{"id":2578,"date":"2016-10-26T04:01:52","date_gmt":"2016-10-26T04:01:52","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2578"},"modified":"2024-09-02T16:37:08","modified_gmt":"2024-09-02T16:37:08","slug":"trabalho-e-energia-resolucao","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/raciocinio-logico\/trabalho-e-energia\/trabalho-e-energia-resolucao\/","title":{"rendered":"Trabalho e Energia – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Trabalho e Energia<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>For\u00e7a\u00a0\"\"\u00a0—\u00a0 W<\/b><\/span><\/span><\/span>F1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.d.cos\u03b8\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>F1<\/b><\/span><\/span><\/sub><\/span>=0,4×2,5.cos\u03b8\u00a0 —\u00a0\u00a0W<\/b><\/span><\/span><\/span>F1<\/b><\/span><\/span><\/sub><\/span>=cos\u03b8=k\u00a0\u00a0 —\u00a0 cos\u03b8=k\u00a0 —\u00a0 For\u00e7a\u00a0\"\"\u00a0\u00a0 —\u00a0\u00a0W<\/b><\/span><\/span><\/span>F2<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.d.cos2\u03b8\u00a0 —W<\/b><\/span><\/span><\/span>F2<\/b><\/span><\/span><\/sub><\/span>=0,4×2,5.cos2\u03b8\u00a0 —\u00a0\u00a0W<\/b><\/span><\/span><\/span>\u00a0F2<\/b><\/span><\/span><\/sub><\/span>=cos2\u03b8\u00a0\u00a0\u00a0—\u00a0 cos2\u03b8\u00a0=cos<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u03b8\u00a0 –\u00a0sen<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u03b8\u00a0 —\u00a0 cos2\u03b8 =k<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 (1 \u2013 cos<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u03b8)\u00a0 —\u00a0 cos2\u03b8\u00a0= k<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 1 + k<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 cos2\u03b8\u00a0= 2k<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 1\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>\u00a0F2<\/b><\/span><\/span><\/sub><\/span>=cos2\u03b8 =2k<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R-D.<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Em todo gr\u00e1fico da for\u00e7a resultante\u00a0\"\"\u00a0na dire\u00e7\u00e3o do deslocamento d e,em fun\u00e7\u00e3o dele, o\u00a0trabalho realizado pela for\u00e7a\u00a0\"\"\u00a0\u00e9 numericamente\u00a0 igual \u00e0 \u00e1rea compreendida entre a reta representativa (linha cheia) e o eixo do deslocamento.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

W<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>(total)= \u00e1rea 1 + \u00e1rea 2 + \u00e1rea3 \u2013 \u00e1rea 4 \u2013 \u00e1rea 5\u00a0 —\u00a0 observe nas figuras abaixo que a \u00e1rea est\u00e1 delimitada entre a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

reta representativa at\u00e9 o eixo d e pode ser calculada pela soma das \u00e1reas do trap\u00e9zio com a \u00e1rea do ret\u00e2ngulo\u00a0 —\u00a0 W=(12 + 4).10\/2 + 12.5\u00a0 —\u00a0\u00a0W=140J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R-C.<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O 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\u00a0<\/b><\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Na 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, voc\u00ea observa que ela transportou na horizontal 1,5 caixas por segundo o que equivale a uma massa de\u00a0\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

–\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 caixa \u2013 80kg\u00a0 —\u00a0 1,5 caixas \u2013\u00a0m\u00a0 —\u00a0 m=80×1,5=120kg\u00a0 —\u00a0 essa \u00e9 a massa que a esteira na parte inclinada deve elevar de 5m em cada 1s com velocidade constante de 1,5m\/s\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

—\u00a0 c\u00e1lculo do trabalho pela altura\u00a0 —\u00a0W=m.g.h=120.10.5\u00a0 —\u00a0 W=6.000J (energia consumida para elevar uma massa de 120kg de 5m).\u00a0 —\u00a0 pot\u00eancia\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u2206t=6.000\/1\u00a0 —\u00a0P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=6.000W\u00a0(energia consumida para elevar uma massa de 120kg de 5m em 1s)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E.<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) 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\u00a0 — 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

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Por dia uma pessoa utiliza 30L\u00a0 —\u00a0 4pessoas\u00a0 —\u00a0 4x30L=12L\u00a0 —\u00a0 em um m\u00eas 4 pessoas utilizar\u00e3o \u2013 V=30.120=3600L\u00a0 —\u00a0 sendo a densidade da \u00e1gua 1kh\/L esse volume de 3600L corresponde a uma massa de m=3600kg=3.600.000g\u00a0 —\u00a0\u00a0m=3.600.000g\u00a0 —\u00a0\u00a0\u00a0c = 4,19 J\/g\u00b0C\u00a0 —\u00a0\u00a0\u2206t=30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C\u00a0 —\u00a0\u00a0equa\u00e7\u00e3o fundamental da calorimetria \u2013 Q=m.c.Dt\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\u00a0 —\u00a0 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 de \u00e1rea de superf\u00edcie de pain\u00e9is solares.<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) 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\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0120.000=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>=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.sen\u03b1=800.10.sen\u03b1\u00a0 —\u00a0 mas, 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=8.000sen\u03b1\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0sen\u03b1=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.sen\u03b8\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\u00a0P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=2.400N\u00a0\u00a0—\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 —\u00a0 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<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=4.400.60\u00a0 —<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=264kW.<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Do 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\u00a0\u2206S=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 \u2013 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>=Wu\/\u2206t\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 —\u00a0Pu\/P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=122.027\/<\/b><\/span><\/span><\/span><\/p>\n

