{"id":1324,"date":"2015-09-15T15:12:16","date_gmt":"2015-09-15T15:12:16","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1324"},"modified":"2024-08-23T13:45:03","modified_gmt":"2024-08-23T13:45:03","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-potencia-de-uma-forca","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/potencia-de-uma-forca\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-potencia-de-uma-forca\/","title":{"rendered":"Pot\u00eancia de uma for\u00e7a – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Pot\u00eancia de uma for\u00e7a<\/b><\/span><\/span><\/span><\/p>\n

\u00a001-<\/b><\/span><\/span><\/span>Maior consumo \u2013 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>W\u00a0 —\u00a0\u00a0\u2206t=2h\u00a0 — P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W(energia consumida)\/\u2206t — 5,10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=W\/2\u00a0 —\u00a0 W=10.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W=10kW R- B<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0Velocidade constante \u2013 a for\u00e7a que est\u00e1 puxando \u00e9 igual ao peso da caixa d\u2019agua\u00a0 —\u00a0 P=m.g=66.10\u00a0 —\u00a0 P=660N\u00a0 —\u00a0 W=P.h=660.18\u00a0 —\u00a0 W=11.800 J\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/Dt=11.800\/120\u00a0 —\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=99W<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0O trabalho do peso independe da trajet\u00f3ria e vamos calcul\u00e1-lo pela altura\u00a0 —\u00a0 h=0,150.60\u00a0 —\u00a0 h=9m\u00a0 —\u00a0 W=m.g.h\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

W=80.10,9=72.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W=7,2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/Dt=7,2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\/120\u00a0 —\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=60,0W\u00a0 R- A<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0Regra de tr\u00eas\u00a0 —\u00a0 1cal \u2013 4,2 J\u00a0 —\u00a0 100.000cal\u00a0 —\u00a0 x J\u00a0 —\u00a0 x=4,2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0J=W (energia)\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u2206t\u00a0 —\u00a0 60=4,2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\/\u2206t\u00a0 — \u2206t=4,2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\/60\u00a0 —\u00a0\u00a0\u2206t=0,07.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>=7.000s\/60=116,7min\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0Na horizontal e na subida a esteira se move com velocidade constante de 1,5m\/s, ou seja, em 1s ela percorre 1,5m e, pela figura,<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observamos que ela transportou, na horizontal e na vertical ela transportou 1,5 caixas por segundo.1 caixa \u2013 80kg\u00a0 —\u00a0 1,5 caixas —\u00a0 m=80.1,5\u00a0 —\u00a0 m=120kg, que \u00e9 a massa que a esteira na parte inclinada deve elevar de 5m em cada 1s com velocidade constante de 1,5m\/s.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

C\u00e1lculo do trabalho pela altura\u00a0 —\u00a0 W=m.g.h=120.10.5\u00a0 —\u00a0 W=6.000J \u2013 energia consumida para elevar uma massa de 120kg de 5m.\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u2206t=6.000\/1\u00a0 —\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=6.000W<\/b><\/span><\/span><\/span>\u00a0– energia consumida para elevar uma massa de 120kg de 5m em 1s..\u00a0<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0com a chave na posi\u00e7\u00e3o inverno \u2013 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u2206t\u00a0 —\u00a0 6.000=W<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\/30.0,5\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=90.000W=<\/b><\/span><\/span><\/span>90kW (<\/b><\/span><\/span><\/span>energia el\u00e9trica consumida) com a chave na posi\u00e7\u00e3o ver\u00e3o \u2013 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/Dt\u00a0 —\u00a0 4.000=W<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>\/15\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>=60.000W=<\/b><\/span><\/span><\/span>60kW(<\/b><\/span><\/span><\/span>energia el\u00e9trica consumida)<\/b><\/span><\/span><\/span><\/p>\n

economia \u2013 90kW \u2013 60kW=\u00a0<\/b><\/span><\/span><\/span>30kW R-B<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0Energia consumida em 1 m\u00eas (30 dias) \u2013 W=30.3.000kcal=9.000.k.(4J)\u00a0 —\u00a0 W=360.000KJ=360.000.000J (energia em Joules, consumida em 1 m\u00eas)\u00a0 —\u00a0 regra de tr\u00eas\u00a0 — 1kWh \u2013 3.600.000J\u00a0 —\u00a0 x kWh \u2013 360.000.000J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>x=100kWh R- C<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

08-<\/b><\/span><\/span><\/span>\u00a0Pot\u00eancia para elevar o peso do elevador vazio \u00e0 altura de 32m em 40s\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>oe<\/b><\/span><\/span><\/sub><\/span>=W\/Dt=m.g.h\/Dt=370.10.32\/40\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>oe<\/b><\/span><\/span><\/sub><\/span>=2.960W<\/b><\/span><\/span><\/span><\/p>\n

Pot\u00eancia para elevar uma pessoa \u00e0 altura de 32m em 40s\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>op<\/b><\/span><\/span><\/sub><\/span>=W\/Dt=m.g.h\/Dt=70.10.32\/40\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>op<\/b><\/span><\/span><\/sub><\/span>=560W<\/b><\/span><\/span><\/span><\/p>\n

