{"id":1333,"date":"2015-09-15T15:17:39","date_gmt":"2015-09-15T15:17:39","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1333"},"modified":"2024-08-23T14:01:34","modified_gmt":"2024-08-23T14:01:34","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-trabalho-de-diversos-tipos-de-forcas","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/trabalho-de-uma-forca\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-trabalho-de-diversos-tipos-de-forcas\/","title":{"rendered":"Trabalho de diversos tipos de for\u00e7as – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Trabalho de diversos tipos de for\u00e7as<\/b><\/span><\/span><\/span><\/p>\n

\u00a001-<\/b><\/span><\/span>\u00a0a) Colocando todas as for\u00e7as que agem sobre o carrinho<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

C\u00e1lculo da intensidade da for\u00e7a de atrito\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=m.N=m.P=m.m.g\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=0,2.10.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=20N<\/b><\/span><\/span><\/p>\n

Se o carrinho est\u00e1 com velocidade constante, a for\u00e7a resultante sobre ele \u00e9 nula e F=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=20N<\/b><\/span><\/span><\/p>\n

Trabalho de cada for\u00e7a:<\/b><\/span><\/span><\/p>\n

W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=F.d.cos\u03b1\u00a0 —\u00a0 W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=20.5.cos0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=100.(1)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=100J (positivo, motor)<\/b><\/span><\/span><\/p>\n

W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>.d.cos \u03b1\u00a0\u00a0 —\u00a0 W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>=20.5.cos 180<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 — W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>=100.(-1)\u00a0 —\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>= -100J (negativo, resistente)<\/b><\/span><\/span><\/p>\n

W<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=W<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=P.d.cos \u03b1\u00a0 \u03b1\u00a0 =P.d.cos90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P.d.0=0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=W<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=0<\/b><\/span><\/span>(perpendiculares ao deslocamento)<\/b><\/span><\/span><\/p>\n

O trabalho da for\u00e7a resultante \u00e9 nulo, pois F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0.<\/b><\/span><\/span><\/p>\n

b) F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=20N\u00a0 —\u00a0 a=3m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F=m.a\u00a0 —\u00a0 F=10.3\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 F=30N<\/b><\/span><\/span><\/p>\n

Trabalho de cada for\u00e7a:<\/b><\/span><\/span><\/p>\n

W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=F.d.cos \u03b1\u00a0\u00a0 —\u00a0 W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=30.5.cos0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=150.(1)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=150J (positivo, motor<\/b><\/span><\/span>)<\/b><\/span><\/span><\/p>\n

W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>.d.cos \u03b1\u00a0\u00a0 —\u00a0 W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>=20.5.cos 180<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 — W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>=100.(-1)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>= -100J (negativo, resistente)<\/b><\/span><\/span><\/p>\n

W<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=W<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=P.d.cos \u03b1\u00a0=P.d.cos90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P.d.0=0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=W<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=0<\/b><\/span><\/span>\u00a0(perpendiculares ao deslocamento)<\/b><\/span><\/span><\/p>\n

Trabalho da for\u00e7a resultante\u00a0 —\u00a0 W<\/b><\/span><\/span>FR<\/b><\/span><\/span><\/sub>=150 \u2013 100 + 0 + 0\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 W<\/b><\/span><\/span>FR<\/b><\/span><\/span><\/sub>=50J<\/b><\/span><\/span>\u00a0ou W<\/b><\/span><\/span>FR<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>.d.cosa=(30 – 20).5.cos0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 W<\/b><\/span><\/span>FR<\/b><\/span><\/span><\/sub>=50.(1)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>FR<\/b><\/span><\/span><\/sub>=50J<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span>\u00a0O trabalho realizado pela for\u00e7a resultante centr\u00edpeta \u00e9 sempre nulo, pois ela \u00e9 sempre perpendicular \u00e0 velocidade\u00a0\"\"(que\u00a0\"\"<\/b><\/span><\/span><\/p>\n

\u00e9 sempre tangente em cada ponto da circunfer\u00eancia) e consequentemente ao deslocamento\u00a0\"\".\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span>\u00a0W=F.d.cos \u03b1\u00a0\u00a0 —\u00a0 W=30.10.1\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W=150J\u00a0 R- C<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span>\u00a0W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=F.d.cos0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0 W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=100.2.(1)\u00a0 —\u00a0 W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=200J e o trabalho da for\u00e7a de atrito \u00e9 negativo (a=180<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>).\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