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>=0,55 (esse n\u00famero representa o rendimento do motor do carro que \u00e9 de 55%)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R-C.<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\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>\u00a0m3\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 —T = 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>\u00a0=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\u00a0\u2206t = 100.T = 100.(2)\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u2206t = 200 s\u00a0 —\u00a0 pot\u00eancia \u00fatil da fonte de rota\u00e7\u00e3o\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=W\/\u2206t=(E<\/b><\/span><\/span><\/span>potencial gravitacional<\/b><\/span><\/span><\/sub><\/span>)\/\u2206t=Mgh\/\u2206t\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>x10x2\/200 =2\/200=1\/100\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>=1,0.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Volume de combust\u00edvel consumido em 1 hora\u00a0 —\u00a0 V=8L=8dm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=8.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 massa de combust\u00edvel consumida em 1 hora\u00a0 —\u00a0 d=m\/V\u00a0 —\u00a0 0,675=m\/8.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 m=5,4.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>g\u00a0 —\u00a0 calor fornecido pela queima dessa massa de combust\u00edvel\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1g \u2013 10000cal\u00a0 —\u00a0 5,4.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>g \u2013 Q cal\u00a0 —\u00a0\u00a0 Q=5,4.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>cal\u00a0 —\u00a0 transformando essa energia em joules\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1cal \u2013 4J\u00a0 —\u00a0 5,4.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>cal\u00a0 —\u00a0 W J\u00a0 —\u00a0 W=4.5,4.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>=2,16.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>J\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=24.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>W\u00a0 —\u00a0 rendimento=pot\u00eancia \u00fatil\/pot\u00eancia total\u00a0 —\u00a0 \u03b7=P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>\/P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=24.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\/6.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0=0,4×100=40%\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) 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 no eixo horizontal, o volume de oxig\u00eanio \u00e9 de 0,2 \u2113\/min\u00a0 —\u00a0 em 12h de repouso ele consumir\u00e1\u00a0 —\u00a0 Vr=720.0,2=<\/b><\/span><\/span><\/span>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 —\u00a0do enunciado, para cada litro de oxig\u00eanio consumido s\u00e3o gastas 5 kcal\u00a0 — \u00a0energia em \u2206t minutos\u00a0 —\u00a0 W=1,6×5=8kcal\u00a0 —\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\u00a0<\/b><\/span><\/span><\/span>\u2206t=70min.<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>N\u00e3o havendo movimento na vertical, nessa dire\u00e7\u00e3o a for\u00e7a resultante sobre o bloco e sobre o carrinho \u00e9 nula\u00a0 —\u00a0 for\u00e7a el\u00e1stica que a mola exerce sobre o bloco de massa m empurrando-o para a direita (figura), quando deformada de x\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>\"\"<\/p>\n

— F<\/b><\/span><\/span><\/span>mola<\/b><\/span><\/span><\/sub><\/span>=kx\u00a0 —\u00a0 n\u00e3o havendo atrito, essa for\u00e7a \u00e9 a resultante sobre o bloco e de intensidade F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 kx=ma<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>=kx\/m (I)\u00a0 —\u00a0 sobre o carrinho de massa M a for\u00e7a resultante \u00e9 a el\u00e1stica que a mola exerce sobre ele empurrando-o para a esquerda (figura), de intensidade F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=kx\u00a0\u00a0 — n\u00e3o havendo atrito, essa for\u00e7a \u00e9 a resultante sobre o bloco e de intensidade F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=M.a<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 como se movem em sentidos contr\u00e1rios a acelera\u00e7\u00e3o relativa entre eles ser\u00e1 a<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span>=a<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span>=kx\/m + kx\/M=kx(1\/m + 1\/M)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span>=kx(m + M)\/mM\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E.<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo a pot\u00eancia mec\u00e2nica constante a pot\u00eancia instant\u00e2nea coincide com a pot\u00eancia m\u00e9dia\u00a0 —\u00a0 P=W\/\u2206t=W\/(t \u2013 0)\u00a0 —\u00a0 P=W\/t\u00a0 —\u00a0 trabalho como varia\u00e7\u00e3o de energia cin\u00e9tica\u00a0 —\u00a0 W=E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>ci\u00a0<\/b><\/span><\/span><\/sub><\/span>= mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 mV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0 sendo V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0 (partiu do repouso)\u00a0 —\u00a0 W=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 P=W\/t=(mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2)\/t\u00a0 —\u00a0 P=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2t\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2Pt\/m\u00a0 —\u00a0\u00a0 mas, veja que, como 2P\/m \u00e9 constante, V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00e9 proporcional a t\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C.<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>1. Observe a figura onde foram colocadas as for\u00e7as\u00a0 a partir do peso do bloco que vale\u00a0 —-\u00a0 P=mg=24.10=240N\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

