMN7ZrAaVtC\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>(Dados: sen 60\u00b0 = 0,87; cos 60\u00b0 = 0,50; g = 10 m\/s<\/strong><strong><sup>2<\/sup><\/strong><strong>)<\/strong><\/p>\n\n\n\n<p><strong>Desprezando a resist\u00eancia do ar, a velocidade de m, quando o fio fica na posi\u00e7\u00e3o vertical, \u00e9, em m\/s:<\/strong><\/p>\n\n\n\n<p>a) 1,0<\/p>\n\n\n\n<p>b) 2,0<\/p>\n\n\n\n<p>c) 3,0<\/p>\n\n\n\n<p>d) 4,0<\/p>\n\n\n\n<p>e) 5,0<\/p>\n\n\n\n<p><strong>RESOLU\u00c7\u00c3O<\/strong><\/p>\n\n\n\n<p>Podemos calcular a velocidade por meio da energia mec\u00e2nica. \u00c9 importante dar um referencial para o objeto. Nesse caso, a altura \u00e9 zero (h = 0). Al\u00e9m disso, damos um nome ao ponto alto (A) e ao ponto baixo (B). Veja:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/63zlt5r-PF9MSvYJMVqENt-CD2BtT9JBUrFFRD6JaxeV_uSdSJx5eV4ONL8UXb00MB1t0m7VrZ0r8wqyOaNuzGHxqkALYTW684ZU4mcRnsqeR1SQANl3G31HAT8hlhN6r-wjgaIS\" alt=\"\"\/><\/figure>\n\n\n\n<p>Assim, podemos dizer que:<\/p>\n\n\n\n<p>Em<sub>B<\/sub> = Em<sub>A<\/sub><\/p>\n\n\n\n<p>Como a velocidade no ponto A \u00e9 nula (v<sub>A<\/sub> = 0), podemos dizer que ela ser\u00e1 igual somente \u00e0 energia potencial gravitacional em A.<\/p>\n\n\n\n<p>J\u00e1 no ponto B, como adotamos que a altura \u00e9 zero, ele s\u00f3 possui energia cin\u00e9tica. Ou seja:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/u8RrnmhVzVsycbpGm4O4QsPbKzzyB1COvH6bDCXlpc8F9SAfWJtAAJgQwFQ8K2QO5hM8hGEns7-5tdMUsA0z5GELlt1axekIOkjz88zLY5MerV5T7_AKRNc5xRdTXhW2bBp3S1RD\" alt=\"\"\/><\/figure>\n\n\n\n<p>O problema \u00e9 que, nessa quest\u00e3o, n\u00e3o sabemos de que altura o corpo cai. E quando n\u00e3o sabemos alguma coisa, devemos recorrer \u00e0 Matem\u00e1tica. Podemos calcular a altura que o objetivo caiu entre A e B. No entanto, \u00e9 melhor calcularmos antes um outro trecho, que chamarei de y. Veja na imagem:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/8rsBShhTw2YrA7Iw13RnO4HVghzjLMpKua9Bqar40iPPGBEC4EiwTXlWc8nDDhjWltSMmIX9qVQvl54jpF6XhDf7HzYNRqnlvqzlb8s0ctkXPIZz6IzkRjB4EAS6fNapJX7XG-ry\" alt=\"\"\/><\/figure>\n\n\n\n<p>Como podemos ver pela \u00e1rea em amarelo, basta utilizarmos o tri\u00e2ngulo ret\u00e2ngulo que foi criado. Assim:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/15kwW8EAWZDJONdWJnmciOJPiknsT_Yk8aNXIGQyCg9RP7cVo3PM4DBYQaANa-7dumuM7unnp9s-VJJKQGo7-CAZbUXB9jDAenLWOR4AXAbpp4ZAvPUo1cIQ1ZdyBI2hJ7WxhtnU\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/7qc7G_oAtPwtc40E92VXARP3M2ue5S4M9p8FhAgQjPNAbdTGl39HXJGRQ7CqN2s_lnZnadphOldvmJHCVOA2Dcp_pQudJqn6FzPzPLLJtQ6fOHcmQv3p8k9XpjWPWefSNKQuCV2A\" alt=\"\"\/><\/figure>\n\n\n\n<p>y = 0,45 m<\/p>\n\n\n\n<p>Se o comprimento inteiro do fio era 0,9 m, a altura de A para B \u00e9 o que sobra, ou seja, h<sub>AB<\/sub> = 0,45 m.