A pot\u00eancia do motor menos a pot\u00eancia para elevar o elevador vazio nos fornece a pot\u00eancia que sobra para elevar o n\u00famero de pessoas\u00a0 —\u00a0 8.500 \u2013 2.960=5.540W\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 pessoa \u2013 560W\u00a0 —\u00a0 n pessoas \u2013 5.540W\u00a0 —\u00a0 n=5.540\/560=9,89\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

n= 9 pessoas\u00a0 R- C<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0d=m\/V\u00a0 —\u00a0 1.000=m\/700\u00a0 —\u00a0 m=700.000kg (massa de \u00e1gua que cai de uma altura de 120m em cada 1s)\u00a0 — P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=m.g.h\/Dt\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=700.000.10.120\/1=840.000.000W\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=840.000kW\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=8,4.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>kW<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=m.g.h\/\u2206t\u00a0 —\u00a0 512.000.000=m.10.120\u00a0 —\u00a0 m=512.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\/12.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 m=42,7.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>=437.000kg\u00a0 + 0,1 x437.000=480.700kg<\/b><\/span><\/span><\/span><\/p>\n

Considerando a densidade da \u00e1gua 1kg\/L, teremos que uma massa de 1kg ocupa um volume de 1L de \u00e1gua\u00a0 — R- E<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0Do 8<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0ao 12<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 4 degraus\u00a0 —\u00a0 h=4.0,18=0,72m\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=m.g.h\/\u2206t=70.10.0,72\/2,8\u00a0 —\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=180W\u00a0 R- A<\/b><\/span><\/span><\/span><\/p>\n

12-V=<\/b><\/span><\/span><\/span>280m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=280.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>dm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=280.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>L=280.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0 m=28.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=m.g.h\/\u2206t=28.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.10.160\/1=2.816.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u00a0 — \u00a0\u00a0P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=448.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0,448.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\u00a0W<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0Considerando que o volume de 1L de \u00e1gua possui massa de 1kg\u00a0 —\u00a0 V=26m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=26.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>dm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=26.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0L\u00a0 —\u00a0 m=26.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0 a transposi\u00e7\u00e3o \u00e9 a eleva\u00e7\u00e3o da \u00e1gua do rio S\u00e3o Francisco da altura de 315m (elevat\u00f3ria 1) at\u00e9 475m (elevat\u00f3ria 4) —\u00a0 h=475 \u2013 315=160m\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>=m.g.h\/Dt=26.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10.160\/1\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=416.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>W=41,6.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>W\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 R- C\u00a0<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0a) uma turbina de pot\u00eancia 680.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>W foi paralisada durante 3h —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u2206t\u00a0 —\u00a0 680.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>=W\/3\u00a0 —\u00a0 W=2.040.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>=2.040.000kWh (energia el\u00e9trica n\u00e3o fornecida por Itaipu durante 3h)\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 domic\u00edlio \u2013 4kWh\u00a0 —\u00a0 N domic\u00edlios \u2013 2.040.000kWh<\/b><\/span><\/span><\/span><\/p>\n

— 4N=2.040.000\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>N=510.000 munic\u00edpios<\/b><\/span><\/span><\/span><\/p>\n

b) d=m\/V\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=m\/600\u00a0 —\u00a0 m=600.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>m=600.000kg<\/b><\/span><\/span><\/span><\/p>\n

c) P=m.g.h=6.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.10.120=720.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P=720MW<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0Por dia uma pessoa utiliza 30L\u00a0 —\u00a0 4pessoas\u00a0 —\u00a0 4.30L\u00a0 —\u00a0 120L\u00a0 —\u00a0 em um m\u00eas 4 pessoas utilizar\u00e3o \u2013 V=30.120=3600L\u00a0 — m=3.600kg\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>m=3.600.000g<\/b><\/span><\/span><\/span>\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>c = 4,19 J\/g\u00b0C<\/b><\/span><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u2206t=30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>equa\u00e7\u00e3o fundamental da calorimetria \u2013 Q=m.c.\u2206t\u00a0 —\u00a0 Q=3.600.000.4,19.30\u00a0 —\u00a0 Q=W=452.520.000J\u00a0 —\u00a0 lembrando que 1kWh=1.000W.3.600s \u00a0—\u00a0 1kWh=3.600.000J\u00a0 — regra de tr\u00eas\u00a0 —\u00a0 1kWh \u2013 3.600.000W\u00a0 —\u00a0 x kWh \u2013 452.520.000\u00a0 —\u00a0 x=W=452.520.000\/3.600.000\u00a0 —\u00a0 W=125,7kWh\u00a0 —\u00a0 Considerando que a \u00e1gua no tanque, de 200 litros, libera 0,30 kWh\/litro, a cada m\u00eas, tem-se que a perda \u00e9 de\u00a0 200×0,30=60 kWh.<\/b><\/span><\/span><\/span><\/p>\n

Assim os pain\u00e9is solares ter\u00e3o que fornecer, por m\u00eas, 125,7 + 60 = 185,7 kWh\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 se 1m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0metro quadrado de painel, no m\u00eas, fornece 60 kWh, 187,5kWh ser\u00e3o fornecidos por x m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 60.x=185,7\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>x=3,1m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0de superf\u00edcie de pain\u00e9is solares.<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0W=(B + b).h\/2 + b.h=(3.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>).100\/2 + 50.3.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=250.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 150.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>W=400.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=4.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>J<\/b><\/span><\/span><\/span><\/p>\n