F=60N.<\/b><\/span><\/span><\/span><\/p>\n

2. Sendo a velocidade constante a massa est\u00e1 em equil\u00edbrio din\u00e2mico e a for\u00e7a resultante sobre ela \u00e9 nula e assim,\u00a0 for\u00e7a\u00a0\"\"\u00e9 a mesma que equilibra o sistema em 1\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0F=60N.<\/b><\/span><\/span><\/span><\/p>\n

3. Sendo o fio inextens\u00edvel, o deslocamento do fio onde a for\u00e7a \u00e9 F \u00e9 4 vezes maior que o deslocamento do fio onde a for\u00e7a \u00e9 P=4F\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>=F.d=60.4d=240d\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>4F=P<\/b><\/span><\/span><\/sub><\/span>=4F.d=4.60.d=240d\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>4F<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>tanto\u00a0\"\"\u00a0como o peso realizam o mesmo trabalho<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O trabalho da for\u00e7a peso (for\u00e7a gravitacional) entre dois pontos A e B, n\u00e3o depende da trajet\u00f3ria, mas apenas da altura (h) vertical entre A e B e \u00e9 fornecido por W<\/b><\/span><\/span><\/span>BA<\/b><\/span><\/span><\/sub><\/span>=+P.h na descida e por W<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>= -P.h na subida por qualquer trajet\u00f3ria (d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0ou d<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>).<\/b><\/span><\/span><\/span>\"\"<\/p>\n

01) Falsa\u00a0 —\u00a0 O trabalho \u00e9 o mesmo nos 3 casos, pois a altura \u00e9 a mesma (do piso t\u00e9rreo a um n\u00edvel superior).<\/b><\/span><\/span><\/span><\/p>\n

02) Correta.<\/b><\/span><\/span><\/span><\/p>\n

04) Falsa\u00a0 —\u00a0 o trabalho \u00e9 o mesmo.<\/b><\/span><\/span><\/span><\/p>\n

08) Falsa\u00a0 —\u00a0 a pot\u00eancia \u00e9 maior, pois P=W\/\u2206t, mas n\u00e3o \u00e9 poss\u00edvel afirmar quantas vezes.<\/b><\/span><\/span><\/span><\/p>\n

16) Correta\u00a0 —\u00a0 \u00a0Porque o tempo no m\u00e9todo 2 \u00e9 menor do que no m\u00e9todo 1 e P=W\/\u2206t.<\/b><\/span><\/span><\/span><\/p>\n

Corretas\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>(02 e 16).<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) \u00c9 relativa, mas pode estar compreendida entre 10cm e 30cm\u00a0 —\u00a0 aproximadamente 15cm.<\/b><\/span><\/span><\/span><\/p>\n

b) Para o avi\u00e3o\u00a0 —\u00a0 (P\/A)\/b=k\u00a0 —\u00a0 (P\/A)=k.60\u00a0 —\u00a0 7200=k.60\u00a0 —\u00a0 k=120\u00a0 —\u00a0 pardal (considerando sua envergadura como b=15cm=0,15m\u00a0 —\u00a0 (P\/A)=kb\u00a0 —\u00a0 (P\/A)=120.0,15\u00a0 —\u00a0 pardal\u00a0 —\u00a0 (P\/A)=18N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 pela express\u00e3o fornecida\u00a0 —\u00a0 (P\/A)=0,37V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 18=0,37V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=48,65\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0V\u22486,97m\/s.<\/b><\/span><\/span><\/span><\/p>\n

c) Energia efetivamente transformada em pot\u00eancia mec\u00e2nica\u00a0 —\u00a0 P=25% de 3,2J\/s=3,2\/4=0,9J\/s\u00a0 —\u00a0 P=F.V\u00a0 —\u00a0 0,9=F.6,97\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0F\u22480,13N.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Para que haja energia potencial gravitacional deve existir diferen\u00e7a de alturas para que a \u00e1gua, na tend\u00eancia de nivelar as alturas, passe atrav\u00e9s da turbina, gerando energia el\u00e9trica\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E.<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Observe atentamente a figura abaixo onde est\u00e3o colocados os dados fornecidos pelo <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

enunciado e pelo gr\u00e1fico\u00a0 —\u00a0\u00a0c\u00e1lculo, de B para A, do trabalho da for\u00e7a peso pelo teorema da varia\u00e7\u00e3o de energia potencial gravitacional\u00a0 — W<\/b><\/span><\/span><\/span>BA<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>PB<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>PA<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>BA<\/b><\/span><\/span><\/sub><\/span>=24000 \u2013 0\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>BA<\/b><\/span><\/span><\/sub><\/span>=24000J=E<\/b><\/span><\/span><\/span>PB<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>PB<\/b><\/span><\/span><\/sub><\/span>=m.g.h<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=P.h<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 24000=P.30\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P=800N<\/b><\/span><\/span><\/span>\u00a0 —\u00a0observe no gr\u00e1fico que a energia el\u00e1stica s\u00f3 come\u00e7a a surgir a partir de 20m, quando a corda come\u00e7a a ser esticada\u00a0 —assim,\u00a0<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=20m<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