<\/p>\n\n\n\n<p>Agora, vamos voltar na nossa f\u00f3rmula:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/vi2BntQ2a5Tv2izdIu8j7qXhh1nLa5LelG5gEVp7I2ZYMRMfghE8dvhASUCKM7otG3T55Ot9dZjIKtTl2v_biGd5o4LudpBQEoKkrKJf_4RVLhYSHzIUnL7RqrcdOgkyKUTxpZK-\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/aMbySKYBRbvI9AfP0ndOMwI5zgT3ncmKEnVn4TS5lt1vlMmmpIVNgRAGAHk219gkRwsv6P6eR9DEB-QidVmpINtcK1nj-1q5DgcuG9wUa1vdlIvYM7Y5kdCUfKuxQ3lCAdXz13wM\" alt=\"\"\/><\/figure>\n\n\n\n<p>v<sub>B<\/sub> = 3 m\/s<\/p>\n\n\n\n<p><strong>RESPOSTA: C<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 2<\/h3>\n\n\n\n<p><strong>(Uece) Uma bola \u00e9 lan\u00e7ada verticalmente para cima,com energia cin\u00e9tica Ec. No ponto mais alto da trajet\u00f3ria,sua energia potencial \u00e9 Ep. Considere que, do lan\u00e7amento ao ponto mais alto, o atrito da bola com o ar tenha causado uma dissipa\u00e7\u00e3o de energia mec\u00e2nica de p % em rela\u00e7\u00e3o ao valor inicial. Assim, p \u00e9 igual a<\/strong><\/p>\n\n\n\n<p>a) 100 [(Ep \/ Ec) &#8211; 1]<\/p>\n\n\n\n<p>b) 100 Ep \/ Ec<\/p>\n\n\n\n<p>c) 100 Ec \/ Ep<\/p>\n\n\n\n<p>d) 100 [1 &#8211; Ep \/ Ec]<\/p>\n\n\n\n<p><strong>RESOLU\u00c7\u00c3O<\/strong><\/p>\n\n\n\n<p>Vamos desenhar:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/wA26xiJnwkM8c9VHnxoNuhUA7X-vBbe_mur2CVxf1PiIkSC24a9qL4s5mXdkDIuP1kEsaGBUJGSdt20xQ7-iVI_2A2QXLFwwMsjPA4GUXxDFfoA4lgLlesvT88jlbGEgAdhM98E4\" alt=\"\"\/><\/figure>\n\n\n\n<p>Lembre-se de que no ponto mais alto a energia cin\u00e9tica \u00e9 nula (Ec = 0). Veja tamb\u00e9m que este n\u00e3o \u00e9 um sistema conservativo. Portanto, <strong>n\u00e3o <\/strong>podemos dizer que a energia mec\u00e2nica no ponto A \u00e9 igual no ponto B (Em<sub>A<\/sub> = Em<sub>B<\/sub>). No ponto A, a energia mec\u00e2nica \u00e9 maior, pois o enunciado nos informa que de A para B houve <strong>dissipa\u00e7\u00e3o <\/strong>de energia.<\/p>\n\n\n\n<p>Ent\u00e3o, nossa sa\u00edda \u00e9 dizer que a energia mec\u00e2nica inicial \u00e9 igual \u00e0 final mais a perda em rela\u00e7\u00e3o ao que t\u00ednhamos no come\u00e7o. Representamos assim:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/xCB4bsmIqclZ2EYIxR-8El7g7BO0yAQOkbDi1O0ENy77v0xsqBvpGmkYLs9UlF48NAdt9YfsH6Hvljjgt-fAT7pdPCcj1Xd6gDb_-wSMHRliJ4rZ_NYb7L-N_Wos6u8BcbITXP5N\" alt=\"\"\/><\/figure>\n\n\n\n<p>Agora, temos que fazer contas, pois devemos lembrar que cada energia mec\u00e2nica dessas \u00e9 a soma da energia cin\u00e9tica com a energia potencial (Em = Ec + Ep). Repare que, no ponto A, s\u00f3 t\u00ednhamos energia cin\u00e9tica, pois o admitimos como nosso referencial. J\u00e1 a energia mec\u00e2nica final, sabemos que \u00e9 nosso Ep, pois B \u00e9 o ponto mais alto e, com isso, a velocidade da bola \u00e9 nula.