{"id":3629,"date":"2018-02-13T12:50:45","date_gmt":"2018-02-13T12:50:45","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3629"},"modified":"2024-08-20T15:55:03","modified_gmt":"2024-08-20T15:55:03","slug":"resolucao-comentada-aplicacoes-das-leis-de-newton-com-atrito","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/aplicacoes-das-leis-de-newton-com-atrito\/resolucao-comentada-aplicacoes-das-leis-de-newton-com-atrito\/","title":{"rendered":"Aplica\u00e7\u00f5es das leis de Newton com atrito – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o Comentada<\/b><\/span><\/span><\/span><\/p>\n

Aplica\u00e7\u00f5es das leis de Newton com atrito<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A velocidade do surfista se torna constante a partir de t=10s quando y\u2248330m (do enunciado e do gr\u00e1fico)\u00a0 —\u00a0 a partir da\u00ed, ele percorre\u00a0\u2206S\u2248535- 330=205m, num intervalo de tempo \u2206t=14 \u2013 10=4s\u00a0 —\u00a0 nesse intervalo de tempo, sua velocidade \u00e9 constante e<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

vale\u00a0 —\u00a0 V= \u2206S\/\u2206t =205\/4\u00a0 —\u00a0 V \u2248 51,25 m\/s\u00a0 (velocidade com que ele chega ao solo, que \u00e9 a velocidade limite)\u00a0 —\u00a0 do enunciado a for\u00e7a de resist\u00eancia do ar \u00e9 fornecida pela express\u00e3o\u00a0 —\u00a0 R=K.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 quando ele atinge a velocidade limite, ele deixa de acelerar e, nesse instante, sua velocidade se torna constante sendo que a condi\u00e7\u00e3o para que isso ocorra \u00e9 que as intensidades da for\u00e7a de resist\u00eancia do ar (vertical e para cima) se iguale \u00e0 intensidade da for\u00e7a peso (vertical e para baixo)\u00a0 —R=P=m.g=75.10=750N\u00a0 —\u00a0 750=K.(50)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0K=0,3Ns<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>As membranas interdigitais das patas funcionam como p\u00e1ra-quedas aumentando a for\u00e7a de resist\u00eancia do ar (devido ao aumento da \u00e1rea de contato com o ar)\u00a0 —\u00a0\u00a0 chega um momento em que a intensidade da for\u00e7a de resist\u00eancia do ar \u00a0fica igual \u00e0 intensidade da for\u00e7a peso\u00a0\u00a0e ele entra em equil\u00edbrio din\u00e2mico (for\u00e7a resultante nula\u00a0\"\") e sua velocidade vertical, nesse instante, \u00e9 chamada velocidade terminal ou velocidade limite\u00a0 —\u00a0 essa velocidade permanece constante at\u00e9 ele chegar ao solo\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Nas figuras abaixo P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 a intensidade da parcela do peso que \u00e9 respons\u00e1vel pela tend\u00eancia de descida do bloco e F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 a intensidade da for\u00e7a de atrito\u00a0 — \u00a0como P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=Psen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0e F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcPcos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0eles s\u00e3o os mesmos nas duas situa\u00e7\u00f5es (fig.1 e fig. 2).\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Figura 1\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=mgsen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=5.10.0,5\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=25N\u00a0 —\u00a0 comparando V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a.t com V=2,0.t, voc\u00ea conclui que V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0 e a=2m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