b) Se o comprimento natural da mola \u00e9 20m e ela foi deformada at\u00e9 30m, a deforma\u00e7\u00e3o \u00e9 x=30 \u2013 20=10m, e para efetuar essa deforma\u00e7\u00e3o a energia potencial el\u00e1stica foi de 24.000J (veja gr\u00e1fico)\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>el\u00e1stica<\/b><\/span><\/span><\/sub><\/span>=K.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 24.000=K.(10)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>K=480N\/m.<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Para atingir a altura desejada de 3m ele deve ganhar de energia potencial a quantidade \u0394E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>, com N idas e vindas\u00a0 —\u00a0 \u0394E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=mgh-800.3\u00a0 —\u00a0 \u0394E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=2.400J\u00a0 —\u00a0 \u0394E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=nE<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2.400=n.600\u00a0 —\u00a0 n=4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura\u00a0 —\u00a0 como o bloco est\u00e1 em equil\u00edbrio, a resultante das for\u00e7as sobre ele \u00e9 nula, ou seja, a for\u00e7a el\u00e1stica \u00e9 equilibrada pela componente tangencial do peso\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 kx=Psen\u03b8\u00a0 —\u00a0 x=10.0,8\/100\u00a0 —\u00a0 x=8,0.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 como centro de massa do bloco est\u00e1 abaixo de plano referencial, sua energia potencial gravitacional \u00e9 negativa\u00a0 —\u00a0 soma pedida das energias\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>pg<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>=P.(-h) + kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=10(-0,1) + 100.(8.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2= -1 + 0,32\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>pg<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>=-0,68J\u00a0 —<\/b><\/span><\/span><\/span>\"\"<\/p>\n

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

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe nas figuras abaixo as for\u00e7as que agem em cada uma das partes:<\/b><\/span><\/span><\/span><\/p>\n

a) Como as partes se movem em movimento retil\u00edneo e uniforme, a resultante das for\u00e7as em cada uma delas \u00e9 nula\u00a0 —\u00a0\u00a0 corpo A\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 \u00a0N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= 5 N\u00a0 —\u00a0 F = F<\/b><\/span><\/span><\/span>atA<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F = \u03bc<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2022\u00a0N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u00a0F = 1,5 N\u00a0 —\u00a0\u00a0 corpo B\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= 10 N\u00a0 —\u00a0 T = F + F<\/b><\/span><\/span><\/span>atB<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 T = F + \u03bc<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2022\u00a0N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>T = 4,5N.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

b) A for\u00e7a T \u00e9 constante\u00a0 –p-\u00a0 trabalho da for\u00e7a T\u00a0 —\u00a0 W = T\u00a0\u2022\u00a0\u0394s\u00a0\u2022\u00a0cos \u03b1\u00a0 —\u00a0 o \u00e2ngulo entre a for\u00e7a e o deslocamento \u00e9 nulo\u00a0 —\u00a0\u00a0 \u00a0cos \u03b1 = 1\u00a0 —\u00a0 como a velocidade dos corpos tamb\u00e9m \u00e9 constante\u00a0 —\u00a0 \u00a0\u0394s = v\u00a0\u2022\u00a0\u0394t\u00a0 —\u00a0 W = T\u00a0\u2022\u00a0v\u00a0\u2022\u00a0\u0394t\u00a0 —\u00a0 W = 4,5\u00a0\u2022\u00a00,1\u00a0\u2022\u00a0120\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W = 54 J.<\/b><\/span><\/span><\/span><\/p>\n

c) Do resultado obtido no item a\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0F = 1,5 N.<\/b><\/span><\/span><\/span><\/p>\n

d) Chamando de \u03b4 a deforma\u00e7\u00e3o da mola\u00a0 —\u00a0 F = k\u00a0\u2022\u00a0\u03b4\u00a0 —\u00a0 1,5 = 10\u00a0\u2022\u00a0\u03b4 \u00a0—\u00a0 \u03b4 = 0,15 m\u00a0 —\u00a0 comprimento da mola\u00a0 —\u00a0 x = xo\u00a0+ \u03b4\u00a0 —\u00a0 x = 0,1 + 0,15\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0x = 0,25 m ou x = 25 cm.<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na figura as alturas dos centros de massa de cada configura\u00e7\u00e3o em rela\u00e7\u00e3o ao apoio horizontal\u00a0 — \u00a0a=10cm<\/b><\/span><\/span><\/span>\"\"<\/p>\n

a=0,1m\u00a0 —\u00a0 M=100g=0,1kg\u00a0 —\u00a0 energia potencial gravitacional em A\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>pA<\/b><\/span><\/span><\/sub><\/span>=m.g.h<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=3×0,1x10x0,05=0,15J\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

