{"id":1295,"date":"2015-09-02T01:27:16","date_gmt":"2015-09-02T01:27:16","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1295"},"modified":"2024-08-23T13:07:20","modified_gmt":"2024-08-23T13:07:20","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-forca-de-resistencia-do-ar","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/forca-de-resistencia-do-ar\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-forca-de-resistencia-do-ar\/","title":{"rendered":"For\u00e7a de resist\u00eancia do ar – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span><\/p>\n

For\u00e7a de resist\u00eancia do ar<\/b><\/span><\/span><\/span><\/p>\n

01<\/b><\/span><\/span>–<\/b><\/span><\/span>\u00a0R- A \u2013 vide teoria<\/b><\/span><\/span><\/p>\n

02<\/b><\/span><\/span>– R- C- vide teoria<\/b><\/span><\/span><\/p>\n

03<\/b><\/span><\/span>– D \u2013 vide teoria<\/b><\/span><\/span><\/p>\n

04<\/b><\/span><\/span>– Como ela cai com velocidade constante, est\u00e1 em equil\u00edbrio din\u00e2mico (for\u00e7a resultante nula). Assim, P=F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=3,2.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=K.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 3,2.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>=8.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=\u00d6(4.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=2.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>m\/s\u00a0 R- B<\/b><\/span><\/span><\/p>\n

05-\u00a0<\/b><\/span><\/span>O peso P do conjunto \u00e9 o mesmo tanto para o p\u00e1ra-quedas aberto como fechado.<\/b><\/span><\/span><\/p>\n

p\u00e1ra-quedas fechado \u2013 velocidade constante \u2013 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=K<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P=K<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 I<\/b><\/span><\/span><\/p>\n

p\u00e1ra-quedas aberto \u2013 velocidade constante \u2013 F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=K<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>P=K<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2\u00a0\u00a0<\/b><\/span><\/span><\/sup>II<\/b><\/span><\/span><\/p>\n

igualando I com II\u00a0 —\u00a0 K<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=K<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 mas K<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=100K<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 K<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=100K<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=1\/100\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=1\/10=0,1<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span>\u00a0a)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) velocidade constante\u00a0 —\u00a0 F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=P=mg=4.10\u00a0 —\u00a0 F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=40N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=K.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 40=2,5.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=4m\/s<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span>\u00a0a) A medida que o conjunto vai caindo a velocidade e a for\u00e7a de resist\u00eancia do ar v\u00e3o aumentando at\u00e9 que F fique igual a P.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A partir da\u00ed o conjunto entra em equil\u00edbrio din\u00e2mico e ele cai com velocidade constante. (a mesma que tinha quando F se igualou a P e a mesma com que ele chega ao solo).<\/b><\/span><\/span><\/p>\n

b) F=P=mg=50.10=500N\u00a0 —\u00a0 F=500N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 F=b.V\u00a0 —\u00a0 500=b.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>b=50N.s\/m<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span>\u00a0Como o exerc\u00edcio n\u00e3o se referiu a desprezar os atritos, existe ar e nesse caso a alternativa correta \u00e9 a B<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span>\u00a0Calculando a constante K pelo gr\u00e1fico\u00a0 —\u00a0 F=K.V\u00a0 —\u00a0 4.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>=K.2\u00a0 —\u00a0 K=2.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>Ns\/m\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Ap\u00f3s ter velocidade constante F=P<\/b><\/span><\/span><\/p>\n

F=mg=10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.10=10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0 F=K.V\u00a0 —\u00a0 10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>=2.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.V\u00a0 —\u00a0 V=10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\/2.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=5m\/s\u00a0 R-C<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span>\u00a0a) Como a caixa cai com velocidade constante, j\u00e1 foi atingido o equil\u00edbrio e F = P =mg=0,2.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=2N<\/b><\/span><\/span><\/p>\n

b) Vamos calcular o intervalo de tempo\u00a0Dt que a caixa demora para, em movimento vertical caie\u00a0DS=2cm=2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m, com velocidade constante\u00a0 V=10m\/s.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

V=DS\/Dt\u00a0 —\u00a0 10=2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/Dt\u00a0 —\u00a0\u00a0Dt =2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/10\u00a0 —\u00a0\u00a0Dt=2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>s. Mas, esse tempo \u00e9 o mesmo que a bala, em movimento horizontal com velocidade V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>, demora para atravessar a caixa e percorrer\u00a0DS=50cm=5.10<\/b><\/span><\/span>=1<\/b><\/span><\/span><\/sup>m.\u00a0 —\u00a0 V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=DS\/Dt\u00a0 —\u00a0 V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=5.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\/2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=250m\/s\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span>\u00a01) Como a velocidade \u00e9 constante a acelera\u00e7\u00e3o \u00e9 nula (equil\u00edbrio din\u00e2mico).<\/b><\/span><\/span><\/p>\n

