{"id":1152,"date":"2015-08-09T22:01:44","date_gmt":"2015-08-09T22:01:44","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1152"},"modified":"2024-08-23T13:53:12","modified_gmt":"2024-08-23T13:53:12","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-o-principio-fundamental-da-dinamica-ou-segunda-lei-de-newton","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/segunda-lei-de-newton-ou-principio-fundamental-da-dinamica\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-o-principio-fundamental-da-dinamica-ou-segunda-lei-de-newton\/","title":{"rendered":"O Princ\u00edpio Fundamental da Din\u00e2mica ou Segunda lei de Newton – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre o Princ\u00edpio Fundamental da Din\u00e2mica ou Segunda lei de Newton<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

\u00a0<\/span><\/p>\n

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

a) Na vertical\u00a0 —\u00a0 35N-20N=15N (para baixo)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Na horizontal\u00a0 —\u00a0 40N-20N=20N (para a direita)<\/b><\/span><\/p>\n

\"\"<\/p>\n

Efetuando a adi\u00e7\u00e3o vetorial<\/b><\/span><\/p>\n

\"\"<\/p>\n

Aplicando Pit\u00e1goras\u00a0 —\u00a0 F<\/b><\/span>R<\/b><\/span><\/sub>2<\/b><\/span><\/sup>\u00a0= (15)<\/b><\/span>2<\/b><\/span><\/sup>\u00a0+ (20)<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span>R<\/b><\/span><\/sub>\u00a0= 25N<\/b><\/span><\/p>\n

F<\/b><\/span>R<\/b><\/span><\/sub>\u00a0= m.a\u00a0 —\u00a0 25=0,5.a\u00a0 —\u00a0 a=50m\/s<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 —\u00a0 a dire\u00e7\u00e3o e sentido de a \u00e9 sempre a mesma que de F<\/b><\/span>R<\/b><\/span><\/sub><\/p>\n

\"\"<\/p>\n

b) Quando as tr\u00eas for\u00e7as tiverem a mesma intensidade e o \u00e2ngulo entre elas for de 120<\/b><\/span>o<\/b><\/span><\/sup>\u00a0a for\u00e7a resultante \u00e9 nula e consequentemente a acelera\u00e7\u00e3o tamb\u00e9m ser\u00e1, ou seja,\u00a0\"\".<\/b><\/span><\/p>\n

c) Somando vetorialmente as duas for\u00e7as de 24N e aplicando a lei dos cossenos:<\/b><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

F<\/b><\/span>1<\/b><\/span><\/sub>2<\/b><\/span><\/sup>\u00a0= (24)<\/b><\/span>2<\/b><\/span><\/sup>\u00a0+ (24)<\/b><\/span>2<\/b><\/span><\/sup>\u00a0+ 2.24.24.cos120<\/b><\/span>o<\/b><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span>1<\/b><\/span><\/sub>2<\/b><\/span><\/sup>\u00a0= (24)<\/b><\/span>2<\/b><\/span><\/sup>\u00a0+ (24)<\/b><\/span>2<\/b><\/span><\/sup>\u00a0+ 2.(24)<\/b><\/span>2<\/b><\/span><\/sup>.(-1\/2)\u00a0 —\u00a0 F<\/b><\/span>1<\/b><\/span><\/sub>=24N<\/b><\/span><\/p>\n

Ent\u00e3o teremos:<\/b><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

Aplicando Pit\u00e1goras\u00a0 —\u00a0 F<\/b><\/span>R<\/b><\/span><\/sub>2<\/b><\/span><\/sup>\u00a0= (24)<\/b><\/span>2<\/b><\/span><\/sup>\u00a0+(10)<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span>R<\/b><\/span><\/sub>=26N<\/b><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

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

\"\"<\/p>\n

03-\u00a0Na vertical\u00a0 —\u00a0 3N para cima.\u00a0 —\u00a0 Na horizontal\u00a0 —\u00a0 4N para a direita\u00a0 —\u00a0 Aplicando Pit\u00e1goras\u00a0 —\u00a0 5N\u00a0 —\u00a0 R- A<\/b><\/span><\/p>\n

