{"id":683,"date":"2015-05-12T17:52:51","date_gmt":"2015-05-12T17:52:51","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=683"},"modified":"2024-09-02T17:46:45","modified_gmt":"2024-09-02T17:46:45","slug":"resolucao-comentada-da-ufla-2","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-ufla-2\/resolucao-comentada-da-ufla-2\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada da UFLA – 2"},"content":{"rendered":"

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

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Na primeira figura est\u00e3o colocadas as for\u00e7as e, considerando a prateleira de densidade homog\u00eanea o peso e todas as outras for\u00e7as est\u00e3o aplicadas no centro (meio) da prateleira e essas for\u00e7as s\u00e3o:<\/span><\/span><\/span><\/p>\n

Peso<\/span><\/span><\/span>\u00a0\"\"<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>— vertical, para baixo e de intensidade P=mg=2.10 —<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>P=20N<\/span><\/span><\/span><\/p>\n

Tens\u00e3o (tra\u00e7\u00e3o)<\/span><\/span><\/span>\u00a0\"\"<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>no cabo aplicada na prateleira pela parede<\/span><\/span><\/span><\/p>\n

Normal<\/span><\/span><\/span>\u00a0\"\"\u00a0<\/span><\/span><\/span>— aplicada pela parede na prateira (horizontal e para a direita)<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na segunda figura a<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>tra\u00e7\u00e3o<\/span><\/span><\/span>\u00a0\"\"\u00a0<\/span><\/span><\/span>foi decomposta em duas parcelas<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>de intensidades<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>T<\/span><\/span><\/span>y<\/span><\/span><\/sub><\/span>=<\/span><\/span><\/span>Tcos60<\/span><\/span><\/span>o<\/span><\/span><\/sup><\/span>=<\/span><\/span><\/span>0,5T<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>e<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>T<\/span><\/span><\/span>x<\/span><\/span><\/sub><\/span>=<\/span><\/span><\/span>Tsen60<\/span><\/span><\/span>o<\/span><\/span><\/sup><\/span>=(\u221a3\/2)T<\/span><\/span><\/span>.<\/span><\/span><\/span><\/p>\n

Na terceira figura, como o sistema est\u00e1 em<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>equil\u00edbrio<\/span><\/span><\/span>, na vertical,<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>T<\/span><\/span><\/span>y<\/span><\/span><\/sub><\/span>=P<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>— 0,5T=20 —<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>T=40N<\/b><\/span><\/span><\/span>.<\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

Obs: Se fosse pedido a rea\u00e7\u00e3o normal<\/span><\/span><\/span>\u00a0\"\"\u00a0<\/span><\/span><\/span>aplicada pela parede na prateira voc\u00ea teria que na horizontal tamb\u00e9m existe equil\u00edbrio e N=F<\/span><\/span><\/span>y<\/span><\/span><\/sub><\/span>=(\u221a3\/2)T=(\u221a3\/2).40 —<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>N=20\u221a3N<\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Trata-se da<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>Primeira lei de Newton (Princ\u00edpio da In\u00e9rcia<\/span><\/span><\/span>).<\/span><\/span><\/span><\/p>\n

R- A<\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se voc\u00ea n\u00e3o domina a teoria, ela est\u00e1 a seguir<\/b><\/span><\/span><\/span>:<\/span><\/span><\/span><\/p>\n

Conserva\u00e7\u00e3o do momento linear (quantidade de movimento) de um sistema<\/span><\/span><\/span>:<\/span><\/span><\/span><\/p>\n

Sistema isolado de for\u00e7as externas<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/span><\/span><\/span>–\u00a0 Um sistema formado por v\u00e1rios corpos ou pontos materiais \u00e9 considerado\u00a0<\/span><\/span><\/span>isolado<\/b><\/span><\/span><\/span>,<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>quando:<\/span><\/span><\/span><\/p>\n

\"\"\u00a0sobre ele n\u00e3o atuam for\u00e7as externas \u2013 exemplo \u2013 um astronauta no espa\u00e7o, fora do campo de a\u00e7\u00e3o gravitacional de qualquer<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

corpo celeste (for\u00e7a peso nula) e no v\u00e1cuo ( n\u00e3o existem for\u00e7as resistivas).<\/span><\/span><\/span><\/p>\n

