{"id":301,"date":"2014-12-17T13:56:53","date_gmt":"2014-12-17T13:56:53","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=301"},"modified":"2024-09-02T17:00:29","modified_gmt":"2024-09-02T17:00:29","slug":"resolucao-comentada-da-unb-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-da-unb-2015\/resolucao-comentada-da-unb-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada da UnB 2015"},"content":{"rendered":"

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

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

\"\"<\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

For\u00e7a resultante<\/b><\/span><\/span><\/span> sobre o conjunto ciclista + bicicleta de massa <\/b><\/span><\/span><\/span>m = 10 + 70 = 80kg<\/b><\/span><\/span><\/span> <\/b>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = m.a =<\/b><\/span><\/span><\/span> 80.2 <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = 160N<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

O t<\/b><\/span><\/span><\/span>rabalho<\/b><\/span><\/span><\/span> \u00e9 fornecido por <\/b><\/span><\/span><\/span>W = F.d <\/b><\/span><\/span><\/span>= 160.4 <\/b><\/span><\/span><\/span>W =<\/b><\/span><\/span><\/span> 640J x 0,1 = <\/b><\/span><\/span><\/span>64J.<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

Observa\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/p>\n

–<\/b><\/span><\/span><\/span> As for\u00e7a na <\/b><\/span><\/span><\/span>dire\u00e7\u00e3o do movimento<\/b><\/span><\/span><\/span>, for\u00e7a de atrito <\/b><\/span><\/span><\/span>(entrando na folha)<\/b><\/span><\/span><\/span> e a for\u00e7a motora <\/b><\/span><\/span><\/span>(saindo da<\/b><\/span><\/span><\/span> <\/b>folha)<\/b><\/span><\/span><\/span> se anulam, pois o sistema, pelo enunciado se move com <\/b><\/span><\/span><\/span>velocidade constante<\/b><\/span><\/span><\/span> (equil\u00edbrio din\u00e2mico, for\u00e7a resultante nula).<\/b><\/span><\/span><\/span><\/p>\n

– <\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>soma vetorial das fo\u00e7as normal e peso <\/b><\/span><\/span><\/span>fornecem a for\u00e7a <\/b><\/span><\/span><\/span>resultante centr\u00edpeta dirigida<\/b><\/span><\/span><\/span> para o <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

centro C<\/b><\/span><\/span><\/span> da circunfer\u00eancia <\/b><\/span><\/span><\/span>de raio R<\/b><\/span><\/span><\/span> conforme voc\u00ea pode observar na figura.<\/b><\/span><\/span><\/span><\/p>\n

10- <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span> 43,2km\/h\/3,6=<\/b><\/span><\/span><\/span>12m\/s.<\/b><\/span><\/span><\/span><\/p>\n

Na vertical a=g=10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b> <\/b>V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span> = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=12.0,5=<\/b><\/span><\/span><\/span>6m\/s<\/b><\/span><\/span><\/span> na <\/b><\/span><\/span><\/span>altura m\u00e1xima<\/b><\/span><\/span><\/span> (ponto mais alto da trajet\u00f3ria) a componente <\/b><\/span><\/span><\/span>vertical da velocidade \u00e9 nula V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span> = 0.<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o de Torricelli <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> = V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> \u2013 2gh <\/b><\/span><\/span><\/span> 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 6<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \u2013 2.10.h h = 36\/20 <\/b><\/span><\/span><\/span>h = 1,8m.<\/b><\/span><\/span><\/span><\/p>\n

Nessa altura <\/b><\/span><\/span><\/span>h=1,8m<\/b><\/span><\/span><\/span>, a <\/b><\/span><\/span><\/span>energia potencial<\/b><\/span><\/span><\/span> (gravitacional) do sistema de massa <\/b><\/span><\/span><\/span>m=80 kg<\/b><\/span><\/span><\/span> vale <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span> <\/b>m.g.h<\/b><\/span><\/span><\/span> = 80.10.1,8 <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = 1440J < 1500J.<\/b><\/span><\/span><\/span><\/p>\n

Afirmativa verdadeira.<\/b><\/span><\/span><\/span><\/p>\n

11- <\/b><\/span><\/span><\/span>Se voc\u00ea <\/b><\/span><\/span><\/span>mudar a rela\u00e7\u00e3o entre os dentes coroa-catraca<\/b><\/span><\/span><\/span>, mantendo o ritmo das pedaladas, voc\u00ea automaticamente estar\u00e1 variando a velocidade da bicicleta, no caso, <\/b><\/span><\/span><\/span>diminuindo.<\/b><\/span><\/span><\/span><\/p>\n

Afirmativa errada.<\/b><\/span><\/span><\/span><\/p>\n

12- <\/b><\/span><\/span><\/span>O <\/b><\/span><\/span><\/span>alcance m\u00e1ximo<\/b><\/span><\/span><\/span> ocorre quando o <\/b><\/span><\/span><\/span>\u00e2ngulo com a horizontal for de 45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>, o que inviabiliza a afirmativa.<\/b><\/span><\/span><\/span><\/p>\n

Afirmativa errada.<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

Confira os exerc\u00edcios<\/a><\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES 01- 02- 03- 04- 05- 06- 07- For\u00e7a resultante sobre o conjunto ciclista + bicicleta de massa m = 10 + 70 = 80kg FR = m.a = 80.2 FR = 160N. O trabalho \u00e9 fornecido por W = F.d = 160.4 W = 640J x 0,1 = 64J. 08- 09- Observa\u00e7\u00f5es: – As for\u00e7a na dire\u00e7\u00e3o do movimento,<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":299,"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-301","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/301","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=301"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/301\/revisions"}],"predecessor-version":[{"id":302,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/301\/revisions\/302"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/299"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=301"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}