17-\u00a0<\/b><\/span><\/span><\/span>Se est\u00e1 com velocidade constante (equil\u00edbrio est\u00e1tico), a intensidade da for\u00e7a\u00a0\"\"\u00a0com que o motor puxa o bloco deve ser igual \u00e0 intensidade da for\u00e7a de atrito\u00a0\"\".<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

F=F<\/b><\/span><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub><\/span>=mN=mP=mmg=0,1.500.10\u00a0 —\u00a0\u00a0\u00a0\u00a0 F=500N\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span><\/span>0<\/b><\/span><\/span><\/span><\/sub><\/span>=F.V=500.0,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>=250N<\/b><\/span><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0a) P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u2206t\u00a0 —\u00a0 300=W\/1800\u00a0 —\u00a0 W=540.000J (energia \u201cgasta\u201d pelo atleta em joules)\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 cal\u00a0 –\u00a0 4,2J\u00a0 —\u00a0 X cal\u00a0 –\u00a0 540.000J\u00a0 —\u00a0 4,2X=540.000\u00a0 —\u00a0 X=12,9.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>=<\/b><\/span><\/span><\/span>1,29.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0cal.<\/b><\/span><\/span><\/span><\/p>\n

b) V=\u2206S\/\u2206t=10.000\/1.800\u00a0 —\u00a0 V=50\/9m\/s\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=F.V\u00a0 —\u00a0 300=F\/50\/9\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 F=54N<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0V=72km\/h\/3,6=20m\/s\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=200.000W\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=F.V\u00a0 —\u00a0 200.000=F.20\u00a0 —\u00a0<\/b><\/span><\/span><\/span>F=10.000N\u00a0 R- A<\/b><\/span><\/span><\/span><\/p>\n

20-\u00a0<\/b><\/span><\/span><\/span>Como a velocidade \u00e9 constante a for\u00e7a motora \u00e9 a for\u00e7a de atrito\u00a0 —\u00a0 F=F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>.V\u00a0 —\u00a0 25.000=F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>.80\/3,6\u00a0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=1.125N\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0Pot\u00eancia utilizada para elevar um bloco de m=75kg a uma altura de 1m\u00a0\u00a0 —\u00a0P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=F.V=P.V =m.g.V =75.10.1\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=750W<\/b><\/span><\/span><\/span><\/p>\n

Pot\u00eancia desenvolvida pelo carro\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>=F.V=P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>.V=mgsenq.V=1.000.10.0,1.15\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>0=<\/b><\/span><\/span><\/sub><\/span>15.000W\u00a0 —\u00a0 regra de tr\u00eas\u00a0 — 1 CV \u2013 750W\u00a0 —\u00a0 X CV \u2013 15.000W\u00a0 —\u00a0 X=15.000\/750\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>X=20CV\u00a0 R- A<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0a) Como ele sobe com velocidade constante (equil\u00edbrio din\u00e2mico \u2013 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=0), F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=P=m.g=5.000.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=5,0.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0N<\/b><\/span><\/span><\/span><\/p>\n

b) Como tem acelera\u00e7\u00e3o\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2\u00ad\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0– P=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 50.000=5.000.5\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=50.000 + 25.000\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=7,5.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>N<\/b><\/span><\/span><\/span><\/p>\n

c) P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.V\u00a0 —\u00a0 150.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=7,5.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V=2,0m\/s<\/b><\/span><\/span><\/span><\/p>\n

d) A velocidade m\u00e1xima ocorre quando a for\u00e7a motora \u00e9 o peso (P=50.000N) e a pot\u00eancia \u00e9 a m\u00e1xima do motor (P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=150.000W)<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=P.V\u00a0 —\u00a0 150.000=50.000.V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V=3,0m\/s.<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0a) Situa\u00e7\u00e3o 1\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=60m\/s \u2013 constante \u2013 equil\u00edbrio din\u00e2mico \u2013 for\u00e7a resultante nula \u2013 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>.V\u00a0 —\u00a0 120.000=F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>.60\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=2.000N\u00a0 —\u00a0 como V=constante (equil\u00edbrio din\u00e2mico) \u2013\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>ar=<\/b><\/span><\/span><\/sub><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=2.000N\u00a0<\/b><\/span><\/span><\/span><\/p>\n

b) Situa\u00e7\u00e3o 2\u00a0 —\u00a0 V constante \u2013 for\u00e7a resultante nula \u2013 motor desligado F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>=m.g.sena=800.10.sena\u00a0 —\u00a0 ms, F<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 a mesma da situa\u00e7\u00e3o 1 (2.000N)\u00a0 —\u00a0 2.000=8.000sena\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>sena=0,25\u00a0<\/b><\/span><\/span><\/span><\/p>\n

c) Situa\u00e7\u00e3o 3\u00a0 —\u00a0 V constante \u2013 for\u00e7a resultante nula<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

For\u00e7as que agem na dire\u00e7\u00e3o do movimento\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=m.g.senq\u00a0– parcela do peso, paralela \u00e0 superf\u00edcie do plano inclinado\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=800.10.0,3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=2.400N\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 F<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 for\u00e7a de resist\u00eancia do ar, constante e igual a 2.000N\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 for\u00e7a motora\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