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

05-<\/b><\/span><\/span>\u00a0I- O trabalho da for\u00e7a de atrito n\u00e3o \u00e9 nulo, \u00e9 negativo \u2013 incorreta\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 II- \u00c9 nulo sim (a=90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>) \u2013 correta\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 III- N\u00e3o \u00e9 nula, \u00e9 igual ao F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>, pois est\u00e1 com velocidade constante e F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0 \u2013 incorreta\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span>\u00a0Como ele n\u00e3o est\u00e1 sendo impulsionado, puxado ou empurrado, a for\u00e7a resultante sobre ele \u00e9 a pr\u00f3pria for\u00e7a de atrito.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>O<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.d\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=8<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0– 2.a.d\u00a0 — \u00a0<\/b><\/span><\/span>a.d= – 32\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>.d.cos0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=m.<\/b><\/span><\/span>a.d<\/b><\/span><\/span>.(1) \u00a0\u00a0—\u00a0 W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>=1.(-32).1<\/b><\/span><\/span><\/p>\n

W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>=-32J\u00a0 R- D<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span>\u00a0F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=m,N=m.P=0,2.2.10\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=4N<\/b><\/span><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F \u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=20 \u2013 4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=16N<\/b><\/span><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span>FR<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>.d.cos0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=16.5.(1)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>FR<\/b><\/span><\/span><\/sub>=80J<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span>\u00a0For\u00e7a F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 W<\/b><\/span><\/span>F1<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>1.<\/b><\/span><\/span><\/sub>.d.cos\u03b8 —\u00a0 W<\/b><\/span><\/span>F1<\/b><\/span><\/span><\/sub>=0,4.2,5.cos\u03b8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>F1<\/b><\/span><\/span><\/sub>=cos\u03b8=k<\/b><\/span><\/span>\u00a0\u00a0 —\u00a0 For\u00e7a F<\/b><\/span><\/span>2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/sub>W<\/b><\/span><\/span>F2<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.d.cos2\u03b8 —\u00a0 W<\/b><\/span><\/span>F2<\/b><\/span><\/span><\/sub>=0,4.2,5.cos2\u03b8 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>F2<\/b><\/span><\/span><\/sub>=cos2\u03b8\u00a0<\/b><\/span><\/span>\u00a0\u00a0—\u00a0 cos2\u03b8<\/b><\/span><\/span>\u00a0=<\/b><\/span><\/span>cos<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u03b8<\/b><\/span><\/span>\u00a0 –<\/b><\/span><\/span>\u00a0sen<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u03b8\u00a0 —\u00a0 cos2\u03b8=k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 (1 \u2013 cos<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u03b8)\u00a0 —\u00a0 cos2\u03b8\u00a0= k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 1 + k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 cos2q\u00a0= 2k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 1\u00a0 —\u00a0 W<\/b><\/span><\/span>F2<\/b><\/span><\/span><\/sub>=cos2q=2k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 1\u00a0 R-D<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span>\u00a0R- D (veja teoria)<\/b><\/span><\/span><\/p>\n

10-\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=mN\u00a0 —\u00a0 W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>.d.cos180<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>=-mNd\u00a0 R- B<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span>\u00a0a) Dist\u00e2ncia percorrida\u00a0DS\u00a0 —\u00a0 V=DS\/Dt\u00a0 —\u00a0 2=DS\/2.400\u00a0 —\u00a0\u00a0DS=4.800m essa seria a dist\u00e2cia que ele percorreria se estivesse andando numa estrada reta, plana e horizontal.<\/b><\/span><\/span><\/p>\n