\"\"<\/p>\n

Aplicando a lei fundamental da din\u00e2mica com a acelera\u00e7\u00e3o para baixo\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a\u00a0 —\u00a0 (P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013F \u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>)=ma\u00a0 —\u00a0 25 \u2013 F \u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=5.2\u00a0 —\u00a0 F + F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=15 (I)\u00a0 —\u00a0 figura 2\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=25N\u00a0 —\u00a0 comparando V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a\u2019t com V=3,0.t voc\u00ea conclui que V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0 e a\u2019=3m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Aplicando a segunda lei da din\u00e2mica com a acelera\u00e7\u00e3o para baixo\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=ma\u2019\u00a0 —\u00a0 (25 + F \u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>)=5.3\u00a0 —\u00a0 F = F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 10 (II)\u00a0 —\u00a0 substituindo (II) em (I)\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 10 + F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0=15\u00a0 —\u00a0 2F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=25\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=12,5N<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 F + F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=15\u00a0 —\u00a0 F + 12,5=15\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F= 2,5N\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcmgcos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 12,5=.\u03bc5.10.\u221a3\/2\u00a0 —\u00a0\u00a0\u03bc=12,5\/25\u221a3\u00a0 —\u00a0\u00a0\u03bc=1\/2\u221a3\u00a0 —\u00a0 \u03bc\u22480,29\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A intensidade da for\u00e7a de atrito est\u00e1tico entre as caixa e a carroceria vale F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>=\u03bcN\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>=\u03bcP\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>=\u03bcmg\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>=0,1.500.10\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>=500N (acima desse valor as caixas se movem).<\/b><\/span><\/span><\/span><\/p>\n

\u00a0 No movimento acelerado\u00a0 —\u00a0 V=Vo\u00a0+ a.t\u00a0 —\u00a0 30=0 + a.20\u00a0 —\u00a0 a=1,5m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=500.1,5\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=750N ( como F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>>F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>, as caixas se movem para tr\u00e1s e ficam juntas na parte traseira do caminh\u00e3o).<\/b><\/span><\/span><\/span><\/p>\n

No movimento uniforme, elas est\u00e3o em equil\u00edbrio din\u00e2mico e F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=0. Assim, elas n\u00e3o se movem e ficam juntas na parte traseira.<\/b><\/span><\/span><\/span><\/p>\n

No movimento retardado\u00a0 —\u00a0 V=Vo\u00a0+ a.t\u00a0 —\u00a0 0=30 \u2013 a.40\u00a0 —\u00a0 a=3\/4m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=500.3\/4\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=375N (menor que 500N, portanto elas n\u00e3o se movem e ficam na parte traseira do caminh\u00e3o)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A for\u00e7a de atrito devido aos pneus independe da largura (\u00e1rea de contato dos mesmos com o solo)\u00a0 —\u00a0 o aerof\u00f3lio, que funciona como uma asa de avi\u00e3o invertida, comprime o carro contra o solo aumentando a intensidade da for\u00e7a normal\u00a0\"\"\u00a0que o carro troca com o solo, aumentando assim a intensidade da for\u00e7a de atrito, pois F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcN\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

Observa\u00e7\u00e3o: Al\u00e9m do aerof\u00f3lio, o formato do carro tamb\u00e9m influi na compress\u00e3o do carro contra o solo (veja figura), aumentando<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

ainda mais a for\u00e7a de atrito fazendo com que o carro efetue curvas com velocidades maiores.<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>No equil\u00edbrio est\u00e1tico o corpo est\u00e1 na imin\u00eancia de movimento\u00a0 —\u00a0 a for\u00e7a de atrito est\u00e1tico \u00e9 <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