 <\/p>\n

04-\u00a0D<\/b><\/span><\/p>\n

05- A\u00a0\u00a0<\/b><\/span><\/p>\n

06- A alternativa a \u00e9 falsa, pois 2F=2m.2a\u00a0 —\u00a0 2F=4ma\u00a0 — F=2ma<\/b><\/span><\/p>\n

A alternativa b \u00e9 correta, pois 2F=2m.a\u00a0 —\u00a0 F=ma<\/b><\/span><\/p>\n

A alternativa c \u00e9 falsa, pois F=2m.2a\u00a0 —\u00a0 F=4ma<\/b><\/span><\/p>\n

A alternativa d \u00e9 falsa, pois 2F=m.a\/2\u00a0 —\u00a0 F=(ma)\/4<\/b><\/span><\/p>\n

A alternativa e \u00e9 falsa, pois 3F=2m.a\/2\u00a0 —\u00a0 F=(ma)\/6<\/b><\/span><\/p>\n

07- Entre 0 e 2s, a velocidade \u00e9 constante e a trajet\u00f3ria reta, portanto trata-se de um MRU (equil\u00edbrio din\u00e2mico) e a for\u00e7a resultante \u00e9 nula. Entre 2s e 4s, o movimento \u00e9 desacelerado e a acelera\u00e7\u00e3o vale\u00a0 —\u00a0 a=(V – V<\/b><\/span>o<\/b><\/span><\/sub>)\/t \u2013 t<\/b><\/span>o<\/b><\/span><\/sub>\u00a0 —\u00a0 a=(0-10)\/(4-2)<\/b><\/span><\/p>\n

—\u00a0 a= – 5m\/s<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 —\u00a0 a for\u00e7a resultante \u00e9 constante e vale F=m.a\u00a0 —\u00a0 F=2.(-5)\u00a0 —\u00a0 F= -10N\u00a0 —\u00a0 como o exerc\u00edcio pede o m\u00f3dulo\u00a0\u00a0 —\u00a0 F=10N.\u00a0 R- A<\/b><\/span><\/p>\n

08- Da equa\u00e7\u00e3o fornecida\u00a0 —\u00a0 a=6m\/s<\/b><\/span>2<\/b><\/span><\/sup>.\u00a0 —\u00a0 F=m.a\u00a0 —\u00a0 F=5.6\u00a0 —\u00a0 F=30N<\/b><\/span><\/p>\n

09- V=V<\/b><\/span>o<\/b><\/span><\/sub>\u00a0+ a.t\u00a0 —\u00a0 0=20 + a.0,10\u00a0 —\u00a0 a= -20\/0,10\u00a0 —\u00a0 a= – 200m\/s<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 —\u00a0 F=m.a\u00a0 —\u00a0 F= -0,40.200\u00a0 —\u00a0 F= -80N\u00a0 —\u00a0 m\u00f3dulo\u00a0 —\u00a0 F=80N<\/b><\/span><\/p>\n

10-\u00a0Observando a figura abaixo, conclu\u00edmos que\u00a0\"\"e que,\u00a0\"\".<\/b><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

Como a for\u00e7a resultante \u00e9 a soma vetorial de todas as for\u00e7as temos\u00a0\"\"\u00a0\u00a0—\u00a0\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/p>\n

\"\"\u00a0 —\u00a0\u00a0\"\"\u00a0\u00a0—\u00a0 F=3.10\u00a0 —\u00a0\u00a0F=30N<\/b><\/span><\/p>\n

11- Decompondo cada for\u00e7a na horizontal\u00a0 —\u00a0\u00a0\"\"—\u00a0\u00a0\"\"\u00a0\u00a0—\u00a0\u00a0\"\"<\/b><\/span><\/p>\n

Resultante na horizontal\u00a0 —\u00a0\u00a0\"\"\u00a0\u00a0—\u00a0 F<\/b><\/span>H<\/b><\/span><\/sub>=4N<\/b><\/span><\/p>\n