\"\"\u00a0sobre ele agem for\u00e7as externas, mas a intensidade da for\u00e7a resultante (soma vetorial de todas as for\u00e7as externas que agem sobre<\/span><\/span><\/span><\/p>\n

ele) \u00e9 nula \u2013 exemplo \u2013 um corpo se deslocando, sem atrito, sobre uma superf\u00edcie plana horizontal e em movimento retil\u00edneo e<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

uniforme e nesse caso n\u00e3o existem for\u00e7as horizontais (MRU) e as verticais\u00a0\"\"\u00a0e\u00a0\"\"\u00a0se anulam.<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Quando um sistema \u00e9 isolado de for\u00e7as externas, sua for\u00e7a resultante tem intensidade nula e consequentemente o impulso dessa for\u00e7a \u00e9 nulo, pois I=F<\/span><\/span><\/span>R<\/span><\/span><\/sub><\/span>.Dt=0.\u00a0Dt=0. Como I=Q<\/span><\/span><\/span>depois<\/span><\/span><\/sub><\/span>\u00a0\u2013 Q<\/span><\/span><\/span>antes<\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0= Q<\/span><\/span><\/span>depois<\/span><\/span><\/sub><\/span>\u00a0\u2013 Q<\/span><\/span><\/span>antes<\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q<\/span><\/span><\/span>antes<\/span><\/span><\/sub><\/span>\u00a0= Q<\/span><\/span><\/span>depois<\/span><\/span><\/sub><\/span>. Sendo a quantidade de movimento grandeza vetorial, a rela\u00e7\u00e3o abaixo representa o\u00a0<\/span><\/span><\/span>princ\u00edpio da conserva\u00e7\u00e3o da quantidade de movimento<\/b><\/span><\/span><\/span>:<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\u00a0<\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>o peso da crian\u00e7a (II) \u00e9 duas vezes maior que o da crian\u00e7a (I),<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>a crian\u00e7a (II), para que haja equil\u00edbrio, deve estar a uma dist\u00e2ncia duas vezes menor que a crian\u00e7a (I)<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>do apoio.<\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Trata-se de um<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>lan\u00e7amento vertical para cima onde a acelera\u00e7\u00e3o \u00e9 a da gravidade<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>g(constante)<\/span><\/span><\/span>, mesmo local) e<\/span><\/span><\/span>, \u00e0 medida que ela sobe, a velocidade diminui<\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Corpos<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>abandonados com V<\/span><\/span><\/span>o<\/span><\/span><\/sub><\/span>=0<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>ou lan\u00e7ados verticalmente para baixo com V<\/span><\/span><\/span>o<\/span><\/span><\/sub><\/span>\u22600<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>de um mesmo local e da mesma altura (desprezando-se os efeitos do ar) demoram o mesmo tempo para chegar ao solo, independente de suas massas.<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>velocidade<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>com que eles chegam ao solo tamb\u00e9m<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>\u00e9 a mesma<\/span><\/span><\/span>.<\/span><\/span><\/span><\/p>\n

Mas, a<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>energia cin\u00e9tica \u00e9<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>diferente<\/span><\/span><\/span>, pois<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>E<\/span><\/span><\/span>c<\/span><\/span><\/sub><\/span>=mV<\/span><\/span><\/span>2<\/span><\/span><\/sup><\/span>\/2 e depende da massa do corpo.<\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Peso de cada esfera<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>— P<\/span><\/span><\/span>1<\/span><\/span><\/sub><\/span>=m<\/span><\/span><\/span>1<\/span><\/span><\/sub><\/span>g=0,3.10 —<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>P<\/span><\/span><\/span>1<\/span><\/span><\/sub><\/span>=3N<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>— P<\/span><\/span><\/span>2<\/span><\/span><\/sub><\/span>=m<\/span><\/span><\/span>2<\/span><\/span><\/sub><\/span>g=0,2.10 —<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>P<\/span><\/span><\/span>2<\/span><\/span><\/sub><\/span>=2N<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>— P<\/span><\/span><\/span>3<\/span><\/span><\/sub><\/span>=m<\/span><\/span><\/span>3<\/span><\/span><\/sub><\/span>g=0,1.10 —<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>P<\/span><\/span><\/span>3<\/span><\/span><\/sub><\/span>=1N<\/span><\/span><\/span><\/p>\n

Observe na<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>figura todas as for\u00e7as que agem sobre cada esfera<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>e, como cada uma est\u00e1<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>em equil\u00edbrio<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>a<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>for\u00e7a resultante sobre cada uma ser\u00e1 nula<\/span><\/span><\/span>, ent\u00e3o voc\u00ea ter\u00e1:<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Esfera 3<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>—<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=1N<\/b><\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>—<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>esfera 2<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>— T<\/span><\/span><\/span>3<\/span><\/span><\/sub><\/span>\u00a0<\/span><\/span><\/span>+ 2 = T<\/span><\/span><\/span>2<\/span><\/span><\/sub><\/span>\u00a0<\/span><\/span><\/span>— 1 + 2=T<\/span><\/span><\/span>2<\/span><\/span><\/sub><\/span>\u00a0<\/span><\/span><\/span>—<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=3N<\/b><\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>—<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>esfera 1<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>— T<\/span><\/span><\/span>1<\/span><\/span><\/sub><\/span>\u00a0<\/span><\/span><\/span>= T<\/span><\/span><\/span>2<\/span><\/span><\/sub><\/span>\u00a0<\/span><\/span><\/span>+ 3 —<\/span><\/span><\/span><\/p>\n