<\/p>\n\n\n\n<p>Assim, trocando em nossa f\u00f3rmula, ficamos com:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/Ja3oVmb35Gp80BTyr_DZgo1W2Zxi4ALiMnl9RwZRfJ6O324naLoRXXwXg7M7pLDkMck6vA2ebjo_V3xgrXxsvHrJ-1GLy8LAcfJY5VQ55H0G45HVqNo9b7FR8L0jIQHGIvzr1n-L\" alt=\"\"\/><\/figure>\n\n\n\n<p>Agora vamos isolar o p:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/FY59nVuLXRN0T77OTvr8gzQoNuUrYFj5kb2BhuOg-xD9_kapRr61V4GbXt36jlUmma2iWa9RITGxuWUoX57PNDDV-9vNCD1N9bWBHTpJpuJ-HPiJjH5btsNko2FNB6vzV5lysF-E\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/adjTWPXBzrXwzcvY-Vf5_-CCYR2lh_Uwt48bpyKlCPfB9P0wYXrSl6NlKfZMQlfyj6obyMbbKE7cAmp_ijXMJ3CXpU7ZarFf-znt08z7eCcnDn_ZyaYS2xM9pcsQPbt9pNNbl4_W\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>RESPOSTA: D<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 3<\/h3>\n\n\n\n<p><strong>(Cesgranrio-RJ) Um bloco de massa = 2,0 kg, apresentado no desenho abaixo, desliza sobre um plano horizontal com velocidade de m\u00f3dulo 10 m\/s. No ponto A, a superf\u00edcie passa a ser curva, com raio de curvatura R = 1,0 m. Suponha que o atrito seja desprez\u00edvel ao longo de toda a trajet\u00f3ria e que g = 10 m\/s<\/strong><strong><sup>2<\/sup><\/strong><strong>. Determine, ent\u00e3o:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/j2VQx8Uj4QIevbwMg6d9YZ3-bGOIRFt42aimpFd01RRXQSIMaHD0A3puhcMNlsNlg7L-cEy6R0gHRZnpaXV4lzd9jGVr5-ncRytTozqF7WjeYrt7TtrCnBLYp6kJuE0S9fOV7BsO\" alt=\"\"\/><\/figure>\n\n\n\n<p>a) a intensidade da rea\u00e7\u00e3o da pista sobre o bloco, no ponto B.<\/p>\n\n\n\n<p>b) a intensidade de rea\u00e7\u00e3o da pista sobre o bloco, no ponto C.<\/p>\n\n\n\n<p><strong>RESOLU\u00c7\u00c3O<\/strong><\/p>\n\n\n\n<p>a) Se o atrito \u00e9 desprez\u00edvel e a for\u00e7a normal n\u00e3o possui trabalho, ent\u00e3o estamos falando de um sistema conservativo. No ponto B, v\u00e3o atuar a normal (para dentro da curva) e o peso (tangencial para baixo). Veja:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/0qQryNq4MVtNbKcpWg02fNfYU60W0DjULwFw232EcZhtkcsLEl1Pl4FhdiWbVf5clLjLv8GygT2GCmNBtKL1HgAOBGNyqFtpkF9P_6VSbWcuZ8OvINLsrhDTDIJvRltOvS_b9cWI\" alt=\"\"\/><\/figure>\n\n\n\n<p>Nesse caso, a normal ser\u00e1 igual \u00e0 resultante centr\u00edpeta (N = Rc), enquanto o peso ser\u00e1 igual \u00e0 resultante tangencial (P = Rt). Mas repare que ele n\u00e3o pediu a resultante tangencial, ent\u00e3o concentre-se apenas em calcular a normal.<\/p>\n\n\n\n<p>Agora veja:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/zyBmu8YVyXBuHWPNHlk27ga-_GCYVhOTRCUYywbHISsr4anVkT4uLPXfkXZMoc58vQvInG3ll2-bxRt-KTMLkzpskQraVBpBQzhPH2EyXQugFHE8LPufXfp9zoKeN8loOjjdk39V\" alt=\"\"\/><\/figure>\n\n\n\n<p>Veja que teremos que calcular a velocidade no ponto B (v<sub>B<\/sub>). E podemos fazer isso pelo teorema da energia cin\u00e9tica ou por energia mec\u00e2nica.<\/p>\n\n\n\n<p>Vamos usar energia mec\u00e2nica: Em<sub>B<\/sub> = Em<sub>i<\/sub>. Para isso, vamos adotar o &#8220;ch\u00e3o&#8221; da figura como referencial e estipular que inicialmente s\u00f3 havia energia cin\u00e9tica no corpo. J\u00e1 em B, h\u00e1 tanto energia cin\u00e9tica quanto energia potencial.