energia potencial gravitacional em B\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>pB<\/b><\/span><\/span><\/sub><\/span>=m.g.h<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=3×0,1x10x0,15=0,45J\u00a0 —\u00a0 \u2206E<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>PB<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>PA<\/b><\/span><\/span><\/sub><\/span>=0,45 \u2013 0,15=0,3J\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\u2206E<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=30.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>J.<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Colocando os pontos A, B e C na figura\u00a0 —\u00a0 Andr\u00e9\u00a0\u2013 ponto A \u2013 como V=0 s\u00f3 tem energia <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

potencial gravitacional\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pA<\/b><\/span><\/span><\/sub><\/span>=m.g.h\u00a0 —\u00a0 ponto C \u2013 como h=0 s\u00f3 tem energia cin\u00e9tica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>= E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 sistema conservativo \u2013 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span>=m.g.h\u00a0 —\u00a0 Daniel\u00a0– \u2013 ponto B \u2013 como V=0 s\u00f3 tem energia potencial gravitacional\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pB<\/b><\/span><\/span><\/sub><\/span>=2m.g.h\/2 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=m.g.h\u00a0 —\u00a0 ponto C \u2013 como h=0 s\u00f3 tem energia cin\u00e9tica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>= E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 sistema conservativo \u2013 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 — E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span>=m.g.h\u00a0 —\u00a0 portanto, ao chegarem ao solo ambos ter\u00e3o\u00a0a mesma energia cin\u00e9tica\u00a0 —\u00a0 como a energia cin\u00e9tica E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0com que ambos chegam ao solo \u00e9 a mesma, assim\u00a0 —\u00a0 Andr\u00e9\u00a0\u2013 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013\u00a0V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=\u221a(2E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\/m)\u00a0 —\u00a0 Daniel\u00a0\u2013 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=2m.V<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>=\u221a(E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\/m)\u00a0 —\u00a0 portanto, ao chegarem ao solo ter\u00e3o velocidades diferentes\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Vamos descrever todas as caracter\u00edsticas deste movimento, desde o in\u00edcio, at\u00e9 quando a velocidade \u00e9 m\u00e1xima\u00a0 — (desprezando-se os atritos):\u00a0<\/b><\/span><\/span><\/span>\"\"<\/p>\n

\u00a0I\u00a0\u2013 a esfera \u00e9 abandonada do repouso (V\u00ad<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00ad=0) e cai sujeita apenas \u00e0 for\u00e7a peso, com velocidade de intensidade V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, que vai aumentando devida apenas \u00e0 acelera\u00e7\u00e3o da gravidade, at\u00e9 atingir a mola.<\/b><\/span><\/span><\/span><\/p>\n

II\u00a0\u2013 ao atingir a mola, come\u00e7ando a comprimi-la, surge a for\u00e7a el\u00e1stica (\"\") que se op\u00f5e ao peso (\"\"). A intensidade de\u00a0\u00a0\"\"\u00a0vai aumentando, diminuindo a intensidade da for\u00e7a resultante que \u00e9 para baixo, fazendo com que V2\u00a0aumente cada vez menos, at\u00e9 chegar \u00e0 situa\u00e7\u00e3o III.<\/b><\/span><\/span><\/span><\/p>\n

III- nesta situa\u00e7\u00e3o as intensidades das for\u00e7as el\u00e1stica e peso tornam-se iguais (P=Fe), quando a velocidade de queda torna-se m\u00e1xima<\/b><\/span><\/span><\/span><\/p>\n

Veja na figura abaixo as condi\u00e7\u00f5es do exerc\u00edcio:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na situa\u00e7\u00e3o final\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=P\u00a0 —\u00a0\u00a0k.x=m.g\u00a0 —\u00a0 200.x=4.10\u00a0 —\u00a0 x=0,2m\u00a0 —\u00a0 toda energia mec\u00e2nica do in\u00edcio (E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=mgh) \u00e9 transformada em E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>cm\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0na situa\u00e7\u00e3o final\u00a0 —\u00a0\u00a0E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mgh=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mV<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —4.10.0,9=200.(0,2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 4.V<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 36= 4 + 2V<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=16\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=4m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B.<\/b><\/span><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Tomando o ponto C como n\u00edvel zero de refer\u00eancia para as alturas, a energia mec\u00e2nica em A, onde a energia cin\u00e9tica \u00e9 zero, ser\u00e1\u00a0E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=m.g.(2 + x)=0,5.10.(2 + x)=10 + 5x\u00a0—\u00a0 considerando o sistema conservativo, a energia mec\u00e2nica em C, onde a energia potencial gravitacional \u00e9 nula (h=0) e a cin\u00e9tica tamb\u00e9m (menor comprimento,V=0) ser\u00e1 somente a energia potencial el\u00e1stica\u00a0 —\u00a0\u00a0E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=k.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=100.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=50x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0—\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 10 + 5x=50x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 10x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013x -10=0\u00a0 —\u00a0 \u2206=B<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 4.A.C=1 + 80=81\u00a0 —\u00a0\u00a0\u221a\u2206=9\u00a0 —\u00a0 X=1\u00b19\/20 \u2013 x<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=10\/20=0,5m \u2013 x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 negativo e n\u00e3o satisfaz\u00a0 —\u00a0 o menor comprimento da mola \u00e9 y=0,8 \u2013 x=0,8 – 0,5\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0y=0,3m.<\/b><\/span><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Energia mec\u00e2nica em A\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh=2V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.0=10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=100J\u00a0 —\u00a0\u00a0energia mec\u00e2nica em B\u00a0 —E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0=m.g.h + mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.10.R + 2.V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=40 + V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=20 + V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0—\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>\u00a0= E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 100=40 + V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=\u221a60m\/s\u00a0 —\u00a0acelera\u00e7\u00e3o centr\u00edpeta \u2013 a<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=(\u221a60)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=30m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