2) Correta, pois como, nesse instante, a velocidade vertical de queda \u00e9 nula a for\u00e7a de resist\u00eancia do ar n\u00e3o existe e a for\u00e7a resultante sobre ele \u00e9 a gravitacional (seu peso), puxando-o para baixo.<\/b><\/span><\/span><\/p>\n

3) Sem ar, trata-se de uma queda livre com a=g=10m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ g.t\u00a0 —\u00a0 V=0 + 10.10\u00a0 —\u00a0 V=100m\/s que multiplicado por 3,6 nos fornece 36km\/h. Correta.<\/b><\/span><\/span><\/p>\n

4) Errada, a for\u00e7a de resist\u00eancia do ar \u00e9 menor que a for\u00e7a da gravidade (peso) e vai aumentando at\u00e9 iguala-la e, a partir da\u00ed cai com velocidade constante.<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span>\u00a0a)A velocidade se torna constante a partir de t=10s quando y=325m (enunciado e gr\u00e1fico). A partir da\u00ed, ele percorre\u00a0DS=525-<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0325=200\u00a0 —\u00a0\u00a0DS=200m num intervalo de tempo\u00a0Dt=14-10\u00a0 —\u00a0\u00a0Dt=4s, com V constante de valor: V=DS\/Dt\u00a0 —\u00a0 V=200\/4\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

V=50m\/s<\/b><\/span><\/span>, que \u00e9 a velocidade com que ele chega ao solo (velocidade limite).<\/b><\/span><\/span><\/p>\n

b) R=K.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R=P=m.g=75.10=750N\u00a0 —\u00a0 750=K.(50)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0K=0,3Ns<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

c) Quando sua velocidade for de 25m\/s \u2013 V=25m\/s e K=0,3Ns<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, a for\u00e7a de resist\u00eancia do ar vale\u00a0 —\u00a0 R=K.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R=0,3.25<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

R=187,5N\u00a0 —\u00a0 P=750N\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 P \u2013 R=ma\u00a0 —\u00a0 750 – 187,5=75.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=7,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

13-<\/b><\/span><\/span>\u00a0R- D vide teoria<\/b><\/span><\/span><\/p>\n

14-\u00a0<\/b><\/span><\/span>As membranas interdigitais das patas funcionam como p\u00e1ra-quedas aumentando a for\u00e7a de resist\u00eancia do ar fazendo com que sua velocidade tenda a um valor limite, a partir da qual cair\u00e1 com velocidade constante\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

15-\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span><\/sub>=KV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 K=F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span><\/sub>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>=m.aV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>=(m.V\/t)\/V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 K=(m\/t)\/V\u00a0 —\u00a0 K=(kg\/s)\/(m\/s)\u00a0 —\u00a0 K=kg\/s x s\/m\u00a0 —\u00a0 K=kg\/m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

16-<\/b><\/span><\/span>Se voc\u00ea n\u00e3o domina a teoria desse exerc\u00edcio, atentamente as informa\u00e7\u00f5es a seguir:\u00a0Quando corpos se movimentam num fluido (ar ou \u00e1gua), al\u00e9m do peso que \u00e9 constante surge tamb\u00e9m uma for\u00e7a, contr\u00e1ria ao movimento, que chamamos de for\u00e7a de resist\u00eancia do ar (\"\"), que depende da velocidade do corpo, de sua forma e da \u00e1rea de sec\u00e7\u00e3o transversal em rela\u00e7\u00e3o \u00e0 dire\u00e7\u00e3o do movimento nesse meio.<\/b><\/span><\/span><\/p>\n

Assim, sobre um p\u00e1ra-quedista no ar, com o para quedas fechado, surgem sempre na dire\u00e7\u00e3o do movimento (vertical), duas for\u00e7as:\u00a0seu peso (\"\") que \u00e9\u00a0 sempre\u00a0constante, para baixo e a for\u00e7a de resist\u00eancia do ar (\"\"), que \u00e9 vari\u00e1vel e sempre para cima.<\/b><\/span><\/span><\/p>\n

Sem p\u00e1ra-quedas ele deve manter sempre o corpo na horizontal para aumentar a resist\u00eancia do ar.<\/b><\/span><\/span><\/p>\n

No in\u00edcio da queda, quando a velocidade vertical \u00e9 nula,\u00a0\"\"=\u00a0\"\"\u00a0e sobre ele age apenas a for\u00e7a peso, acelerando-o para baixo (figura 1).<\/b><\/span><\/span><\/p>\n

\"\"\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\"\"<\/p>\n

A partir da\u00ed, sendo\u00a0\"\">\"\"\u00a0ele cai acelerando e sua velocidade vai aumentando e\u00a0\"\"\u00a0tamb\u00e9m, pois quanto maior a velocidade maior ser\u00e1\u00a0\"\"\u00a0(figura 2).<\/b><\/span><\/span><\/p>\n