Como a resultante \u00e9 nula \u2013 F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>\u00a0+ F<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=2.400 + 2.000\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=4.400N\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>.V=4.400.60\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=264kW<\/b><\/span><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0Colocando as for\u00e7as que agem sobre os sistema:<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 peso total do elevador\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=(200 + 420).10\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=6.200N\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>cp<\/b><\/span><\/span><\/sub><\/span>=220.10\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>cp<\/b><\/span><\/span><\/sub><\/span>=2.200N\u00a0 —\u00a0 Como tanto elevador como contra peso est\u00e3o com velocidade constante, est\u00e3o em equil\u00edbrio din\u00e2mico e a for\u00e7a resultante sobre eles \u00e9 nula\u00a0 —\u00a0 contra peso\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>cp<\/b><\/span><\/span><\/sub><\/span>=T\u00a0 —\u00a0 T=2.200N\u00a0 —\u00a0 elevador\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=T + F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 6.200=2.200 + F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=4.000N\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>.V\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=4.000.0,5\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=2.000W=2kW<\/b><\/span><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0Colocando as for\u00e7as:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Elevador \u2013 peso total – P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=1.300.10\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=13.000N\u00a0 —\u00a0 contra peso B \u2013 P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=m.g=800.10=8.000N\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=8.000N\u00a0<\/b><\/span><\/span><\/span><\/p>\n

a) ambos est\u00e3o em equil\u00edbrio din\u00e2mico com for\u00e7a resultante nula\u00a0 —\u00a0 contra peso \u2013 P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=T<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>—\u00a0 T=8.000N\u00a0 —\u00a0 elevador \u2013 Pt=F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0+ T\u00a0 —\u00a0 13.000=F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 8.000\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=5.000N\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>.V\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=5.000×1\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=5kW<\/b><\/span><\/span><\/span><\/p>\n

b) contra peso B \u2013 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 T=m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.a\u00a0 —\u00a0 8.000 \u2013 T=800.0,5\u00a0 —\u00a0 T=7.600N\u00a0 —\u00a0 elevador – F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>.a\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0+ T- P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=1.300.0,5\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 7.600 \u2013 13.000=650\u00a0 —\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=6.050N\u00a0<\/b><\/span><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0Trata-se de uma fun\u00e7\u00e3o de 1<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0grau (gr\u00e1fico \u00e9 uma reta) de equa\u00e7\u00e3o F=F<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ k.V, onde k \u00e9 a constante de proporcionalidade\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=50N (constante)\u00a0 —\u00a0 F=50 + k.V\u00a0 —\u00a0 quando V=30m\/s \u2013 F=20N\u00a0 —\u00a0 20=50 + k.30\u00a0 —\u00a0 k= -1N.s\/m (constante)\u00a0 —\u00a0 a fun\u00e7\u00e3o \u00e9 F=50 \u2013 1.V\u00a0 —\u00a0 quando\u00a0<\/b><\/span><\/span><\/span>V=20m\/s<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 F=50 -1.20 –\u00a0<\/b><\/span><\/span><\/span>F=30N<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=F.V=30.20\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=600W\u00a0 R-B<\/b><\/span><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=1.000W\u00a0 —\u00a0 a pot\u00eancia \u00fatil (P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>) \u00e9 aquela que realiza trabalho \u2013 P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>\/\u2206t=3200\/4=800\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=800W\u00a0 —\u00a0\u00a0h=P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>\/P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=800\/1000=0,8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>h=80 %\u00a0 R-D<\/b><\/span><\/span><\/span>\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><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=8.640\/60=144W\u00a0 —\u00a0\u00a0h=144\/200=0,72=<\/b><\/span><\/span><\/span>72 %<\/b><\/span><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0a) velocidade constante,\u00a0 a for\u00e7a motora \u00e9 igual \u00e0 for\u00e7a peso e o trabalho do peso, que \u00e9 uma for\u00e7a conservativa independe da trajet\u00f3ria e vamos calcular esse trabalho em fun\u00e7\u00e3o da altura, no deslocamento de C at\u00e9 B.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=h\/15\u00a0 —\u00a0 h=7,5m\u00a0 —\u00a0 W=m.g.h=80.10.7,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W=6.000J<\/b><\/span><\/span><\/span><\/p>\n