<\/p>\n\n\n\n<p>Ent\u00e3o ficamos com:<\/p>\n\n\n\n<p>Em<sub>B<\/sub> = Em<sub>i<\/sub><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/UgAp0dzbYw5zlMG_-k1jUDSLMbNXC6pO4lpWdsXVzOXHPzrx42TF0Z2yIR4vtoVm29HrgiF2bW6XijkIYoO1V08ZdAhDu4BHInP7DPoKBXze0FvwkffFfM_RIw9XFZDR1muJXU8O\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/ogTiik7k7d5PKH9H97xMQ5cj3TNLeS-NC9HBwp9MWJRJZsqJAhhx5p29zz4b0tz-GJGYcF34uAbZBHbjyZGWmPCQwmyI9Nr2xiH84eC7aFCoHor86MPd-SAm-6FkTlPbmYMuGBak\" alt=\"\"\/><\/figure>\n\n\n\n<p>V<sub>B<\/sub><sup>2<\/sup> = 80 m<sup>2<\/sup>\/s<sup>2<\/sup><\/p>\n\n\n\n<p>Agora podemos calcular a normal:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/zyBmu8YVyXBuHWPNHlk27ga-_GCYVhOTRCUYywbHISsr4anVkT4uLPXfkXZMoc58vQvInG3ll2-bxRt-KTMLkzpskQraVBpBQzhPH2EyXQugFHE8LPufXfp9zoKeN8loOjjdk39V\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/ZjUxDDMUp5iHbDDataLSJ71OcQ7rNKIupRvaHJ409WUI1fhebYVwCc62BxqEG3elT3TVHSYkC_9KnHD7ZbnjeLo2unO5qm2nz5YBtYuiuYTskYC856w7Ecyd62ztbfLbrZ02OkcX\" alt=\"\"\/><\/figure>\n\n\n\n<p>N<sub>B<\/sub> = 160 N<\/p>\n\n\n\n<p>b) No ponto C, tanto normal quanto o peso do corpo v\u00e3o atuar para baixo. Portanto:<\/p>\n\n\n\n<p>Rc = N<sub>C<\/sub> + P<\/p>\n\n\n\n<p>Teremos que calcular a velocidade em C:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/zANr_nc_xRPOGe0_91U54uXX3pJ9YbRiFgiIHHsgqzHwentj6wZckHvq6R4nnaMZu2_rmspSErNVaX2PyYHFEWB9wC-DHftto1xuPDoEa9_Ln1r4Vya1BHgO5lWn6jLq2hnSaZAS\" alt=\"\"\/><\/figure>\n\n\n\n<p>N\u00e3o temos elementos suficientes para fazer essa conta. Segure essa f\u00f3rmula a\u00ed. Agora temos que igualar \u00e0 energia mec\u00e2nica:<\/p>\n\n\n\n<p>Em<sub>C<\/sub> = Em<sub>i<\/sub><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/ofT1378pMDuFu_wY_vSJ3_pzk1wFT5gFQ67ZOUZzSU0six-hLmBszeP2XH90kiywhZUs03ks0ROFMgy8LWrUGn96bLN8w3GiFKBFXfPHFjVLw4fn9YbThdep_PBDkijgkMYwdepQ\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/MkdC51tSOH_kh7w1cnnXu3tK_bHFn0Qmokuw-QxdS1zzZIHw7rg0NnGlpiJdoRD18qgqC7t_wJWw_auX4eWqaDET_jl55LfXijFkxjqS5DV6RqmcbtBPoeJNDzODvX-7dPbJRZCG\" alt=\"\"\/><\/figure>\n\n\n\n<p>V<sub>c<\/sub><sup>2<\/sup> = 60 m<sup>2<\/sup>\/s<sup>2<\/sup><\/p>\n\n\n\n<p>Agora sim voltamos \u00e0 nossa equa\u00e7\u00e3o inicial:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/zANr_nc_xRPOGe0_91U54uXX3pJ9YbRiFgiIHHsgqzHwentj6wZckHvq6R4nnaMZu2_rmspSErNVaX2PyYHFEWB9wC-DHftto1xuPDoEa9_Ln1r4Vya1BHgO5lWn6jLq2hnSaZAS\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/i2MQ25WwBsvcBZuzqE0s2_1zgP-CWpWzZoyA0RPgmk4iM_T9uEtlmKtcbAO8LFj-z3IC9jXL3dxO6fsjNULPef9UQiaHEHKFFdRiZQOVOZ_4LdAknjob_T0CrGhHROpzKWyly0fQ\" alt=\"\"\/><\/figure>\n\n\n\n<p>N<sub>C<\/sub> = 100 N<\/p>\n\n\n\n<hr class=\"wp-block-separator is-style-wide\"\/>\n\n\n\n<p>Espero que voc\u00ea tenha entendido um pouco melhor <strong>como resolver exerc\u00edcios de energia mec\u00e2nica<\/strong>. Para conferir a resolu\u00e7\u00e3o dessas e de muitas outras quest\u00f5es sobre essa mat\u00e9ria, assista \u00e0 minha live:<\/p>\n\n\n\n<figure class=\"wp-block-embed-youtube wp-block-embed is-type-video is-provider-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<iframe loading=\"lazy\" title=\"Live #14 \u25ba ENERGIA MEC\u00c2NICA - Professor Pinguim\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/xzY1GmmCkZg?