b) Energia mec\u00e2nica em C \u2013 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=m.g.2R + m.V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=2.10.4 + 2V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=80 + V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=80 + V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 100=80 + V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=\u221a(20) m\/s\u00a0 —\u00a0 quando o carrinho passa por C com velocidade de\u00a0\u221a20m\/s, duas for\u00e7as para baixo<\/b><\/span><\/span><\/span>\"\"<\/p>\n

agem sobre ele, seu peso\u00a0\"\"\u00a0e a rea\u00e7\u00e3o da parte superior da superf\u00edcie\u00a0\"\"\u00a0\u00a0— for\u00e7a resultante centr\u00edpeta em C \u2013 \u00a0F<\/b><\/span><\/span><\/span>RC<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

=m.V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 N + P= m.V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 N + 2.10=2V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 N + 20=(\u221a20)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 N=20 \u2013 20\u00a0\u00a0—\u00a0<\/b><\/span><\/span><\/span>\u00a0N=0.<\/b><\/span><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Vamos supor que seja no ponto X que a formiga perca contato com a bola\u00a0 —\u00a0 nesse ponto, a rea\u00e7\u00e3o normal\u00a0\"\"\u00a0deixa de existir agindo sobre a formiga somente seu peso\u00a0\"\", que \u00e9 a for\u00e7a resultante centr\u00edpeta, dirigida para o centro da circunfer\u00eancia e de intensidade F<\/b><\/span><\/span><\/span>RC<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 observe no tri\u00e2ngulo menor – cos\u03b1=F<\/b><\/span><\/span><\/span>RC<\/b><\/span><\/span><\/sub><\/span>\/P\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>RC<\/b><\/span><\/span><\/sub><\/span>=P.cos\u03b1=mgcos\u03b1\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

–\u00a0\u00a0 mgcos\u03b1=mV<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0\u00a0—\u00a0 gcos\u03b1=V<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R \u00a0(I)\u00a0 —\u00a0\u00a0observe no tri\u00e2ngulo maior –\u00a0cos\u03b1=h\/R, que substitu\u00eddo em (I)\u00a0 —\u00a0 g.h\/R=V<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=g.h.\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

b) Chamando a posi\u00e7\u00e3o inicial da formiga de Y e considerando a horizontal que passa por X como n\u00edvel zero de altura\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 E<\/b><\/span><\/span><\/span>mY<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>CY<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pY\u00a0<\/b><\/span><\/span><\/sub><\/span>= 0 + mg.(R-h)\u00a0\u00a0—\u00a0 E<\/b><\/span><\/span><\/span>mY<\/b><\/span><\/span><\/sub><\/span>=m.g.(R-h)\u00a0\u00a0—\u00a0 E<\/b><\/span><\/span><\/span>mX<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cX<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pX<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 0\u00a0\u00a0—\u00a0 Em<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mY<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mX<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 m.g.(R-h)=m.V<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2.g.(R-h)\u00a0 —\u00a0 que substitu\u00eddo em V<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=g.h\u00a0 —\u00a0 g.h=2.g.R \u2013 2.g.h\u00a0 —\u00a0 3h=2R\u00a0\u00a0—\u00a0<\/b><\/span><\/span><\/span>\u00a0h=2R\/3.<\/b><\/span><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Observe no gr\u00e1fico que a energia potencial da bola varia com a altura sendo ambas fun\u00e7\u00e3o do tempo\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>(t) =<\/b><\/span><\/span><\/span><\/p>\n

=m.g.H(t)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>(t)=0,6.10.H(t)\u00a0 —\u00a0\u00a0\u00a0E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>(t)=6.H(t)\u00a0 —\u00a0 colocando valores para H, constru\u00edmos o gr\u00e1fico E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0X t:<\/b><\/span><\/span><\/span><\/p>\n

t=0 \u2013 H=1,6m \u2013 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=6.1,6=9,6J\u00a0 —\u00a0 t=5 – H=0,6m \u2013 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=6.5=3J\u00a0 —\u00a0 t=t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 H=0 \u2013 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 t=9 \u2013 H=0,4m \u2013 E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=6.0,4=<\/b><\/span><\/span><\/span><\/p>\n