b) P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/Dt=6.000\/30\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=200W<\/b><\/span><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 c)\u00a0h=P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>\/P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=200\/400=0,5=50 %\u00a0 —\u00a0<\/b><\/span><\/span><\/span>h=50 %<\/b><\/span><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=P.h\/Dt\u00a0 —\u00a0 125=50.10\/Dt\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>Dt=4,0s\u00a0 R- E<\/b><\/span><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0R- A \u2013 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0C\u00e1lculo do trabalho \u00fatil\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=F.d.cosa=1800.400.1\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=7,2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —h=W<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>\/W<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0,45=7,2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\/W<\/b><\/span><\/span><\/span>t\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>—\u00a0 W<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=7,2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\/0,45\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=1,6MJ\u00a0 R-E<\/b><\/span><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0Do enunciado, a pot\u00eancia total P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 dada por P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=299cv x 740=221.260W\u00a0 —\u00a0 c\u00e1lculo da acelera\u00e7\u00e3o do carro quando sua velocidade passa de 0 a 28m\/s em 7,5s \u2013 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a.t\u00a0 —\u00a0 28=0 + a.7,5 \u2013 a=3,7m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 c\u00e1lculo do deslocamento\u00a0DS=d=v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t + a.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0 + 3,7.(7,5)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 d=104m\u00a0 —\u00a0 c\u00e1lculo da for\u00e7a motora – F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=m.a=2.400.3,7\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=8.880N\u00a0 —\u00a0 c\u00e1lculo do trabalho \u00fatil realizado pelo motor nesse deslocamento \u2013 W<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>.d.cos0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=8.800.104.1\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=915.200J\u00a0 —\u00a0 c\u00e1lculo da pot\u00eancia \u00fatil \u2013 P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>\/Dt\u00a0 —\u00a0P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=915.200\/7,5\u00a0 —\u00a0\u00a0 P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=122.027W\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>\/P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=122.027\/221.260\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>\/P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=0,<\/b><\/span><\/span><\/span>55\u00a0 R-C<\/b><\/span><\/span><\/span><\/p>\n

34-\u00a0a) Considerando a densidade da \u00e1gua 1kg\/L – 1m3\u00a0= 103L = 103\u00a0kg\u00a0 —\u00a0 m=4.105kg ( a usina utiliza essa massa de \u00e1gua por segundo<\/b><\/span><\/span>)<\/b><\/span><\/span><\/p>\n

Pot\u00eancia el\u00e9trica total recebida pela usina em 1s\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=m.g.h\/Dt=4.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.10.9\/1\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=36.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>W\u00a0 —\u00a0 Pot\u00eancia \u00fatil=90 % de P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=36.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u00a0x 0,9\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=32,4.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>W=32,4 MW<\/b><\/span><\/span><\/span><\/p>\n

b) c\u00e1lculo da energia el\u00e9trica \u00fatil produzida pela usina hidrel\u00e9trica em u m\u00eas=720h\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>\/Dt\u00a0 —\u00a0 32,4.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>=W<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>\/720\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

W<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=23.328.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>Wh\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 habitante \u2013 36.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>Wh\u00a0 —\u00a0 N habitantes \u2013 23.328.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>Wh\u00a0 —\u00a0 N=23.328.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\/36.10<\/b><\/span><\/span><\/span>4\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0—<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup><\/span><\/p>\n

N=64.800 habitantes<\/b><\/span><\/span><\/span><\/p>\n

35-\u00a0<\/b><\/span><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0= 300 kWh = 300.10<\/b><\/span><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup><\/span>.(J\/s).3600 s = 1.080.000×10<\/b><\/span><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0J = 1,08×10<\/b><\/span><\/span><\/span><\/span>9<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0J\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span><\/span>C<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0= 180×30.000 = 5. <\/b><\/span><\/span><\/span><\/span>400 .000 kJ = 5,4×10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0J\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/E<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=5,4.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\/1,08.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>=5\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

36-\u00a0<\/b><\/span><\/span><\/span>a) No trecho em que a for\u00e7a de tra\u00e7\u00e3o \u00e9 constante, calculando a \u00e1rea do ret\u00e2ngulo (figura), voc\u00ea est\u00e1 calculando o trabalho<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

realizado\u00a0 —\u00a0 trabalho = \u00e1rea=base.altura\u00a0 —\u00a0 W=(50 \u2013 10).800\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0W=32.000J<\/b><\/span><\/span><\/span><\/p>\n

b) O trabalho realizado foi de 444.000 J no intervalo de tempo de 40 s\u00a0 —\u00a0 como pot\u00eancia \u00e9 a raz\u00e3o entre o trabalho realizado e o intervalo de tempo\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= (Trabalho)\/Dt = 444.000\/40 = 11.100 W\u00a0 —\u00a0 em\u00a0 para cavalo-vapor\u00a0 —\u00a0 \u00a011.100\/740 \u00a0\u00a0—\u00a0<\/b><\/span><\/span><\/span>\u00a0P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>= 15 CV<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

37-\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 quantidade de an\u00e9is\u00a0 — \u00a0n = 100\u00a0 —\u00a0 volume de \u00e1gua em cada anel\u00a0 —\u00a0 V = 1 cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>= 10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 densidade da \u00e1gua\u00a0 —\u00a0 \u00a0d = 1 g\/cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0= 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 altura de eleva\u00e7\u00e3o\u00a0 —\u00a0 \u00a0h = 2 m\u00a0 —\u00a0 per\u00edodo de rota\u00e7\u00e3o do eixo\u00a0 —\u00a0 \u00a0T = 2 s\u00a0 —\u00a0 volume total de \u00e1gua contido nos 100 an\u00e9is\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0= n V = 100.(10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>) m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0= 10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u00a0m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 esse volume representa uma massa de\u00a0 —\u00a0 M = d V<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.(10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>) = 10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0kg\u00a0 — \u00a0tempo de eleva\u00e7\u00e3o dessa massa de \u00e1gua\u00a0 —\u00a0\u00a0Dt = 100.T = 100.(2)\u00a0 —\u00a0\u00a0\u00a0\u00a0Dt = 200 s.<\/b><\/span><\/span><\/span><\/p>\n