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div><\/figure>\n\n\n\n<hr class=\"wp-block-separator is-style-wide\"\/>\n\n\n\n<p>E se quiser ajuda para melhorar seu n\u00edvel de F\u00edsica em outras mat\u00e9rias,<a href=\"https:\/\/professorpinguim.com.br\/\" target=\"_blank\" rel=\"noreferrer noopener\"> <strong>entre em contato comigo e escolha o curso de F\u00edsica mais adequado para voc\u00ea<\/strong><\/a>!<\/p>\n\n\n\n<p>Tem d\u00favidas de como funciona a minha plataforma e a minha metodologia de ensino? <strong><a rel=\"noreferrer noopener\" href=\"https:\/\/conteudo.professorpinguim.com.br\/experimente-plataforma\" target=\"_blank\">Clique aqui<\/a> <\/strong>e acesse gratuitamente os m\u00f3dulos de <strong>Cinem\u00e1tica <\/strong>e <strong>\u00d3ptica<\/strong> do meu <strong>&nbsp;SEMIEXTENSIVO exclusivo para FUVEST, UNICAMP, UNESP E UNIFESP. <\/strong>Tenho certeza que vai mudar a sua vida.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/professorpinguim.com.br\/blog\/wp-content\/uploads\/2020\/07\/BannerBlog_728x90.png\" alt=\"O atributo alt desta imagem est\u00e1 vazio. O nome do arquivo \u00e9 BannerBlog_728x90.png\"\/><\/figure>\n\n\n\n<p>Me acompanhe nas redes sociais: curta a minha&nbsp;<a href=\"https:\/\/www.facebook.com\/professorpinguim\/\"><strong>p\u00e1gina no Facebook<\/strong><\/a><strong>,<\/strong>&nbsp;me&nbsp;siga no&nbsp;<a href=\"https:\/\/www.instagram.com\/professorpinguim\/\"><strong>Instagram<\/strong><\/a><strong>,<\/strong>&nbsp;se inscreva no&nbsp;<a href=\"https:\/\/www.youtube.com\/c\/PinguimV%C3%ADdeoF%C3%ADsica\/?sub_confirmation=1\"><strong>Youtube<\/strong><\/a>&nbsp;e participe do meu canal oficial no&nbsp;<strong><a href=\"http:\/\/professorpinguim.com.br\/telegram\">Telegram<\/a><\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Fala, pessoal, tudo belezinha? Sem enrola\u00e7\u00e3o, vamos direto para a resolu\u00e7\u00e3o de alguns exerc\u00edcios de Energia Mec\u00e2nica. Bons estudos!<\/p>\n","protected":false},"author":1,"featured_media":3426,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[355],"tags":[667,648],"class_list":["post-3425","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mecanica","tag-exercicios-fisica","tag-mecanica-fisica"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.12 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Energia mec\u00e2nica: exerc\u00edcios resolvidos - Professor Pinguim | F\u00edsica<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/professorpinguim.com.br\/blog\/energia-mecanica-exercicios-resolvidos\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Energia mec\u00e2nica: exerc\u00edcios resolvidos - Professor Pinguim | F\u00edsica\" \/>\n<meta property=\"og:description\" content=\"Fala, pessoal, tudo belezinha? 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