Decompondo cada for\u00e7a na vertical<\/b><\/span><\/p>\n

\"\"<\/p>\n

Resultante na vertical:<\/b><\/span><\/p>\n

\"\"<\/p>\n

Ent\u00e3o teremos:\u00a0<\/b><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

Aplicando Pit\u00e1goras\u00a0 —\u00a0\u00a0F<\/b><\/span>R<\/b><\/span><\/sub>=5N<\/b><\/span><\/p>\n

12-1<\/b><\/span>o<\/b><\/span><\/sup>\u00a0movimento\u00a0 —\u00a0 F=ma\u00a0 —\u00a0 m=F\/a\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2<\/b><\/span>o<\/b><\/span><\/sup>\u00a0movimento\u00a0 —\u00a0 4F=ma<\/b><\/span>1<\/b><\/span><\/sub>\u00a0 —\u00a0 m=4F\/a<\/b><\/span>1<\/b><\/span><\/sub>\u00a0 —\u00a0 F\/a=4F\/a<\/b><\/span>1<\/b><\/span><\/sub>\u00a0 —\u00a0\u00a0a<\/b><\/span>1<\/b><\/span><\/sub>=4a<\/b><\/span><\/p>\n

1<\/b><\/span>o<\/b><\/span><\/sup>\u00a0movimento\u00a0 —\u00a0 V=V<\/b><\/span>o<\/b><\/span><\/sub>\u00a0+ a.t\u00a0 —\u00a0 5=0 + a.t\u00a0 —\u00a0\u00a0a.t=5<\/b><\/span><\/p>\n

2<\/b><\/span>o<\/b><\/span><\/sup>\u00a0movimento\u00a0 —\u00a0 V<\/b><\/span>1<\/b><\/span><\/sub>=V<\/b><\/span>o<\/b><\/span><\/sub>\u00a0+ a<\/b><\/span>1<\/b><\/span><\/sub>.t\u00a0 —\u00a0 V<\/b><\/span>1<\/b><\/span><\/sub>=5 + 4.a.t\u00a0 —\u00a0 V<\/b><\/span>1<\/b><\/span><\/sub>=5 + 4.5\u00a0 —\u00a0\u00a0V<\/b><\/span>1<\/b><\/span><\/sub>=25m\/s\u00a0\u00a0\u00a0R- C<\/b><\/span><\/p>\n

13-\u00a0F=ma\u00a0 —\u00a0 3.000=m.1,5\u00a0 —\u00a0 m=2.000kg\u00a0 —\u00a0 I \u2013 correta\u00a0\u00a0 —\u00a0\u00a0 II – falsa<\/b><\/span><\/p>\n

V=V<\/b><\/span>o<\/b><\/span><\/sub>+a.t\u00a0 —\u00a0 V=0+1,5.4\u00a0 —\u00a0 V=6m\/s\u00a0 —\u00a0 III – correta<\/b><\/span><\/p>\n

DS=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t + a.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0\u00a0DS=0.2 + 1,5.4\/2\u00a0 —\u00a0\u00a0DS=3m\u00a0 —\u00a0 IV – falsa<\/b><\/span><\/span><\/p>\n

V \u2013 correta\u00a0 —\u00a0 m=F\/a=constante\u00a0 — F e a s\u00e3o diretamente proporcionais<\/b><\/span><\/p>\n

R \u2013 C<\/b><\/span><\/p>\n

14- Observe na segunda lei de Newton que: F=m.a\u00a0 —\u00a0 a=F\/m\u00a0 —\u00a0 a e m s\u00e3o inversamente proporcionais\u00a0 —\u00a0 R- A<\/b><\/span><\/p>\n