Pot\u00eancia \u00fatil da fonte de rota\u00e7\u00e3o\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/Dt=E<\/b><\/span><\/span><\/span>potencial<\/b><\/span><\/span><\/sub><\/span>\/Dt=Mgh\/Dt\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>.10.2\/200=2\/200=1\/100\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=1,0.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>W\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

38-\u00a0<\/b><\/span><\/span><\/span>A pot\u00eancia (em watt) \u00e9 a raz\u00e3o entre a energia transformada (em joule) e o tempo (em segundo)\u00a0 —\u00a0 1W=1J\/1s\u00a0 — 4.800W=4.800J\/1s\u00a0 —\u00a0 4.800W=4.800J\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

\u00a039-\u00a0<\/b><\/span><\/span><\/span>a) Dados\u00a0 — \u00a0n = 10 voltas\u00a0 — \u00a0Dt = 20 s\u00a0 —\u00a0\u00a0 freq\u00fc\u00eancia\u00a0 —\u00a0 f=n\/\u0394t=10\/20\u00a0 —\u00a0 f=0,5Hz\u00a0 —\u00a0 velocidade angular\u00a0 —\u00a0 W=2\u03c0f=2\u03c0.0,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W=\u03c0 rad\/s<\/b><\/span><\/span><\/span><\/p>\n

b) Dado\u00a0 — \u00a0F = P = 450 N\u00a0 — \u00a0\u2206S = h = 0,5 m\u00a0 —\u00a0 o trabalho (W) realizado pela enfermeira \u00e9 igual ao trabalho para levantar o peso\u00a0 —\u00a0 W=P.h=450.0,5\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0W=225J<\/b><\/span><\/span><\/span><\/p>\n

c) Calculando a pot\u00eancia\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u0394t=225\/20\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=11,25W<\/b><\/span><\/span><\/span><\/p>\n

40-\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 — \u00a01 cal\u00a0\"\"\u00a04J\u00a0 —\u00a0 1 telha = 1,5 kg\u00a0 —\u00a0 \u00a0g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

a) S\u00e3o dados\u00a0 —\u00a0 n = 20\u00a0 —\u00a0 \u00a0m = 1,5 kg\u00a0 —\u00a0\u00a0 g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 \u00a0h = 3 m\u00a0 —\u00a0\u00a0 v = 0,1 m\/s\u00a0 —\u00a0\u00a0 r = 20%\u00a0 —\u00a0 \u00a0\u2206t = 2 h = 7.200 s\u00a0 —\u00a0\u00a0 1 cal = 4 J\u00a0 —\u00a0 como o transporte \u00e9 feito com velocidade constante, a for\u00e7a exercida pelo trabalhador em cada levantamento tem a mesma intensidade do peso de 20 telhas\u00a0 —\u00a0 a pot\u00eancia (P) \u00e9 dada pelo produto da intensidade da for\u00e7a (F) pelo m\u00f3dulo da velocidade (v) e o trabalho realizado (W) \u00e9 igual ao produto da pot\u00eancia pelo tempo (\u2206t)\u00a0 —\u00a0 P=F.V\u00a0 —\u00a0 W=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u0394t\u00a0 —\u00a0 W=F.V.\u0394t\u00a0 —\u00a0 W=20mgV\u0394t=20.1,5.10.0,1.7200\u00a0 —\u00a0 W=216.000J\u00a0 —\u00a0 esse trabalho corresponde a 15% da energia (E) a ser ingerida\u00a0 —\u00a0 0,15E=216.000\u00a0 —\u00a0 E=216.000\/0,15\u00a0 —\u00a0 E=1.440kJ\u00a0 —\u00a0 em quilocalorias\u00a0 —\u00a0 E=1.440\/4\u00a0 —\u00a0<\/b><\/span><\/span><\/span>E=360kcal<\/b><\/span><\/span><\/span><\/p>\n

b) Seja N o n\u00famero de telhas transportadas em 2 h\u00a0 —\u00a0 W = N m g h\u00a0 —\u00a0 \u00a0N = W\/mgh\u00a0 —\u00a0 N=216.000\/1,5.10.3\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>N=4.800 telhas<\/b><\/span><\/span><\/span><\/p>\n

41-\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>\/\u0394t\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>\/\u0394t\u00a0 —\u00a0 o tempo de realiza\u00e7\u00e3o de um trabalho \u00e9 inversamente proporcional \u00e0 pot\u00eancia empregada\u00a0 —\u00a0 isso significa que quanto menor \u00e9 o tempo, maior \u00e9 a pot\u00eancia\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

42-\u00a0<\/b><\/span><\/span><\/span>A pot\u00eancia corresponde \u00e0 \u00a0medida da rapidez com que se transfere energia\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u0394t\u00a0 —\u00a0 observe nessa express\u00e3o que o mais eficiente (maior pot\u00eancia) \u00e9 aquele que fornece maior quantidade de energia em menos tempo\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