O deslocamento\u00a0\u00a0\"\"\u00a0que \u00e9 uma grandeza vetorial \u00e9 nulo em rela\u00e7\u00e3o a um referencial (qualquer ponto) fixo no solo.<\/b><\/span><\/span><\/p>\n

b)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Consumo de calorias em joules\u00a0 —\u00a0 1 cal \u2013 4J\u00a0 —\u00a0 300.000cal \u2013 xJ\u00a0 —\u00a0 x=4X300.000\u00a0 —\u00a0 x=1,2.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0 esse consumo de calorias \u00e9 o pr\u00f3prio trabalho\u00a0 —\u00a0 W=F.DS.cos0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 1,2.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>=F.4.400.(1)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=250N<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span>\u00a0Observe nas figuras abaixo que podemos decompor\u00a0\"\"\u00a0em\u00a0\u00a0\"\"\u00a0e\u00a0\u00a0\"\", sendo que quem realmente realiza trabalho \u00e9 F<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=Fcos\u03b1,<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

pois \u00e9 a parcela de\u00a0\"\"\u00a0que efetivamente influi no deslocamento, tendo a dire\u00e7\u00e3o e o sentido dele. J\u00e1 a parcela\u00a0\"\"\u00a0que \u00e9 perpendicular ao deslocamento n\u00e3o realiza trabalho, pois n\u00e3o influi no deslocamento.<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span>\u00a0Decompondo a for\u00e7a\u00a0\"\"\u00a0em\u00a0\u00a0\"\"\u00a0e em\u00a0\"\"\u00a0\u00a0—\u00a0 F<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=Fcos\u03b8\u00a0 —\u00a0 F<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=Fsen\u03b8\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0 F + N = P \u00a0—\u00a0 N=P – F<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/p>\n

 <\/p>\n

\"\"<\/p>\n

N = P – Fsen\u03b8\u00a0 —\u00a0 e W<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=0 (\u03b8=90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>)\u00a0 R- C<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span>\u00a0W=b.h\/2\u00a0 —\u00a0 W=1.10\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0W=5J\u00a0 R- C<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>15-<\/b><\/span><\/span>\u00a0W=b.h\/2\u00a0 —\u00a0 W=x\u2019.F\u2019\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span>1- W=b.h\/2=1.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.15\/2\u00a0 —\u00a0 W=7,5.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>J\u00a0 verdadeira\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2- F \u00e9 constante e vale F=15N\u00a0 —\u00a0 F=ma\u00a0 —\u00a0 15=1,5.a\u00a0 —\u00a0 a=10m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

Verdadeira\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3- correta, se F=0 ele est\u00e1 em equil\u00edbrio din\u00e2mico (MRU)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4- W=1.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.15\/2 + 1.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.15 + 2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.15\/2 \u2013<\/b><\/span><\/span><\/p>\n

2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.12\/2\u00a0 —\u00a0 W=25,5.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>J\u00a0 Falsa\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

17<\/b><\/span><\/span>– W=1.2\/2 + 1.2 + 1.2\/2 \u2013 2.2\/2 + 0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W=2J<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span>\u00a0W=(12 + 4).10\/2 + 12.5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W=140J\u00a0 R-C<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span>\u00a0R- B\u00a0 (veja teoria)<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span>\u00a0W=m.g.h=70.10.2\u00a0 —\u00a0 W=1.400J\u00a0 —\u00a0 ordem de grandeza\u00a0 —\u00a0 1,4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0J\u00a0 —\u00a0 1,4 est\u00e1 abaixo de 5 e se transforma em 1\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

W=1.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0J\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span>\u00a0a) peso do pil\u00e3o manual\u00a0 —\u00a0 P=m.g=5.10=50N\u00a0 —\u00a0 W=P.h=50.0,6\u00a0 —\u00a0 W=30J<\/b><\/span><\/span><\/p>\n

b) peso do monjolo\u00a0 —\u00a0 P=m.g=30.10=300N\u00a0 —\u00a0 trabalho em uma queda\u00a0 —\u00a0 W=P.h=300.2\u00a0 —\u00a0 W=600J<\/b><\/span><\/span><\/p>\n

Considerando esse intervalo de tempo como 1s\u00a0 —\u00a0 manual- W=15J\u00a0 —\u00a0 monjolo \u2013 W=150J<\/b><\/span><\/span><\/p>\n