15-\u00a0m<\/b><\/span>1<\/b><\/span><\/sub>=3m<\/b><\/span>2<\/b><\/span><\/sub>\/4\u00a0 —\u00a0 a<\/b><\/span>2<\/b><\/span><\/sub>=a<\/b><\/span>1<\/b><\/span><\/sub>=4m\/s<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 — \u00a0F<\/b><\/span>2<\/b><\/span><\/sub>=m<\/b><\/span>2<\/b><\/span><\/sub>.a<\/b><\/span>2<\/b><\/span><\/sub>\u00a0 —\u00a0 m<\/b><\/span>2<\/b><\/span><\/sub>=8\/4\u00a0 —\u00a0 m<\/b><\/span>2<\/b><\/span><\/sub>=2kg\u00a0 —\u00a0 m<\/b><\/span>1<\/b><\/span><\/sub>=(3.2)\/4\u00a0 —\u00a0 m<\/b><\/span>1<\/b><\/span><\/sub>=1,5kg\u00a0 —\u00a0 F<\/b><\/span>o<\/b><\/span><\/sub>=m<\/b><\/span>1.<\/b><\/span><\/sub>a<\/b><\/span>1\u00a0<\/b><\/span><\/sub>\u00a0—\u00a0 F<\/b><\/span>o<\/b><\/span><\/sub>=1,5.4\u00a0 —\u00a0\u00a0F<\/b><\/span>o<\/b><\/span><\/sub>=6N.<\/b><\/span><\/p>\n

16-\u00a0C\u00e1lculo da acelera\u00e7\u00e3o do cubo de lado L\u00a0 —\u00a0 \u0394S=V<\/b><\/span>o<\/b><\/span><\/sub>t + at<\/b><\/span>2<\/b><\/span><\/sup>\/2\u00a0 —\u00a0 80=0 + a.10<\/b><\/span>2<\/b><\/span><\/sup>\/2\u00a0 —\u00a0 a=1,6m\/s<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 —\u00a0 como o novo cubo \u00e9 id\u00eantico ao anterior, eles possuem a mesma densidade\u00a0 —\u00a0 d=m\/v\u00a0 —\u00a0 d=m\/(L\/2)<\/b><\/span>3<\/b><\/span><\/sup>\u00a0 —\u00a0 observe na express\u00e3o anterior que, se o volume fica 8 vezes menor, a massa tamb\u00e9m ficar\u00e1 8 vezes menor, pois a densidade \u00e9 constante\u00a0 —\u00a0 F=ma\u00a0 —\u00a0<\/b><\/span><\/p>\n

Sendo a for\u00e7a F a mesma, se a massa fica 8 vezes menor a acelera\u00e7\u00e3o dever\u00e1 ficar 8 vezes maior\u00a0 —\u00a0 nova acelera\u00e7\u00e3o\u00a0 —\u00a0 a=8×1,6\u00a0 —\u00a0 a=12,8m\/s<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 —\u00a0 \u0394S\u2019=V<\/b><\/span>o<\/b><\/span><\/sub>t + at<\/b><\/span>2<\/b><\/span><\/sup>\/2=0 + 12,8.10<\/b><\/span>2<\/b><\/span><\/sup>\/2\u00a0 —\u00a0 \u0394S\u2019=640m\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/p>\n

17- R- A\u00a0 —\u00a0 Princ\u00edpio da in\u00e9rcia<\/b><\/span><\/p>\n

18-\u00a0I. Correta\u00a0 —\u00a0 Princ\u00edpio da In\u00e9rcia<\/b><\/span><\/p>\n

II. Correta\u00a0 —\u00a0 se ele sobe a for\u00e7a resultante sobre ele tem que ser para cima.<\/b><\/span><\/p>\n

III. Correta\u00a0 —\u00a0 Correta\u00a0 —\u00a0 quanto maior a for\u00e7a de compress\u00e3o com o solo, maior ser\u00e1 a for\u00e7a de atrito<\/b><\/span><\/p>\n

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

19-\u00a0Se existem for\u00e7as agindo sobre um objeto ele n\u00e3o est\u00e1 necessariamente acelerado o que ocorre somente se a intensidade da for\u00e7a resultante for diferente de zero\u00a0 —\u00a0 mas, se essas for\u00e7as se anularem ele estar\u00e1 em repouso ou em MRU\u00a0 —\u00a0 R- C<\/b><\/span><\/p>\n