43-\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 \u00a0m = 400 kg\u00a0 —\u00a0 \u00a0g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u00a01 litro de O<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0libera 4,9 kcal\u00a0 —\u00a0 fazendo a convers\u00e3o da energia liberada por 2,5 L de O<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u00a0E<\/b><\/span><\/span><\/span>liberada<\/b><\/span><\/span><\/sub><\/span>\u00a0= 2,5(4,9)(4.200) = 51.450 J\u00a0 —\u00a0 essa energia corresponde a energia potencial adquirida pelo bloco\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>potencial<\/b><\/span><\/span><\/sub><\/span>\u00a0= E<\/b><\/span><\/span><\/span>liberada<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mgh = E<\/b><\/span><\/span><\/span>liberada<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 400(10)h = 51.450\u00a0 —\u00a0 h\u00a0\u224812,9 m.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

44-\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 — \u00a01 L = 1.000 mL de O<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0libera 4,9 kcal = 4,9\u00b44.200 = 20.580 J\u00a0 —\u00a0\u00a0Dt = 1 min = 60 s\u00a0 —\u00a0 c\u00e1lculo da quantidade de energia liberada (E) pela absor\u00e7\u00e3o de 50 mL\u00a0 —\u00a0 E=(50mL).(20.580J)\/(1.000mL)=1.029J\u00a0 —\u00a0 pot\u00eancia liberada por kg de massa da pessoa\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=E\/\u0394t=1.029\/60\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=17,15W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

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

45-<\/b><\/span><\/span><\/span>W=F.d.cos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=25.20.0,5\u00a0 —\u00a0 W=250J\u00a0 —\u00a0 P=W\/\u2206T=250\/5\u00a0 —\u00a0 P=50W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

46-\u00a0\u00a0Energia (Trabalho) necess\u00e1ria\u00a0 para elevar uma massa m de \u00e1gua \u00e0 altura de h=80cm=0,8m\u00a0 —\u00a0 W=m.g.h=m.10.0,8\u00a0 —\u00a0 W=8.m\u00a0 —\u00a0 massa de \u00e1gua que essa bomba de pot\u00eancia Po=20W consegue elevar com essa energia W=8.m em \u2206t=1s\u00a0 —\u00a0 Po=W\/\u2206t\u00a0 —\u00a0 20=8m\/1\u00a0 —\u00a0 m=20\/8=2,5kg\u00a0 —\u00a0 como o enunciado afirma que 1kg de \u00e1gua equivale ao volume de 1L\u00a0 —\u00a0 V=2,4L\u00a0 —\u00a0\u00a0R- B\u00a0<\/b><\/span><\/span><\/p>\n

47–\u00a0Po=W\/\u2206t\u00a0 —\u00a0 W=Po.\u2206t\u00a0 —\u00a0 l\u00e2mpadas\u00a0 —\u00a0 W1=0,1.10=1kWh\u00a0 —\u00a0 televis\u00e3o\u00a0 —\u00a0 W2=0,1.8=0,8kWh\u00a0 —\u00a0 geladeira\u00a0 —\u00a0 W3=0,3.24=7,2kWh\u00a0 —\u00a0 ventilador\u00a0 —\u00a0 W4=0,125.8=1,0kWh\u00a0 —\u00a0 Wtotal=1,0 + 0,8 + 7,2 + 1,0=10,0kWh\u00a0 —\u00a0 1 painel \u2013 0,5kWh\u00a0 —\u00a0 n pain\u00e9is \u2013 10,0kWh\u00a0 —\u00a0 N=20 pain\u00e9is\u00a0 —\u00a0\u00a0R-<\/b><\/span><\/span><\/p>\n

48-Volume de combust\u00edvel consumido em 1 hora\u00a0 —\u00a0 V=8L=8dm3=103cm3\u00a0 —\u00a0 massa de combust\u00edvel consumida em 1 hora\u00a0 —\u00a0 d=m\/V\u00a0 —\u00a0 0,675=m\/8.103\u00a0 —\u00a0 m=5,4.103g\u00a0 —\u00a0 calor fornecido pela queima dessa massa de combust\u00edvel\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1g \u2013 10000cal\u00a0 —\u00a0 5,4.103g \u2013 Q cal\u00a0 —\u00a0\u00a0 Q=5,4.107cal\u00a0 —\u00a0 transformando essa energia em joules\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1cal \u2013 4J\u00a0 —\u00a0 5,4.107cal\u00a0 —\u00a0 W J\u00a0 —\u00a0 W=4.5,4.107=2,16.108J\u00a0 —\u00a0 pot\u00eancia gerada em 1 hora=3600s\u00a0 —\u00a0 P=W\/\u2206t=2,16.108\/3600\u00a0 —\u00a0 P=6.104W\u00a0 —\u00a0 a pot\u00eancia desenvolvida pelo carro \u00e9 a pot\u00eancia \u00fatil=24kW=24.103W\u00a0 —\u00a0 rendimento=pot\u00eancia \u00fatil\/pot\u00eancia total\u00a0 —\u00a0 \u03b7=Pu\/Pt=24.103\/6.104\u00a0=0,4×100=40%\u00a0 —R- C\u00a0<\/b><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span>\u00a0a) Observe no gr\u00e1fico que eles v\u00e3o ter a mesma velocidade no ponto onde a reta (andando) e a curva (correndo) se interceptam que \u00e9 aproximadamente de V=8,5km\/h (eixo horizontal-velocidade)\u00a0 — observe tamb\u00e9m no eixo vertical (volume) que a partir dessa velocidade o consumo de oxig\u00eanio do atleta que est\u00e1 correndo \u00e9 menor do que o do atleta que est\u00e1 andando.<\/b><\/span><\/span><\/span><\/p>\n