Regra de tr\u00eas\u00a0 —\u00a0 15J \u2013 1 pessoa\u00a0 —\u00a0 150J \u2013 n pessoas\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>n=10 pessoas<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span>\u00a0R- D (veja teoria)<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span>\u00a0O trabalho realizado (energia transferida) para elevar o mesmo balde a uma mesma altura com velocidade constante \u00e9 o mesmo para Oscarito e Ankito, mas a for\u00e7a m\u00ednima exercida por Oscarito \u00e9 menor, j\u00e1 que o deslocamento \u00e9 maior\u00a0 (no caso, a polia m\u00f3vel dobra o deslocamento e reduz a for\u00e7a \u00e0 metade)\u00a0 —\u00a0 R- D<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span>\u00a0a) densidade do produto=massa do produto\/volume do produto\u00a0 —\u00a0 d=1.000\/(1.312,5 \u2013 62,5)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>d=0,8g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/p>\n

b) massa total de cada caixa \u2013 m=10 + 1.000 + 100=1.110g=1,11kg\u00a0 —\u00a0 peso de cada caixa \u2013 P=m.g=1,11.10\u00a0 —\u00a0 P=11,1N<\/b><\/span><\/span><\/p>\n

W=P.h\u00a0 —\u00a0 W=11,1.1,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W=16,65J<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span>\u00a0Trabalho do peso<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=h\/12\u00a0 —\u00a0 1\/2=h\/12\u00a0 —\u00a0 h=6m\u00a0 —\u00a0 W<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=P.h=m.g.h=4.10<\/b><\/span><\/span>3.<\/b><\/span><\/span><\/sup>.10.6\u00a0 —\u00a0 W<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=240kJ<\/b><\/span><\/span><\/p>\n

Trabalho da resultante —\u00a0\u00a0\u2206S=d=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t + a.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 12=0.4 + a.16\/2\u00a0 —\u00a0 a=1,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a=4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.1,5\u00a0 —<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0 W<\/b><\/span><\/span>FR<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>.d.cos0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.12.(1)\u00a0 —\u00a0 W<\/b><\/span><\/span>FR<\/b><\/span><\/span><\/sub>=72kJ\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span>01- Correta \u2013 a polia da direita (m\u00f3vel) divide a for\u00e7a aplicada pela metade<\/b><\/span><\/span><\/p>\n

02- Correta \u2013 como a velocidade \u00e9 constante o corpo est\u00e1 em equil\u00edbrio\u00a0 —\u00a0 fig. I\u00a0 —\u00a0 F=P\u00a0\u00a0 —\u00a0\u00a0 fig. 2\u00a0 —\u00a0 F=Psen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P\/2<\/b><\/span><\/span><\/p>\n

04- Correta \u2013 O trabalho da for\u00e7a peso independe da trajet\u00f3ria<\/b><\/span><\/span><\/p>\n

08- Falsa \u2013 vide 04<\/b><\/span><\/span><\/p>\n

Soma das corretas 07<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span>\u00a0De A para B\u00a0 —\u00a0 W<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>= -m.g.h=- 70.10.0,3\u00a0 —\u00a0 W<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>= – 2.10 J\u00a0 —\u00a0 De B para C\u00a0 —\u00a0 W<\/b><\/span><\/span>BC<\/b><\/span><\/span><\/sub>=+m.g.h=+70.10.0,5\u00a0 —\u00a0 W<\/b><\/span><\/span>BC<\/b><\/span><\/span><\/sub>= 350 J\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

W<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>= – 2.10 + 3.50\u00a0 —\u00a0 W<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>= 1.40 J\u00a0\u00a0<\/b><\/span><\/span>ou\u00a0 W<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>= 1,4.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0J<\/b><\/span><\/span><\/p>\n

ou W<\/b><\/span><\/span>AC<\/b><\/span><\/span><\/sub>=m.g.(h<\/b><\/span><\/span>BC<\/b><\/span><\/span><\/sub>\u00a0\u2013 h<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>)=70.10.(0,5 \u2013 0,3)\u00a0 —\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>AC<\/b><\/span><\/span><\/sub>=1,4.10<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>J\u00a0 R- D<\/b><\/span><\/span><\/p>\n