2,4J e assim por diante.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) Como a energia total, que \u00e9 a energia mec\u00e2nica da bola se conserva no intervalo entre os choques (dados do exerc\u00edcio), nesses intervalos E<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 constante e igual \u00e0 energia potencial gravitacional nas alturas m\u00e1ximas, onde a energia cin\u00e9tica \u00e9 nula (V=0)\u00a0 —\u00a0 nos pontos mais altos \u2013 E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=constante<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

c) Para o choque que ocorre no instante t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, a energia mec\u00e2nica na altura m\u00e1xima E<\/b><\/span><\/span><\/span>p1<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>m1<\/b><\/span><\/span><\/sub><\/span>=m.g.H<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0,6.10.1,6\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>m1<\/b><\/span><\/span><\/sub><\/span>=9,6J \u00e9 igual \u00e0 energia cin\u00e9tica no choque com o solo\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>m1<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>c1<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0,6.V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>m1<\/b><\/span><\/span><\/sub><\/span>=0,3.V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 igualando, pois durante a queda a energia mec\u00e2nica se conserva\u00a0 —\u00a0 9,6=0,3V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=\u221a32\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=4\u221a2m\/s\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

da mesma maneira, na segunda queda\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>m2<\/b><\/span><\/span><\/sub><\/span>=mgH=0,6.10.0,4=2,4J=m.V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0,3V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0\u00a0—\u00a0 2,4=0,3V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=2\u221a2m\/s\u00a0 —\u00a0 coeficiente de restitui\u00e7\u00e3o\u00a0 —\u00a0 CR=V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2\u221a2\/4\u221a2\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span>CR=0,5.<\/b><\/span><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=mgh=60.10.4\u00a0\u00a0—\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=2.400J\u00a0 —\u00a0\u00a0E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0=mV<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.h=60V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 60.10.0,8=\u00a030V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 480\u00a0—\u00a0 2.400=30V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 480\u00a0 —\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=8m\/s.<\/b><\/span><\/span><\/span><\/p>\n

b) Trata-se de um lan\u00e7amento obliquo com velocidade V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>, no ponto B. Nele, a velocidade de lan\u00e7amento (inicial) V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0possui duas componentes V<\/b><\/span><\/span><\/span>BX<\/b><\/span><\/span><\/sub><\/span>\u00a0horizontal e constante durante todo o movimento e V<\/b><\/span><\/span><\/span>BY<\/b><\/span><\/span><\/sub><\/span>, vertical e vari\u00e1vel<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(lan\u00e7amento vertical)\u00a0 —\u00a0 no ponto mais alto da trajet\u00f3ria, ponto C, a componente vertical V<\/b><\/span><\/span><\/span>BY<\/b><\/span><\/span><\/sub><\/span>\u00a0se anula (V<\/b><\/span><\/span><\/span>BY<\/b><\/span><\/span><\/sub><\/span>=0),\u00a0 — restando apenas a componente constante V<\/b><\/span><\/span><\/span>BY<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>BX<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.cos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=8.\u221a3\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>BY<\/b><\/span><\/span><\/sub><\/span>=4\u221a3m\/s\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>CY<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>BY<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>CY<\/b><\/span><\/span><\/sub><\/span>=60.16.3\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>CY<\/b><\/span><\/span><\/sub><\/span>=1.440J.<\/b><\/span><\/span><\/span>\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

c) E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>=\u00a0E<\/b><\/span><\/span><\/span>cC<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pC<\/b><\/span><\/span><\/sub><\/span>=60.(4\u221a3)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 60.10.h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2.440=1.440 + 600h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0<\/b><\/span><\/span><\/span>\u00a0h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=1,6m.<\/b><\/span><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A for\u00e7a de 1,5N provoca um deslocamento de 1\/4 da circunfer\u00eancia de valor\u00a0 —\u00a0 \u2206S=2\u03c0R\/4=2×3,1×4.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>4=6,2.\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 essa energia de ativa\u00e7\u00e3o corresponde ao trabalho realizado por essa for\u00e7a nesse deslocamento\u00a0 —\u00a0 W=F.d\u00a0 — W= =1,5.6,2.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>=9,3.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0J\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E.<\/b><\/span><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Energia mec\u00e2nica inicial da mola (ponto P) \u2013 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pei<\/b><\/span><\/span><\/sub><\/span>=0 + k.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>= 10.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 observe <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

na figura abaixo que x=(2\u221a2m), comprimento da mola deformada\u00a0 — (1m), comprimento natural da mola\u00a0 —\u00a0 deforma\u00e7\u00e3o inicial da mola \u2013 x=(2\u221a2 \u2013 1)m\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=10.(2\u221a2-1)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=5(2.1,4 \u2013 1)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=5.1,82\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=16,2J\u00a0 —\u00a0 energia mec\u00e2nica na situa\u00e7\u00e3o final da mola (ponto T) \u2013 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pef<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 observe na figura abaixo que a deforma\u00e7\u00e3o da mola na situa\u00e7\u00e3o final \u00e9 x<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=comprimento da mola <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