b) Do gr\u00e1fico, quando ele est\u00e1 parado, velocidade nula eixo horizontal, o volume de oxig\u00eanio \u00e9 de 0,2 \u2113\/min\u00a0 —\u00a0 em 12h de repouso ele consumir\u00e1\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span>=720.0,2=144\u2113 de oxig\u00eanio.<\/b><\/span><\/span><\/span><\/p>\n

c) Pelo gr\u00e1fico, quando sua velocidade \u00e9 de 15km\/h ele consome 3,6\u2113\/min\u00a0 — do enunciado, para cada litro de oxig\u00eanio consumido s\u00e3o gastas 5 kcal\u00a0 —\u00a0 energia (W) consumida quando V=15km\/h\u00a0 —\u00a0 W=3,6.5=18kcal\u00a0 —\u00a0 1 cal= 4 J\u00a0 —\u00a0 W=18.000calx4=72.000J\u00a0 — essa energia \u00e9 consumida em 1min=60s\u00a0 —\u00a0 P=W\/\u2206t=72kJ\/60s\u00a0 —\u00a0 P=1,kW=1.200W.<\/b><\/span><\/span><\/span><\/p>\n

W=72kJ.<\/b><\/span><\/span><\/span><\/p>\n

d) O gr\u00e1fico fornece que a cada 7km\/h, andando, o consumo de oxig\u00eanio \u00e9 de 1,6 \u2113\/min\u00a0 —\u00a0 energia em \u2206t minutos\u00a0 —\u00a0 W=1,6.5=8kcal\u00a0 — barra de chocolate=560kcal\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 min \u2013 8kcal\u00a0 —\u00a0 \u2206t min \u2013 560kcal\u00a0 —\u00a0 \u2206t=560\/8=70 min\u00a0 —\u00a0 \u2206t=60min.<\/b><\/span><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span>a) Pelo enunciado, a pot\u00eancia \u00fatil\u00a0 P (W) desse sistema para bombeamento de \u00e1gua pode ser obtida\u00a0 pela express\u00e3o P = 0,1xAxv<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

P=0,1.2.A=2m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0e v=5m\/s\u00a0 —\u00a0 P=0,1.2.5=25W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P=25W.<\/b><\/span><\/span><\/span><\/p>\n

b) O trabalho (quantidade de energia) para elevar uma massa m de \u00e1gua a uma altura vertical de h=7,5m \u00e9 dado por W=m.g.h\u00a0 —\u00a0 d=m\/V\u00a0 —\u00a0 1kg\/L=1000g\/1000cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=1g\/cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 d=1kg\/L\u00a0 —\u00a0 1kg\/L=m\/1L\u00a0 —\u00a0 m=1kg\u00a0 —\u00a0 W=1.10.7,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W=75J.<\/b><\/span><\/span><\/span><\/p>\n

c) P=W\/\u2206t\u00a0 —\u00a0 25=mgh\/\u2206t\u00a0 —\u00a0 25=d.V.g.h\/\u2206t\u00a0 ===\u00a0 25=1.V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.10.7,5\/\u2206t\u00a0 —\u00a0 25\/75=V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/\u2206t\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/\u2206t=V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/1=volume de \u00e1gua bombeado em 1s=25\/75=1\/3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=(1\/3) L<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

d) Analisando a express\u00e3o fornecida pelo enunciado P = 0,1xAxv<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0voc\u00ea\u00a0 observa que a press\u00e3o P \u00e9 diretamente proporcional ao cubo da velocidade v\u00a0 —\u00a0 se v for reduzida \u00e0 metade a pot\u00eancia P ficar\u00e1 2<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=8 vezes menor, o que implica que a vaz\u00e3o (V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=(1\/3)L tamb\u00e9m ficar\u00e1 dividida por 8\u00a0 — V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=(1\/3.8)L=(1\/24) L\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=(1\/24) L.<\/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 dos exerc\u00edcios de vestibulares sobre Pot\u00eancia de uma for\u00e7a \u00a001-Maior consumo \u2013 Po=5.103W\u00a0 —\u00a0\u00a0\u2206t=2h\u00a0 — Po=W(energia consumida)\/\u2206t — 5,103=W\/2\u00a0 —\u00a0 W=10.103W\u00a0 —\u00a0\u00a0W=10kW R- B 02-\u00a0Velocidade constante \u2013 a for\u00e7a que est\u00e1 puxando \u00e9 igual ao peso da caixa d\u2019agua\u00a0 —\u00a0 P=m.g=66.10\u00a0 —\u00a0 P=660N\u00a0 —\u00a0 W=P.h=660.18\u00a0 —\u00a0 W=11.800 J\u00a0 —\u00a0 Po=W\/Dt=11.800\/120\u00a0 —\u00a0Po=99W 03-\u00a0O trabalho do peso independe da trajet\u00f3ria<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1320,"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-1324","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1324","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=1324"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1324\/revisions"}],"predecessor-version":[{"id":10855,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1324\/revisions\/10855"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1320"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1324"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}