20-\u00a0a) As for\u00e7as que agem na massa pendular s\u00e3o o peso e a tra\u00e7\u00e3o.<\/b><\/span><\/p>\n

\"\"<\/p>\n

b) Como o movimento \u00e9 retil\u00edneo, a componente vertical da resultante \u00e9 nula\u00a0 —\u00a0\u00a0 T<\/b><\/span>y<\/b><\/span><\/sub>=P\u00a0 —\u00a0 A resultante \u00e9 ent\u00e3o na dire\u00e7\u00e3o<\/b><\/span><\/p>\n

\"\"<\/p>\n

horizontal\u00a0 —\u00a0 F<\/b><\/span>R<\/b><\/span><\/sub>= T<\/b><\/span>X<\/b><\/span><\/sub>\u00a0 —\u00a0\u00a0 como o vag\u00e3o parte do repouso, ele acelera no sentido da resultante, ou seja, para a direita\u00a0 —\u00a0<\/b><\/span><\/p>\n

\"\"<\/p>\n

c) Do princ\u00edpio fundamental da din\u00e2mica\u00a0 —\u00a0 F<\/b><\/span>R<\/b><\/span><\/sub>\u00a0= m a \u00a0—\u00a0 T<\/b><\/span>X<\/b><\/span><\/sub>\u00a0= m a<\/b><\/span>max<\/b><\/span><\/sub>\u00a0 —\u00a0 como, na vertical, a componente da resultante \u00e9 nula\u00a0 —\u00a0 T<\/b><\/span>y<\/b><\/span><\/sub>\u00a0= P = m g\u00a0 —\u00a0 tg14<\/b><\/span>o<\/b><\/span><\/sup>=T<\/b><\/span>X<\/b><\/span><\/sub>\/T<\/b><\/span>y<\/b><\/span><\/sub>=m.a<\/b><\/span>Max<\/b><\/span><\/sub>\/m.g\u00a0 —\u00a0 0,25=a<\/b><\/span>max<\/b><\/span><\/sub>\/10\u00a0 —\u00a0 a<\/b><\/span>max<\/b><\/span><\/sub>=10.0,25\u00a0 —\u00a0\u00a0a<\/b><\/span>max<\/b><\/span><\/sub>=2,5m\/s<\/b><\/span>2<\/b><\/span><\/sup><\/p>\n

21-\u00a0(I) Incorreta\u00a0 —\u00a0 o trailer \u00e9 uniformemente acelerado apenas no intervalo 0 a t<\/b><\/span>1<\/b><\/span><\/sub>, onde a resultante tem intensidade constante.<\/b><\/span><\/p>\n

(II) Correta\u00a0 —\u00a0 at\u00e9 o instante t<\/b><\/span>4<\/b><\/span><\/sub>\u00a0h\u00e1 uma for\u00e7a resultante acelerando o trailer, fazendo sua velocidade aumentar.<\/b><\/span><\/p>\n

(III) Correta\u00a0 —\u00a0 se a resultante \u00e9 nula, o movimento retil\u00edneo e uniforme.\u00a0<\/b><\/span><\/p>\n

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

22-\u00a0 Intensidade da resultante dessas for\u00e7as\u00a0 —\u00a0 F<\/b><\/span>R<\/b><\/span><\/sub>2<\/b><\/span><\/sup>=F<\/b><\/span>1<\/b><\/span><\/sub>2<\/b><\/span><\/sup>\u00a0+ F<\/b><\/span>2<\/b><\/span><\/sub>2<\/b><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span>R<\/b><\/span><\/sub>=\u221a(2<\/b><\/span>2<\/b><\/span><\/sup>\u00a0+ 1,5<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span>R<\/b><\/span><\/sub>=2,5N\u00a0 —\u00a0 pelo princ\u00edpio fundamental da din\u00e2mica\u00a0 —\u00a0 F<\/b><\/span>R<\/b><\/span><\/sub>=ma\u00a0 —\u00a0 2,5=2.a\u00a0 —\u00a0\u00a0 a=1,25m\/s<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/p>\n