28-\u00a0<\/b><\/span><\/span>O trabalho realizado \u00e9 numericamente igual \u00e0 \u00e1rea\u00a0\u00a0 —\u00a0 W=b.h\/2=2.100\/2\u00a0 —\u00a0<\/b><\/span><\/span>W=100 J\u00a0 R- B<\/b><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span>\u00a0a) 8cm<\/b><\/span><\/span><\/p>\n

b) A- F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=k.X=20.0 —\u00a0 F<\/b><\/span><\/span>eA<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 F<\/b><\/span><\/span>eB<\/b><\/span><\/span><\/sub>=20.0,08\u00a0 —\u00a0 Fe<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=1,6N\u00a0\u00a0 —\u00a0 F<\/b><\/span><\/span>eC<\/b><\/span><\/span><\/sub>=20.0,16\u00a0 —\u00a0 F<\/b><\/span><\/span>eC<\/b><\/span><\/span><\/sub>=3,2N<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

c) Como a for\u00e7a \u00e9 vari\u00e1vel\u00a0 —\u00a0 \u00a0W<\/b><\/span><\/span>Fe\u00ad<\/b><\/span><\/span><\/sub>=\u00e1rea do trap\u00e9zio=(B + b).h\/2=(3,2 + 1,6).(8.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>)\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>Fe<\/b><\/span><\/span><\/sub>= – 0,192 J<\/b><\/span><\/span>\u00a0(negativo, for\u00e7ado, F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u00e9 contr\u00e1rio ao deslocamento)<\/b><\/span><\/span><\/p>\n

W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>\u00a0tem o mesmo valor,\u00a0 mas \u00e9 positivo (a favor do deslocamento)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=0,192 J.<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span>\u00a0V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=7,2\/3,6=2m\/s\u00a0 —\u00a0 V=0\u00a0 —\u00a0\u00a0DS=X=1m\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.(-a).X\u00a0 —\u00a0 0 = 4 +2.a\u00a0 —\u00a0 a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=7,0.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>.2\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=14.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=k.x\u00a0 —\u00a0 14.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>=k.1\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0k=14.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span>Fe<\/b><\/span><\/span><\/sub>=k.X<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=14.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>.1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 W<\/b><\/span><\/span>Fe<\/b><\/span><\/span><\/sub>=7.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>J<\/b><\/span><\/span><\/p>\n

31<\/b><\/span><\/span>– Sobre o peixe agem duas for\u00e7as\u00a0 —\u00a0 for\u00e7a de tra\u00e7\u00e3o (que o pescador aplica para erguer o peixe, vertical e para cima) e o peso do peixe (vertical e para baixo)\u00a0 —\u00a0 como o deslocamento \u00e9 para cima e o peso do peixe \u00e9 para baixo, eles possuem sentidos contr\u00e1rios e o trabalho da for\u00e7a peso \u00e9 resistente (contr\u00e1rio ao deslocamento)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span>\u00a0a) O trabalho da for\u00e7a de atrito \u00e9 dado por\u00a0 —\u00a0 W = F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>.d.cos180\u00b0\u00a0 —\u00a0 W =\u00a0m.N.d.(-1)\u00a0 —\u00a0 W = – 0,60.3.2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W = – 3,6.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0J<\/b><\/span><\/span><\/p>\n

b) Pela leitura do gr\u00e1fico dado temos que na tens\u00e3o de 20 kV = 2.104 V tem-se uma press\u00e3o de 2.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>\u00a0N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 — press\u00e3o =for\u00e7a\/\u00e1rea<\/b><\/span><\/span><\/p>\n

For\u00e7a = press\u00e3o.\u00e1rea\u00a0 —\u00a0 For\u00e7a = 2.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>.0,25.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0\u00a0\u00a0—\u00a0<\/b><\/span><\/span>\u00a0F= 50 N<\/b><\/span><\/span><\/p>\n

33<\/b><\/span><\/span>–\u00a0\u00a0(01) Correta\u00a0 —\u00a0 entre 0 e 4s\u00a0 —\u00a0 a=(V \u2013 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\/(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=(45 \u2013 0)\/(4 \u2013 0)\u00a0 —\u00a0 a<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>=11,25km\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 entre 7s e 9s\u00a0 —\u00a0 a=(V \u2013 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\/(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=(0 \u2013 45)\/(9 \u2013 7)\u00a0 —\u00a0 a<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>=22,5km\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 como a massa \u00e9 a mesma (mesmo carro) e F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 F<\/b><\/span><\/span>RI<\/b><\/span><\/span><\/sub>=m.11,25\u00a0 —\u00a0 F<\/b><\/span><\/span>RII<\/b><\/span><\/span><\/sub>=m.22,5\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>RII<\/b><\/span><\/span><\/sub>=2F<\/b><\/span><\/span>RI<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