dada por F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>=\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>N, e se equilibra com o peso P, ou seja, F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>=P\u00a0 —\u00a0 na situa\u00e7\u00e3o do exerc\u00edcio, a for\u00e7a normal N \u00e9 tal que N = F\u00a0 —\u00a0 assim, \u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>N = P\u00a0 —\u00a0 \u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>F=P\u00a0 — \u00a0\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u00a0= P\/F\u00a0 —\u00a0 o fato de \u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u00a0< 1 implica que\u00a0 P\/F<1\u00a0 —\u00a0 0 < P < F\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Colocando todas as for\u00e7as que se relacionam com a pessoa, a parede e o solo.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 for\u00e7as trocadas entre a pessoa e a parede de intensidade 120N\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0e F<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0 for\u00e7as de atrito trocadas entre os p\u00e9s da pessoa e o solo\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 peso da pessoa trocada com o centro da Terra\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sub><\/span>, for\u00e7a normal que a pessoa troca com o solo.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Como a pessoa est\u00e1 em repouso, a for\u00e7a resultante sobre ela \u00e9 nula e as for\u00e7as verticais e horizontais se anulam\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcN, e como \u03bc \u00e9 constante, a for\u00e7a de atrito \u00e9 diretamente proporcional \u00e0 intensidade da for\u00e7a normal (trocada entre a caixa e o solo)\u00a0 —\u00a0 quanto menor N, menor ser\u00e1 as for\u00e7a que a caixa troca com o solo, tornando-a mais \u201cleve\u201d e consequentemente menor a for\u00e7a de atrito — das alternativas, a \u00fanica for\u00e7a capaz de diminuir a\u00a0 for\u00e7a de compress\u00e3o N da caixa com o ch\u00e3o \u00e9 a C\u00a0 <\/b><\/span><\/span><\/span>—<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Colocando as for\u00e7as e decompondo\u00a0\"\"em\u00a0\"\"\u00a0e\u00a0\"\"<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0P=mg=42.10\u00a0 —\u00a0\u00a0P=420N\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>=Fcos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=F.0,5\u00a0 —\u00a0\u00a0F<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>=0,5F\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>=Fsen60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=F.\u221a3\/2\u00a0 —\u00a0\u00a0F<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>=0,85F.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0No equil\u00edbrio est\u00e1tico temos, na horizontal F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>\u00a0e na vertical, P=N + F<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 N=P – F<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 \u03bcN=0,5F\u00a0 —\u00a0 \u03bc(P – F<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>)=0,5F\u00a0 —\u00a0 0,4(420 – 0,85F)=0,5F\u00a0 —\u00a0 168 – 0,34F=0,5F\u00a0 —\u00a0 0,84F=168\u00a0 —\u00a0 F=200N (qualquer valor acima de 200N desloca o bloco)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

10- <\/b><\/span><\/span><\/span>Vamos chamar de L\u2019, comprimento total do pano e de L\u2019 \u2013 L o comprimento da parte pendente\u00a0 —\u00a0 como o pano \u00e9 de constitui\u00e7\u00e3o homog\u00eanea, sua densidade \u00e9 a mesma.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

d=m<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 d (densidade) \u2013 m<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0(massa total) \u2013 V<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0(volume total)\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=d.V<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=d.(A.L\u2019), sendo A (\u00e1rea se se\u00e7\u00e3o transversal, espessura do pano) e L\u2019 (comprimento total do pano)\u00a0 —\u00a0 massa apoiada\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u2013 m<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=d.A.L e massa pendente \u2013 m<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=d.A.(L\u2019 \u2013 L)\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"– peso da parte pendente\u00a0 —\u00a0 P=m<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>g (m<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>, massa da parte pendente)\u00a0 —\u00a0 a for\u00e7a de atrito entre o pano de comprimento L e a mesa est\u00e1 segurando o peso da parte pendente de comprimento (L\u2019 \u2013 L), ou seja, F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=P\u00a0 —\u00a0 \u03bcm<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>g= m<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>g\u00a0 —\u00a0 \u03bc.d.A.L.g=d.A.(L\u2019 \u2013 L).g\u00a0 —\u00a0\u00a0 0,5.L=(60 \u2013 L)\u00a0 —\u00a0 1,5L=60\u00a0 —\u00a0\u00a0L=40cm\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A for\u00e7a de atrito pode servir como for\u00e7a motora no deslocamento de um ve\u00edculo\u00a0 — \u00a0observe a <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

figura\u00a0 —\u00a0 se voc\u00ea quiser acelerar o carro para a esquerda, o motor do mesmo deve fazer o eixo e consequentemente a roda girar no sentido anti-hor\u00e1rio\u00a0 —\u00a0 o pneu empurra o solo para tr\u00e1s (\"\") e o solo reage sobre o pneu e consequentemente sobre o carro (\"\"), movendo-o para a esquerda\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