23-\u00a0F<\/b><\/span>1<\/b><\/span><\/sub>=m<\/b><\/span>1<\/b><\/span><\/sub>.a<\/b><\/span>1<\/b><\/span><\/sub>\u00a0 —\u00a0 \u00a0F<\/b><\/span>1<\/b><\/span><\/sub>\u00a0= 2 (3)\u00a0 —\u00a0 F<\/b><\/span>1<\/b><\/span><\/sub>\u00a0= 6 N\u00a0 —\u00a0 F<\/b><\/span>1<\/b><\/span><\/sub>=F<\/b><\/span>2<\/b><\/span><\/sub>=F = m<\/b><\/span>2<\/b><\/span><\/sub>\u00a0a<\/b><\/span>2<\/b><\/span><\/sub>\u00a0 —\u00a0 6 = 1 a<\/b><\/span>2<\/b><\/span><\/sub>\u00a0 —\u00a0 a<\/b><\/span>2<\/b><\/span><\/sub>\u00a0= 6 m\/s<\/b><\/span>2<\/b><\/span><\/sup>\u00a0—\u00a0\u00a0R- D<\/b><\/span><\/p>\n

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

24-\u00a0A dist\u00e2ncia percorrida em um diagrama de velocidade versus tempo \u00e9 dada pela \u00e1rea sob a linha de gr\u00e1fico\u00a0 —\u00a0 aproximando esta figura para um tri\u00e2ngulo ret\u00e2ngulo\u00a0 —\u00a0 d=base.altura\/2=1,4.1\/2\u00a0 —\u00a0 d=0,7m\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/p>\n

25-\u00a0a=(V \u2013 V<\/b><\/span>o<\/b><\/span><\/sub>)\/(t \u2013 t<\/b><\/span>o<\/b><\/span><\/sub>)=(0,4 \u2013 0,8)\/0,8 \u2013 0,4 \u00a0—\u00a0 a=-1m\/s<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 —\u00a0 pela segunda lei de Newton\u00a0 —\u00a0 F=ma=2.(-1)\u00a0 —\u00a0 F=-2N\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/p>\n

26-\u00a0O vetor velocidade\u00a0\"\"\u00e9 tangente \u00e0 trajet\u00f3ria e \u00e9 vertical e para cima na subida e vertical e para baixo na descida\u00a0 —\u00a0 a acelera\u00e7\u00e3o da gravidade\u00a0\"\"\u00a0tem sempre dire\u00e7\u00e3o vertical e sentido para baixo\u00a0 —\u00a0 a for\u00e7a de atrito\u00a0\"\"\u00a0tem dire\u00e7\u00e3o do movimento e sentido contr\u00e1rio a ele, ou seja, ao vetor velocidade\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/p>\n

27-\u00a0A for\u00e7a resultante sobre o elevador \u00e9 nula (P=T) se ele estiver subindo ou descendo em movimento retil\u00edneo e uniforme ou em repouso\u00a0 —\u00a0R- D<\/b><\/span><\/p>\n

28- Se o avi\u00e3o acelera para frente, por in\u00e9rcia, o corpo pendurado no fio tende a ficar parado em rela\u00e7\u00e3o \u00e0 pista e, portanto, vai para tr\u00e1s em rela\u00e7\u00e3o ao avi\u00e3o\u00a0 —\u00a0 o fio inclina-se para a esquerda\u00a0 —\u00a0 a acelera\u00e7\u00e3o do avi\u00e3o\u00a0ser\u00e1 dada por\u00a0 —\u00a0 a = g.tan 25<\/b><\/span>o<\/b><\/span><\/sup>\u00a0= 10.0,47\u00a0 —\u00a0<\/b><\/span><\/p>\n

a = 4,7 m\/s\u00b2\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/p>\n

 <\/p>\n

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

Em todo gr\u00e1fico V x t a dist\u00e2ncia percorrida \u00e9 numericamente igual \u00e0 \u00e1rea hachurada, entre 0 e 5s, da figura abaixo\u00a0\u00a0\"\"<\/b><\/span><\/p>\n