(02) Correta\u00a0 —\u00a0 \u00a0no gr\u00e1fico v\u00a0\u00b4\u00a0t a dist\u00e2ncia percorrida \u00e9 obtida pela \u201c\u00e1rea\u201d entre a linha do gr\u00e1fico e o eixo do tempo\u00a0 —\u00a0 \u00e1rea do trap\u00e9zio\u00a0 —\u00a0 \u0394S=(B + b).h\/2=(9 + 3)\/2.(45\/3,6)\u00a0 —\u00a0 \u0394S=75m.
\n(04) Errada\u00a0 —\u00a0 o trabalho total (entenda-se: trabalho da resultante) \u00e9 igual \u00e0 varia\u00e7\u00e3o da energia cin\u00e9tica\u00a0 —\u00a0 como a velocidade final \u00e9 igual \u00e0 inicial, ambas nulas, o trabalho total \u00e9 nulo.<\/b><\/span><\/span><\/p>\n

(08) Errada\u00a0 —\u00a0 no gr\u00e1fico, a \u201c\u00e1rea\u201d do tri\u00e2ngulo de 0 a 4 s \u00e9 o dobro da \u201c\u00e1rea\u201d do tri\u00e2ngulo de 7 a 9 s\u00a0 —\u00a0 portanto, a dist\u00e2ncia percorrida de 0 a 4 s \u00e9 o dobro da dist\u00e2ncia percorrida de 7 a 9 s.<\/b><\/span><\/span><\/p>\n

(16) Correta\u00a0 —\u00a0 \u00a0D = v\u00a0Dt\u00a0 — 75=(45\/3,6).\u0394t\u00a0 —\u00a0\u00a0 \u00a0Dt = 6s.\u00a0<\/b><\/span><\/span><\/p>\n

R- (01+ 02 + 16)= 19<\/b><\/span><\/span><\/p>\n

34-\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 F = 30 N\u00a0 — \u00a0DS = 800 m\u00a0 —\u00a0 o trabalho (W) de uma for\u00e7a constante\u00a0\"\"\u00a0\u00e9 dado pela express\u00e3o\u00a0 —\u00a0 W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=F.d.cos\u03b1\u00a0 —\u00a0 sendo a for\u00e7a paralela ao deslocamento\u00a0 —\u00a0\u00a0a\u00a0= 0\u00b0, cos\u00a0a\u00a0= 1 \u00a0—\u00a0 W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=30.(800) = 24.000 J = 24 kJ\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

35-\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 m = 1.140 ton = 1,14\u00a0\u00b4\u00a010<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0kg\u00a0 —\u00a0 \u00a0h = 710 m\u00a0 —\u00a0 g = 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=m g h = (1,14.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>) (10) (710) = 8,094.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>\u00a0J =\u00a0 8.094.000.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0J\u00a0 —\u00a0 W<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=8.094.000 kJ\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

36-\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 F = 5 N\u00a0 — \u00a0d = 2 m\u00a0 — \u00a0a\u00a0= 15\u00b0\u00a0 —\u00a0 o enunciado nos permite construir a figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

O trabalho de uma for\u00e7a \u00e9 dado pelo trabalho de sua componente paralela ao deslocamento\u00a0 —\u00a0 T = F d cos\u00a0a.<\/b><\/span><\/span><\/p>\n

Por\u00e9m,\u00a0a\u00a0e\u00a0q\u00a0s\u00e3o complementares\u00a0 —\u00a0 sen\u00a0q\u00a0= cos \u00ada\u00a0 —\u00a0 T = F d cos\u00a0a\u00a0= F d sen\u00a0q\u00a0 —\u00a0 substituindo os valores dados\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

T = 5.2.sen 75\u00b0\u00a0 —\u00a0 assim,\u00a0q\u00a0= 75\u00b0.\u00a0 \u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

37-\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= 6 kg\u00a0 —\u00a0 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 1,5 kg\u00a0 —\u00a0 g = 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u00a0DS = 30 m\u00a0 — \u00a0a\u00a0= 60\u00b0\u00a0 —\u00a0 se a for\u00e7a\u00a0\"\"\u00a0\u00e9 a necess\u00e1ria para levantar o corpo de massa m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0com velocidade constante, ent\u00e3o a intensidade dessa for\u00e7a \u00e9\u00a0 —\u00a0 F = P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0g = 15 N\u00a0 — \u00a0trabalho realizado (W) para arrastar a mochila\u00a0 —\u00a0 W = F.DS.cos 60\u00b0 = (15).(30).(0,5)\u00a0 —\u00a0 W = 225 J\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