Chega um momento em que a intensidade de\u00a0\"\"\u00a0fica igual \u00e0 intensidade da for\u00e7a peso\u00a0\"\"\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 (que permanece a mesma at\u00e9 ele abrir o p\u00e1ra-quedas). Essa \u00e9 a primeira velocidade limite . Observe que ela \u00e9 maior para a pessoa mais pesada, pois para ela, como seu peso \u00e9 maior, ele demora mais tampo para equilibrar a for\u00e7a de resist\u00eancia do ar e sua velocidade ser\u00e1 maior .\u00a0Quando ele abre o p\u00e1ra-quedas, a \u00e1rea de contato com o ar aumenta, aumentando tamb\u00e9m a for\u00e7a de resist\u00eancia do ar\u00a0\"\"\u00a0que fica maior que o peso\u00a0\"\"\u00a0(figura 4).<\/b><\/span><\/span><\/p>\n

Como, agora,\u00a0\"\">\"\"\u00a0ele desacelera diminuindo\u00a0\"\"\u00a0at\u00e9 que novamente eles se igualem\u00a0\"\"=\"\"\u00a0 e o p\u00e1ra-quedista come\u00e7a a cair novamente com velocidade constante (figura 5), que ser\u00e1 novamente maior para a pessoa mais pesada, pelo mesmo motivo que o anterior.<\/b><\/span><\/span><\/p>\n

\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\"\"<\/p>\n

Essa segunda velocidade limite que \u00e9 a velocidade com que ele chega ao solo, sendo maior para o corpo mais pesado\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E.<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span>\u00a0Primeira explica\u00e7\u00e3o\u00a0 — verdadeira —\u00a0 observe no gr\u00e1fico que, entre t<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0a velocidade \u00e9 constante e, consequentemente a acelera\u00e7\u00e3o tamb\u00e9m ser\u00e1 nula, assim como a for\u00e7a resultante, pois a for\u00e7a de resist\u00eancia do ar anula a for\u00e7a peso.<\/b><\/span><\/span><\/p>\n

Segunda explica\u00e7\u00e3o\u00a0 —\u00a0 verdadeira\u00a0 —\u00a0 quanto maior a velocidade, maior ser\u00e1 a intensidade da for\u00e7a de resist\u00eancia do ar.<\/b><\/span><\/span><\/p>\n

Terceira explica\u00e7\u00e3o\u00a0 —\u00a0 falsa\u00a0 —\u00a0 como existe atrito com o ar o sistema n\u00e3o \u00e9 conservativo e a energia mec\u00e2nica n\u00e3o se conserva.<\/b><\/span><\/span><\/p>\n

Quarta explica\u00e7\u00e3o\u00a0 —\u00a0 falsa\u00a0 —\u00a0 se a for\u00e7a resultante fosse para cima, a intensidade da for\u00e7a de resist\u00eancia seria maior que a for\u00e7a peso e o p\u00e1ra-quedista subiria.<\/b><\/span><\/span><\/p>\n

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

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

a) Observe a figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) Durante a queda, a velocidade do macaco\u00a0 vai aumentando e, consequentemente a intensidade da for\u00e7a de resist\u00eancia do ar (\"\") tamb\u00e9m, enquanto que o peso(\"\")\u00e9 sempre o mesmo\u00a0 —\u00a0 a intensidade m\u00e1xima da for\u00e7a de resist\u00eancia do ar ocorre quando ela e o peso (\"\") se anulam e a for\u00e7a resultante sobre o macaco fica nula (\"\")\u00a0 —\u00a0 a partir da\u00ed o macaco cai sempre com velocidade constante, com P=F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 . P=m.g=1.10\u00a0 —\u00a0 P=10N\u00a0 —\u00a0 P=F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=10N<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

c) Quando o macaco atinge a vegeta\u00e7\u00e3o ele tem velocidade de v<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=30m\/s e depois, ele p\u00e1ra (v<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=0)\u00a0 —\u00a0 como o enunciado pede para desprezar o trabalho do peso (for\u00e7a gravitacional), a dissipa\u00e7\u00e3o de energia mec\u00e2nica se refere apenas \u00e0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

varia\u00e7\u00e3o de energia cin\u00e9tica\u00a0 —\u00a0 \u2206E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=\u2206E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2 \u2013 m.v<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2=1.0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 \u2013 1.30<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 \u2206E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>= – 450J\u00a0 —\u00a0 energia dissipada\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u2206E<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=450J<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre For\u00e7a de resist\u00eancia do ar 01–\u00a0R- A \u2013 vide teoria 02– R- C- vide teoria 03– D \u2013 vide teoria 04– Como ela cai com velocidade constante, est\u00e1 em equil\u00edbrio din\u00e2mico (for\u00e7a resultante nula). Assim, P=Fr=3,2.10-7N Fr=K.V2\u00a0 —\u00a0 3,2.10-7=8.10-6.V2\u00a0 —\u00a0 V=\u00d6(4.10-2)\u00a0 —\u00a0\u00a0V=2.10-1m\/s\u00a0 R- B 05-\u00a0O peso P do conjunto \u00e9 o mesmo tanto<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1291,"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-1295","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1295","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=1295"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1295\/revisions"}],"predecessor-version":[{"id":10826,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1295\/revisions\/10826"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1291"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1295"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}