em rodas com tra\u00e7\u00e3o a for\u00e7a de atrito sobre o m\u00f3vel \u00e9 a favor do movimento e, em rodas sem tra\u00e7\u00e3o, \u00e9 contr\u00e1ria ao movimento\u00a0 — <\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados: m = 120 kg; g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>; m = 0,8; F\/A= 30 N\/cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>; N = P = 1.200 N\u00a0 —\u00a0 para colocar a caixa em movimento o operador tem que vencer a for\u00e7a de atrito\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>oper\u00e1rio<\/b><\/span><\/span><\/sub><\/span>\u00a0= F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0= \u03bcN=0,8 (1.200) = 960 N\u00a0 —\u00a0 como ele est\u00e1 usando os dois bra\u00e7os, cada bra\u00e7o aplica ent\u00e3o\u00a0 —\u00a0 \u00a0F = F<\/b><\/span><\/span><\/span>oper\u00e1rio<\/b><\/span><\/span><\/sub><\/span>\/2=480 N\u00a0 —\u00a0 F\/A=30\u00a0 —\u00a0 480\/A=30\u00a0 —\u00a0 A=16cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0—\u00a0 uma das prote\u00ednas respons\u00e1veis pela contra\u00e7\u00e3o das miofibrilas \u00e9 a actina\u00a0 —<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u00a0\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0,60\u00a0 —\u00a0 \u00a0\u03bc<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0,80\u00a0 —\u00a0 \u00a0m = 1.200 kg\u00a0 —\u00a0 \u00a0for\u00e7a que a pista exerce no ve\u00edculo tem duas componentes\u00a0 —\u00a0 normal\u00a0\"\"\u00a0\u00a0e \u00a0atrito\u00a0\"\"\u00a0\u00a0\u00a0—\u00a0 supondo que a frenagem ocorra em pista horizontal, a componente normal\u00a0\"\"\u00a0da for\u00e7a que a pista aplica no ve\u00edculo tem intensidade igual \u00e0 do seu peso\u00a0\"\"\u00a0\u00a0—\u00a0 N = P = m g = 12.000 N\u00a0 —\u00a0 \u00a0componente de atrito est\u00e1tico m\u00e1xima\u00a0 —\u00a0 \u00a0F<\/b><\/span><\/span><\/span>atmax<\/b><\/span><\/span><\/sub><\/span>= \u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>N = 0,8 (12.000)\u00a0 —\u00a0 \u00a0\u00a0F<\/b><\/span><\/span><\/span>atmax<\/b><\/span><\/span><\/sub><\/span>\u00a0= 9.600 N\u00a0 —\u00a0 a componente de atrito cin\u00e9tico: F<\/b><\/span><\/span><\/span>atc<\/b><\/span><\/span><\/sub><\/span>\u00a0 = \u03bc<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0N = 0,6. (12.000)\u00a0 —\u00a0 \u00a0F<\/b><\/span><\/span><\/span>atc<\/b><\/span><\/span><\/sub><\/span>\u00a0= 7.200 N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0<\/b><\/span><\/span><\/span><\/p>\n

Observa\u00e7\u00f5es: As figuras abaixo ilustram algumas das utilidades dos freios ABS<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A for\u00e7a de tra\u00e7\u00e3o que a corda aplica nas m\u00e3os da garota \u00e9 a for\u00e7a que est\u00e1 puxando-as para a esquerda\u00a0 —\u00a0 a for\u00e7a de atrito \u00e9 paralela ao apoio horizontal e para a direita, pois \u00e9 sempre contr\u00e1rio ao movimento ou \u00e0 sua tend\u00eancia\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o Comentada Aplica\u00e7\u00f5es das leis de Newton com atrito 01-\u00a0A velocidade do surfista se torna constante a partir de t=10s quando y\u2248330m (do enunciado e do gr\u00e1fico)\u00a0 —\u00a0 a partir da\u00ed, ele percorre\u00a0\u2206S\u2248535- 330=205m, num intervalo de tempo \u2206t=14 \u2013 10=4s\u00a0 —\u00a0 nesse intervalo de tempo, sua velocidade \u00e9 constante e vale\u00a0 —\u00a0 V= \u2206S\/\u2206t =205\/4\u00a0 —\u00a0 V \u2248 51,25<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3627,"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-3629","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3629","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=3629"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3629\/revisions"}],"predecessor-version":[{"id":10628,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3629\/revisions\/10628"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3627"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3629"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}