38-\u00a0<\/b><\/span><\/span>O trabalho de for\u00e7a constante \u00e9 calculado pela express\u00e3o\u00a0 —\u00a0 T = F d\u00a0 cos\u00a0a\u00a0 —\u00a0 ela mostra que o trabalho (T) de for\u00e7a constante \u00e9 diretamente proporcional ao deslocamento (d)\u00a0 —\u00a0 portanto, o gr\u00e1fico T = f (d) \u00e9 uma reta que passa pela origem\u00a0 —\u00a0 para os valores fornecidos\u00a0 —\u00a0 T = 4 (1) cos 60\u00b0 = 4 (0,5)\u00a0 —\u00a0 T = 2 J\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

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

39-\u00a0<\/b><\/span><\/span>Analisemos cada uma das afirma\u00e7\u00f5es:<\/b><\/span><\/span><\/p>\n

(I) Errada\u00a0 —\u00a0 W=F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>.d.cos\u03b1\u00a0 —\u00a0 se \u03b1=90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 cos90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=0\u00a0 —\u00a0 W=0<\/b><\/span><\/span><\/p>\n

(II) Errada\u00a0 —\u00a0 A 3\u00aa lei de Newton afirma tamb\u00e9m que essas for\u00e7as agem em corpos diferentes, portanto n\u00e3o se equilibram.<\/b><\/span><\/span><\/p>\n

(III) Correta\u00a0 —\u00a0 \u00c9 exatamente o que afirma o teorema da energia cin\u00e9tica: \u201co trabalho da resultante das for\u00e7as que agem num corpo \u00e9 igual \u00e0 varia\u00e7\u00e3o de sua energia cin\u00e9tica.\u201d Portanto, se o trabalho \u00e9 nulo, a energia cin\u00e9tica se mant\u00e9m constante.\u00a0<\/b><\/span><\/span><\/p>\n

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

40-\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 \u00a0E = 94.684.781 MWh\u00a0@\u00a09,5\u00a0\u00b4\u00a010<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>\u00a0MWh = 9,5\u00a0\u00b4\u00a010<\/b><\/span><\/span>13<\/b><\/span><\/span><\/sup>\u00a0Wh\u00a0 — \u00a0L = 0,45\u00a0\u00b4\u00a010<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>\u00a0J\/kg\u00a0 —\u00a0 transformando essa quantidade de energia em joules\u00a0 —\u00a0 E = 9,5\u00a0\u00b4\u00a010<\/b><\/span><\/span>13<\/b><\/span><\/span><\/sup>\u00a0Wh = (9,5\u00a0\u00b4\u00a010<\/b><\/span><\/span>13<\/b><\/span><\/span><\/sup>\u00a0W)\u00a0\u00b4(3,6\u00a0\u00b4\u00a010<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0s) = 3,4 \u00ad\u00b4\u00a010<\/b><\/span><\/span>17<\/b><\/span><\/span><\/sup>\u00a0J\u00a0 —\u00a0 para o petr\u00f3leo\u00a0 —\u00a0 E = m L\u00a0 —\u00a0 m =E\/L=3,4×10<\/b><\/span><\/span>17<\/b><\/span><\/span><\/sup>\/0,45×10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>\u00a0 = 7,5\u00a0\u00b4\u00a010<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>\u00a0kg = 7,5\u00a0\u00b4\u00a010<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0toneladas \u00a0—\u00a0 m = 7,5 milh\u00f5es de toneladas\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

41-\u00a0<\/b><\/span><\/span>O enunciado exige menor impacto ambiental. J\u00e1 que a incid\u00eancia solar na regi\u00e3o \u00e9 alta, a melhor forma para obten\u00e7\u00e3o de energia \u00e9 a fotovoltaica\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

42-\u00a0<\/b><\/span><\/span>Reservat\u00f3rio de energia\u00a0 —\u00a0 bateria e meio ambiente;<\/b><\/span><\/span><\/p>\n

