{"id":613,"date":"2015-04-12T13:48:23","date_gmt":"2015-04-12T13:48:23","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=613"},"modified":"2024-09-02T17:13:49","modified_gmt":"2024-09-02T17:13:49","slug":"resolucao-comentada-do-iff-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-do-iff-2015\/resolucao-comentada-do-iff-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada do IFF – 2015"},"content":{"rendered":"

RESOLU\u00c7\u00d5ES<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

02- <\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span><\/span> Dados: \u03bc = 0,3; m=200g=0,2kg; <\/b><\/span><\/span><\/span>P=mg<\/b><\/span><\/span><\/span> = 0,2.10 <\/b><\/span><\/span><\/span>= 2N = N<\/b><\/span><\/span><\/span> e d<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> = 50cm = 0,5m <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>Fat<\/b><\/span><\/span><\/sub><\/span> = F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span><\/b><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>N = <\/b><\/span><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/span> = <\/b><\/span><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>mg = 0,3.0,2.10 <\/b><\/span><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub><\/span> = 0,6 N<\/b><\/span><\/span><\/span><\/span> W<\/b><\/span><\/span><\/span><\/span>Fat<\/b><\/span><\/span><\/span><\/sub><\/span> = 0,6.0,5 <\/b><\/span><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span><\/span>Fat<\/b><\/span><\/span><\/span><\/sub><\/span> = 0,3 J<\/b><\/span><\/span><\/span><\/span><\/p>\n

b) Entre P e A n\u00e3o existe atrito e a energia mec\u00e2nica se conserva E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> mgh + kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + E<\/b><\/span><\/span><\/span>cP<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> 0,2.10.0,15 + 100.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 0,2.V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> 0,3 + 50x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 0,2.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = 0,3 + 50x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Como existe atrito entre A e B a energia mec\u00e2nica em B ser\u00e1 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> \u2013 W<\/b><\/span><\/span><\/span>Fat<\/b><\/span><\/span><\/sub><\/span> mv<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = 0,3 + 50x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \u2013 0,3 0,2V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = 50x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 500x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = \u221a(500x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>) (velocidade horizontal com que o objeto abandona a mesa no ponto B).<\/b><\/span><\/span><\/span><\/p>\n

Agora, trata-se de um lan\u00e7amento horizontal com V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = \u221a(500x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>) que, na horizontal \u00e9 um MRU com velocidade constante V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = \u221a(500x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>) e tempo de queda t que \u00e9 o tempo que ele demora para percorrer o alcance horizontal A:<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = A\/t \u221a(500x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>) = A\/t t = A\/\u221a(500x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>) (tempo de queda)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na vertical, rata-se de uma <\/b><\/span><\/span><\/span>queda livre de equa\u00e7\u00e3o h = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span> <\/b>+ gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span> <\/b>1 = 0.t + 5.A\/\u221a(500x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> 1 = 5.A\/500x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> 1 = 5.A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/\u221a(500x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> 500X<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 5.A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>A (x) = 10X.<\/b><\/span><\/span><\/span><\/p>\n

c)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

03- <\/span><\/span><\/span>a) Observe na <\/b><\/span><\/span><\/span>sequ\u00eancia abaixo<\/b><\/span><\/span><\/span> o c\u00e1lculo da <\/b><\/span><\/span><\/span>resist\u00eancia do resistor equivalente <\/b><\/span><\/span><\/span>e, onde podemos curto-circuitar o amper\u00edmetro (ideal) e retirarmos o volt\u00edmetro (ideal):<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = U (\u03b5)\/i \"\" 12 = 36\/i \"\" <\/b><\/span><\/span><\/span>i = 3 A.<\/b><\/span><\/span><\/span><\/p>\n

Observe<\/b><\/span><\/span><\/span> detalhadamente na figura abaixo <\/b><\/span><\/span><\/span>como foi determinada <\/b><\/span><\/span><\/span>a indica\u00e7\u00e3o do amper\u00edmetro que \u00e9<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

de <\/b><\/span><\/span><\/span>I = 0,5 A.<\/b><\/span><\/span><\/span><\/p>\n

Observe<\/b><\/span><\/span><\/span> detalhadamente na figura abaixo <\/b><\/span><\/span><\/span>como foi determinada <\/b><\/span><\/span><\/span>a indica\u00e7\u00e3o do volt\u00edmetro que \u00e9 de<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

U=18V<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

b)<\/b><\/span><\/span><\/span> As pot\u00eancias pedidas est\u00e3o indicadas na figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

c) O <\/b><\/span><\/span><\/span>resistor<\/b><\/span><\/span><\/span> que dissipa <\/b><\/span><\/span><\/span>maior pot\u00eancia<\/b><\/span><\/span><\/span> \u00e9 o de <\/b><\/span><\/span><\/span>maior resist\u00eancia<\/b><\/span><\/span><\/span> (4\u2126) percorrido por <\/b><\/span><\/span><\/span>maior corrente<\/b><\/span><\/span><\/span> (3 A) \"\" P = R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 4.3<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 4.9 \"\" <\/b><\/span><\/span><\/span>P=36W.<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os Exerc\u00edcios<\/span><\/a><\/h3>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES 01- 02- a) Dados: \u03bc = 0,3; m=200g=0,2kg; P=mg = 0,2.10 = 2N = N e dAB = 50cm = 0,5m WFat = Fat.dAB Fat = \u03bcN = \u03bcP = \u03bcmg = 0,3.0,2.10 Fat = 0,6 N WFat = 0,6.0,5 WFat = 0,3 J b) Entre P e A n\u00e3o existe atrito e a energia mec\u00e2nica se conserva EmP<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":611,"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-613","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/613","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=613"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/613\/revisions"}],"predecessor-version":[{"id":615,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/613\/revisions\/615"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/611"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=613"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}