T<\/span><\/span><\/span>1<\/span><\/span><\/sub><\/span>=3 + 3 —<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>T<\/span><\/span><\/span>1<\/span><\/span><\/sub><\/span>=6N<\/span><\/span><\/span><\/p>\n

R- A<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>No<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>ponto mais alto A<\/span><\/span><\/span>, nos dois casos elas possuem a<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>mesma velocidade inicial V<\/span><\/span><\/span>o<\/span><\/span><\/sub><\/span>\u00a0<\/span><\/span><\/span>e mesma altura h<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>— assim a energia<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>mec\u00e2nica em A para as duas bolas<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>vale<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>E<\/span><\/span><\/span>mA<\/span><\/span><\/sub><\/span>=mV<\/span><\/span><\/span>o<\/span><\/span><\/sub><\/span>2<\/span><\/span><\/sup><\/span>\/2 + mgh.<\/span><\/span><\/span><\/p>\n

No solo, chegam com velocidade V na altura h=0<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>e de energia mec\u00e2nica — E<\/span><\/span><\/span>mB<\/span><\/span><\/sub><\/span>=mV<\/span><\/span><\/span>2<\/span><\/span><\/sup><\/span>\/2 + m.g.0 —<\/span><\/span><\/span><\/p>\n

E<\/span><\/span><\/span>mB<\/span><\/span><\/sub><\/span>=mV<\/span><\/span><\/span>2<\/span><\/span><\/sup><\/span>\/2.<\/span><\/span><\/span><\/p>\n

Pelo princ\u00edpio da conserva\u00e7\u00e3o da energia mec\u00e2nica E<\/span><\/span><\/span>mA<\/span><\/span><\/sub><\/span>=E<\/span><\/span><\/span>mB<\/span><\/span><\/sub><\/span>\u00a0<\/span><\/span><\/span>— mV<\/span><\/span><\/span>o<\/span><\/span><\/sub><\/span>2<\/span><\/span><\/sup><\/span>\/2 + mgh= mV<\/span><\/span><\/span>2<\/span><\/span><\/sup><\/span>\/2 —<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>V=\u221a(V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/span>+ 2gh)<\/b><\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>—<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>como V<\/span><\/span><\/span>o<\/span><\/span><\/sub><\/span>, g e h s\u00e3o os mesmos elas chegam ao solo com a<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>mesma velocidade<\/span><\/span><\/span>.<\/span><\/span><\/span><\/p>\n

Observe na figura que as<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>trajet\u00f3rias<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>e consequentemente<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>dist\u00e2ncia percorrida<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>pelas bolas \u00e9<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>diferente<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>e, assim, elas chegam ao solo em<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>instantes diferentes<\/b><\/span><\/span><\/span>. Comprova-se que o tempo que a bola permanece no ar est\u00e1 relacionado\u00a0 com a altura\u00a0 —\u00a0 maior altura, maior tempo de perman\u00eancia no ar, conforme figura abaixo\u00a0\u00a0<\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Para calcular o deslocamento vetorial<\/span><\/span><\/span>\u00a0\"\"<\/span><\/span><\/span>pelo m\u00e9todo da linha poligonal basta voc\u00ea unir I com F (figura).<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Para calcular a intensidade de<\/span><\/span><\/span>\u00a0\"\"\u00a0<\/span><\/span><\/span>voc\u00ea aplica Pit\u00e1goras no tri\u00e2ngulo hachurado — d<\/span><\/span><\/span>2<\/span><\/span><\/sup><\/span>=3<\/span><\/span><\/span>2<\/span><\/span><\/sup><\/span>\u00a0<\/span><\/span><\/span>+ 4<\/span><\/span><\/span>2<\/span><\/span><\/sup><\/span>\u00a0<\/span><\/span><\/span>— d=5×10=50cm — V=d\/t=50\/200 — V=0,25m\/s.<\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/h2>\n

Voltar para exerc\u00edcios.<\/span><\/a><\/h2>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES 01-\u00a0Na primeira figura est\u00e3o colocadas as for\u00e7as e, considerando a prateleira de densidade homog\u00eanea o peso e todas as outras for\u00e7as est\u00e3o aplicadas no centro (meio) da prateleira e essas for\u00e7as s\u00e3o: Peso\u00a0\u00a0— vertical, para baixo e de intensidade P=mg=2.10 —\u00a0P=20N Tens\u00e3o (tra\u00e7\u00e3o)\u00a0\u00a0no cabo aplicada na prateleira pela parede Normal\u00a0\u00a0— aplicada pela parede na prateira (horizontal e para a<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":681,"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-683","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/683","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=683"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/683\/revisions"}],"predecessor-version":[{"id":684,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/683\/revisions\/684"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/681"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=683"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}