Transformador de energia\u00a0 —\u00a0 bateria e l\u00e2mpada;<\/b><\/span><\/span><\/p>\n

Modo de transfer\u00eancia de energia\u00a0 —\u00a0 realizando trabalho el\u00e9trico sobre os el\u00e9trons-livres no interior do fio condutor<\/b><\/span><\/span><\/p>\n

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

43-\u00a0<\/b><\/span><\/span>A tabela mostra a energia gasta em cada uma das atividades, dentro do tempo especificado em cada uma das proposi\u00e7\u00f5es, destacando aquela que mais se aproxima do gasto de 304 quilocalorias.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0\u00a0\u00a0R- A<\/b><\/span><\/span><\/p>\n

44-\u00a0<\/b><\/span><\/span>A parcela de F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0que influi no deslocamento horizontal vale F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u2019=F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>cos (90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0\u2013 30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u2019=10.1\/2=5N \u00a0—\u00a0 para que o trabalho seja fornecido pela \u00e1rea, no eixo vertical a for\u00e7a deve ser a for\u00e7a resultante, cujos valores s\u00e3o fornecidos no gr\u00e1fico abaixo\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

O trabalho total realizado pela for\u00e7a resultante \u00e9 numericamente igual \u00e0 soma das \u00e1reas\u00a0 —\u00a0 W=(1.11) + ((13 + 11).1\/2 + ((1.13) + (13 + 9).1\/2\u00a0 —\u00a0 W<\/b><\/span><\/span>FR<\/b><\/span><\/span><\/sub>=47J\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

 <\/p>\n

45-<\/b><\/span><\/span>O trabalho da for\u00e7a resultante com ela atuando na mesma dire\u00e7\u00e3o e sentido do deslocamento \u00e9 numericamente igual \u00e0 \u00e1rea\u00a0 entre a reta representativa e o eixo do deslocamento\u00a0 —\u00a0 W=\u00e1rea do tri\u00e2ngulo=b.h\/2=26.12\/2\u00a0 —\u00a0 W=156J\u00a0 —\u00a0 R- D<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

46-<\/b><\/span><\/span>O trabalho realizado pela for\u00e7a peso da pessoa ao subir as escadas independe da trajet\u00f3ria e \u00e9, numericamente igual \u00e0 varia\u00e7\u00e3o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

de sua energia potencial gravitacional\u00a0 —\u00a0 W=\u2206E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=m.g.\u2206h=80.10.450\u00a0 —\u00a0 W=360 000J=360kJ\u00a0 —\u00a0 s\u00e3o ingeridas 4 por\u00e7\u00f5es e o rendimento de cada uma \u00e9 25% (1\/4), logo, a energia qu\u00edmica de uma por\u00e7\u00e3o \u00e9 convertida em trabalho\u00a0 —\u00a0 W=360kJ\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

47-<\/b><\/span><\/span>\u00a0Considerando o n\u00edvel da jusante (parte mais baixa da \u00e1gua) nulo, o da montante (parte mais elevada da \u00e1gua) ser\u00e1<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0de 25m\u00a0 —\u00a0 \u2206E<\/b><\/span><\/span>pg<\/b><\/span><\/span><\/sub>=m.g.\u2206h=1,2.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>.10.(25 \u2013 0)\u00a0 —\u00a0\u00a0 \u2206E<\/b><\/span><\/span>pg<\/b><\/span><\/span><\/sub>=300.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>J=3,0.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Trabalho de diversos tipos de for\u00e7as \u00a001-\u00a0a) Colocando todas as for\u00e7as que agem sobre o carrinho C\u00e1lculo da intensidade da for\u00e7a de atrito\u00a0 —\u00a0 Fat=m.N=m.P=m.m.g\u00a0 —\u00a0 Fat=0,2.10.10\u00a0 —\u00a0\u00a0Fat=20N Se o carrinho est\u00e1 com velocidade constante, a for\u00e7a resultante sobre ele \u00e9 nula e F=Fat\u00a0 —\u00a0\u00a0F=20N Trabalho de cada for\u00e7a: WF=F.d.cos\u03b1\u00a0 —\u00a0 WF=20.5.cos0o\u00a0 —\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1329,"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-1333","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1333","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=1333"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1333\/revisions"}],"predecessor-version":[{"id":10868,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1333\/revisions\/10868"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1329"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1333"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}