— \u2206S=\u00e1rea\u00a0 —\u00a0 \u2206S=(B + b).h\/2=(20 + 10).5\/2\u00a0 —\u00a0 \u2206S=75m\u00a0 —\u00a0 c\u00e1lculo da acelera\u00e7\u00e3o do carro pelo gr\u00e1fico\u00a0 —\u00a0 a=(V \u2013 V<\/b><\/span>o<\/b><\/span><\/sub>)\/(t \u2013 t<\/b><\/span>o<\/b><\/span><\/sub>)=(0 \u2013 20)\/(10 \u2013 0)\u00a0 —\u00a0 a=- 2m\/s<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 —\u00a0 for\u00e7a resultante\u00a0 —\u00a0 F<\/b><\/span>R<\/b><\/span><\/sub>=m.a=1000.(-2)\u00a0 —\u00a0 F<\/b><\/span>R<\/b><\/span><\/sub>=- 2000N em m\u00f3dulo F<\/b><\/span>R<\/b><\/span><\/sub>=2000N\u00a0 —\u00a0\u00a0R- C.<\/b><\/span><\/p>\n

 <\/p>\n

 <\/p>\n

30-\u00a0Para calcular a intensidade da for\u00e7a resultante que age sobre a part\u00edcula c\u00f3smica voc\u00ea pode decompor as for\u00e7as nas dire\u00e7\u00f5es norte e leste\u00a0 —\u00a0 observe na sequ\u00eancia abaixo que a intensidade da for\u00e7a resultante \u00e9 de 1N no sentido leste\u00a0 —<\/b><\/span><\/p>\n

\"\"<\/p>\n

Como a velocidade inicial da part\u00edcula tem intensidade V<\/b><\/span>o<\/b><\/span><\/sub>=1200m\/s do norte para o sul e a for\u00e7a resultante sobre ela tem intensidade 1N do oeste para leste, o movimento da part\u00edcula tem as caracter\u00edsticas de composi\u00e7\u00e3o de dois movimentos, um no sentido leste e outro no sentido sul (veja figura)\u00a0 —\u00a0 no sentido leste, a proje\u00e7\u00e3o da velocidade inicial \u00e9 nula V<\/b><\/span>oL<\/b><\/span><\/sub>=0 e ela se desloca sob a\u00e7\u00e3o de uma for\u00e7a resultante de valor F<\/b><\/span>R<\/b><\/span><\/sub>=1N e com acelera\u00e7\u00e3o\u00a0 —\u00a0 F<\/b><\/span>R<\/b><\/span><\/sub>=m.a\u00a0 —\u00a0 1=2.10<\/b><\/span>-3<\/b><\/span><\/sup>.a\u00a0 —\u00a0 a= 500m\/s<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 —\u00a0 sua velocidade nessa dire\u00e7\u00e3o ap\u00f3s t=1s ter\u00e1 intensidade\u00a0 —\u00a0 V<\/b><\/span>L<\/b><\/span><\/sub>=V<\/b><\/span>oL<\/b><\/span><\/sub>\u00a0+ a<\/b><\/span>L<\/b><\/span><\/sub>.t=0 + 500.1\u00a0 —\u00a0 V<\/b><\/span>L<\/b><\/span><\/sub>=500m\/s\u00a0 —\u00a0 no sentido sul ela ser\u00e1 lan\u00e7ada para baixo com V<\/b><\/span>oS<\/b><\/span><\/sub>=1200m\/s, acelerando com acelera\u00e7\u00e3o da gravidade g=10m\/s<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 —\u00a0 ap\u00f3s t=1s, sua velocidade nessa dire\u00e7\u00e3o ser\u00e1\u00a0 —\u00a0 V<\/b><\/span>S<\/b><\/span><\/sub>=V<\/b><\/span>oS<\/b><\/span><\/sub>\u00a0+ g.t=1200 + 10.1=1210m\/s\u00a0 —<\/b><\/span><\/p>\n