deformada (2m) \u2013 comprimento natural da mola (1m)\u00a0 —\u00a0 x<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=2 \u2013 1=1m\u00a0 —\u00a0E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=10.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 1.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=5 + V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=5 + V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 16,2=5 + V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V\u2248\u221a22,4\u00a0\u2248\u00a04,7m\/s.<\/b><\/span><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Nos dois casos, a deforma\u00e7\u00e3o da mola \u00e9 a mesma (x), armazenando as duas molas mesma energia potencial el\u00e1stica\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>pel<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 — \u00a0energia potencial gravitacional em rela\u00e7\u00e3o \u00e0 linha da mola n\u00e3o deformada\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>pg<\/b><\/span><\/span><\/sub><\/span>= = m g x\u00a0 —\u00a0 pela conserva\u00e7\u00e3o da energia, a velocidade v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0de lan\u00e7amento de um dardo \u00e9\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pel<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pg<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=mgx + kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2\/m(mgx + kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=\u221a2\/(mgx + kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 observe que, como a massa m aparece no denominador, o dardo de maior massa \u00e9 o que tem menor velocidade inicial, ou seja, o dardo I, que tem um peda\u00e7o de chumbo grudado nele\u00a0 —\u00a0 ap\u00f3s sair dos canos dos brinquedos, desprezando a resist\u00eancia do ar, os dados ficam sujeitos exclusivamente \u00e0 for\u00e7a peso, tendo, portanto, a mesma acelera\u00e7\u00e3o g\u00a0 —\u00a0 por isso os gr\u00e1ficos s\u00e3o retas paralelas, como mostrado na op\u00e7\u00e3o A\u00a0 —<\/b><\/span><\/span><\/span>R- A.<\/b><\/span><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pela conserva\u00e7\u00e3o da quantidade de movimento\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=m.V + 2m.0\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=m.V\u00a0 —\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Q<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span>=(m + 2m).V\u2019\u00a0 —\u00a0Q<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span>\u00a0 — \u00a0m.V=3m.V\u2019\u00a0 —\u00a0 V\u2019=V\/3\u00a0 —\u00a0 supondo que a energia mec\u00e2nica do sistema de massa 3m se conserve, como a altura n\u00e3ovaria, toda energia \u00a0cin\u00e9tica antes do contato <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

com a mola se transforma em energia potencial<\/b><\/span><\/span><\/span><\/p>\n

el\u00e1stica ap\u00f3s a composi\u00e7\u00e3o parar e a mola for comprimida de x\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=3mV\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 —\u00a0 E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>pe\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 3mV\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 (3m\/2)\/(V\/3)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= kx2\/2\u00a0 —\u00a0 x = v \u00b7\u221a m\/ (3k)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E.<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

Sendo a o\u00a0atrito com o ar desprezado a energia mec\u00e2nica em cada ponto \u00e9 constante\u00a0— em A,\u00a0E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span> <\/b>MgH\u00a0— num\u00a0ponto B de altura h onde E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=3E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=3Mgh\u00a0—\u00a0E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>=Mgh + 3Mgh=4Mgh\u00a0—\u00a0E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span>\u00a0—<\/b><\/span><\/span><\/span><\/p>\n

MgH = 4Mgh —\u00a0h=H\/4.<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

 <\/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 Trabalho e Energia 01-\u00a0For\u00e7a\u00a0\u00a0—\u00a0 WF1=F1.d.cos\u03b8\u00a0 —\u00a0 WF1=0,4×2,5.cos\u03b8\u00a0 —\u00a0\u00a0WF1=cos\u03b8=k\u00a0\u00a0 —\u00a0 cos\u03b8=k\u00a0 —\u00a0 For\u00e7a\u00a0\u00a0\u00a0 —\u00a0\u00a0WF2=F2.d.cos2\u03b8\u00a0 —WF2=0,4×2,5.cos2\u03b8\u00a0 —\u00a0\u00a0W\u00a0F2=cos2\u03b8\u00a0\u00a0\u00a0—\u00a0 cos2\u03b8\u00a0=cos2\u03b8\u00a0 –\u00a0sen2\u03b8\u00a0 —\u00a0 cos2\u03b8 =k2\u00a0\u2013 (1 \u2013 cos2\u03b8)\u00a0 —\u00a0 cos2\u03b8\u00a0= k2\u00a0\u2013 1 + k2\u00a0 —\u00a0 cos2\u03b8\u00a0= 2k2\u00a0\u2013 1\u00a0 —\u00a0 W\u00a0F2=cos2\u03b8 =2k2\u00a0\u2013 1\u00a0 —\u00a0\u00a0R-D. 02-\u00a0Em todo gr\u00e1fico da for\u00e7a resultante\u00a0\u00a0na dire\u00e7\u00e3o do deslocamento d e,em fun\u00e7\u00e3o dele, o\u00a0trabalho realizado<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2576,"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-2578","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2578","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=2578"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2578\/revisions"}],"predecessor-version":[{"id":10989,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2578\/revisions\/10989"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2576"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2578"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}