\"\"<\/p>\n

\u00a0observe na figura que essas duas velocidades s\u00e3o perpendiculares e, aplicando Pit\u00e1goras voc\u00ea obter\u00e1 V<\/b><\/span>2<\/b><\/span><\/sup>\u00a0= V<\/b><\/span>L<\/b><\/span><\/sub>2<\/b><\/span><\/sup>\u00a0+ V<\/b><\/span>S<\/b><\/span><\/sub>2<\/b><\/span><\/sup>=(500)<\/b><\/span>2<\/b><\/span><\/sup>\u00a0+ (1210)<\/b><\/span>2<\/b><\/span><\/sup>\u00a0 —\u00a0 V=\u221a(1714100)\u00a0 —\u00a0 V=1309m\/s=1,3km\/s\u00a0 —\u00a0\u00a0R- A.<\/b><\/span><\/p>\n

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

Pelo princ\u00edpio da in\u00e9rcia, se as for\u00e7as deixarem de atura, a for\u00e7a resultante sobre ela ser\u00e1 nula e, ap\u00f3s esse innstante, por in\u00e9rcia, ela seguir\u00e1 em MRU com velocidade constante de 1,3km\/s\u00a0 —\u00a0 observe na resolu\u00e7\u00e3o do exerc\u00edcio anterior (08) que, antes de 1s a trajet\u00f3ria era parab\u00f3lica\u00a0 —\u00a0<\/b><\/span>\u00a0<\/b><\/span>R- D.<\/b><\/span><\/p>\n

 <\/p>\n

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

Se o deslocamento ocorresse com velocidade constante a for\u00e7a resultante sobre a caixa seria nula\u00a0 —\u00a0 nesse caso, a<\/b><\/span><\/p>\n

\"\"<\/p>\n

for\u00e7a que a pessoa exerce sobre a caixa tem que ter a mesma intensidade que a for\u00e7a de atrito\u00a0 —\u00a0 note que a for\u00e7a que a pessoa exerce sobre a caixa\u00a0 tem a mesma intensidade que a for\u00e7a que a caixa exerce sobre a pessoa (princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o)\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/p>\n

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

Nesse caso, como existe acelera\u00e7\u00e3o a resultante das for\u00e7as sobre a caixa \u00e9 diferente de zero\u00a0 —\u00a0 F<\/b><\/span>R<\/b><\/span><\/sub>=m.a\u00a0 — como a<\/b><\/span><\/p>\n

\"\"<\/p>\n

caixa se desloca na mesma dire\u00e7\u00e3o e sentido que F<\/b><\/span>p<\/b><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span>p<\/b><\/span><\/sub>\u00a0\u2013 F<\/b><\/span>a<\/b><\/span><\/sub>=m.a\u00a0 —\u00a0 F<\/b><\/span>p<\/b><\/span><\/sub>\u00a0><\/b><\/span>\u00a0<\/b><\/span><\/sub>F<\/b><\/span>a<\/b><\/span><\/sub>\u00a0 —\u00a0 note que a for\u00e7a que a pessoa exerce sobre a caixa\u00a0 tem a mesma intensidade que a for\u00e7a que a caixa exerce sobre a pessoa (princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o)\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/p>\n

 <\/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 o Princ\u00edpio Fundamental da Din\u00e2mica ou Segunda lei de Newton \u00a0 \u00a0 01- a) Na vertical\u00a0 —\u00a0 35N-20N=15N (para baixo)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Na horizontal\u00a0 —\u00a0 40N-20N=20N (para a direita) Efetuando a adi\u00e7\u00e3o vetorial Aplicando Pit\u00e1goras\u00a0 —\u00a0 FR2\u00a0= (15)2\u00a0+ (20)2\u00a0 —\u00a0 FR\u00a0= 25N FR\u00a0= m.a\u00a0 —\u00a0 25=0,5.a\u00a0 —\u00a0 a=50m\/s2\u00a0 —\u00a0 a dire\u00e7\u00e3o e sentido de a<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1148,"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-1152","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1152","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=1152"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1152\/revisions"}],"predecessor-version":[{"id":10860,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1152\/revisions\/10860"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1148"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1152"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}