{"id":3623,"date":"2018-02-13T12:43:55","date_gmt":"2018-02-13T12:43:55","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3623"},"modified":"2024-08-20T15:56:30","modified_gmt":"2024-08-20T15:56:30","slug":"resolucao-comentada-aplicacoes-das-leis-de-newton-sem-atrito","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/aplicacoes-das-leis-de-newton-sem-atrito\/resolucao-comentada-aplicacoes-das-leis-de-newton-sem-atrito\/","title":{"rendered":"Aplica\u00e7\u00f5es das leis de Newton sem Atrito – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o Comentada \u00a0<\/b><\/span><\/span><\/span><\/p>\n

Aplica\u00e7\u00f5es das leis de Newton sem Atrito<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>As leis de Kepler n\u00e3o justificam a afirma\u00e7\u00e3o do astronauta porque elas versam sobre forma da \u00f3rbita, per\u00edodo da \u00f3rbita e \u00e1rea varrida na \u00f3rbita\u00a0 —\u00a0 a afirma\u00e7\u00e3o explica-se pelo Princ\u00edpio Fundamental da Din\u00e2mica, pois o que est\u00e1 em quest\u00e3o s\u00e3o a massa e o peso do telesc\u00f3pio\u00a0 —\u00a0 como o astronauta e o telesc\u00f3pio est\u00e3o em \u00f3rbita, est\u00e3o sujeitos apenas \u00e0 for\u00e7a peso, e, consequentemente, \u00e0 mesma acelera\u00e7\u00e3o (centr\u00edpeta), que \u00e9 a da gravidade local, tendo peso <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

APARENTE nulo\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=P\u00a0 —\u00a0 \u00a0m a = m g\u00a0 —\u00a0 a = g\u00a0 —\u00a0a acelera\u00e7\u00e3o de queda do telesc\u00f3pio e dos astronautas \u00e9 a mesma\u00a0 —\u00a0 \u00e9 por esse motivo que os astronautas \u00a0flutuam dentro e fora de uma nave e diz-se nesse caso que os corpos est\u00e3o em estado de imponderabilidade\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

Observa\u00e7\u00e3o\u00a0 —\u00a0 considerando R = 6.400 km o raio da Terra, \u00e0 altura h = 540 km, o raio da \u00f3rbita do telesc\u00f3pio \u00e9 r = R + h = 6.400 + 540 = 6.940 km\u00a0 —\u00a0 de acordo com a lei de Newton da gravita\u00e7\u00e3o, a intensidade do campo gravitacional num ponto da \u00f3rbita \u00e9 \u00a0\u00a0—\u00a0 g = g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.(R\/r)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>, sendo g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>= 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(acelera\u00e7\u00e3o da gravidade na superf\u00edcie da Terra)\u00a0 —\u00a0 g=10.(6.400\/6.940)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 g=8,5m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 ou seja, o peso REAL do telesc\u00f3pio na \u00f3rbita n\u00e3o \u00e9 pequeno, \u00e9 85% do seu peso na superf\u00edcie terrestre.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Peso de cada objeto\u00a0 —\u00a0 P=m.g\u00a0 —\u00a0 P=0,050.10\u00a0 —\u00a0 P=0,5N\u00a0 —\u00a0 n=peso total (que o fio suporta sem se romper)\/peso de cada objeto=5,80\/05\u00a0 —\u00a0 n=11,6\u00a0 —\u00a0 n=12 objetos\u00a0 —\u00a0 o fio se rompe entre o suporte e o primeiro objeto, onde a for\u00e7a de tra\u00e7\u00e3o \u00e9 maior\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

03-\u00a0<\/b><\/span><\/span><\/span>Observe na figura abaixo que a for\u00e7a\u00a0\"\"\u00a0aplicada pelo comerciante sobre o prato foi decomposta em duas parcelas\u00a0\"\"\u00a0e\u00a0\"\".\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A parcela\u00a0\"\"\u00a0horizontal e para a direita n\u00e3o influi na indica\u00e7\u00e3o da balan\u00e7a\u00a0 —\u00a0 \u00a0a for\u00e7a adicional aplicada \u00e9 a parcela vertical\u00a0\"\"\u00a0de intensidade\u00a0 —\u00a0 sen37<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=F<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>\/F\u00a0 —\u00a0 0,6=FY\/5\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>=3N\u00a0 —\u00a0 assim, uma massa de 1,5kg que pesa P=mg\u00a0 —P=1,5.10=15N, passar\u00e1 a pesar 18N e ter\u00e1 massa de m=18\/10=1,8kg\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como despreza-se os atritos, sobre ele n\u00e3o surgem for\u00e7as horizontais, apenas as verticais que s\u00e3o o peso e a for\u00e7a de contato entre ele e a superf\u00edcie horizontal\u00a0 —\u00a0 essas for\u00e7as tem sentidos contr\u00e1rios com a parte esquerda do brinquedo descendo e a direita<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0Subindo\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Calculo do peso de uma formiga\u00a0 —\u00a0 massa de uma formiga\u00a0 —\u00a0 m=20.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>g=2.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0 P=mg=2.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>.10\u00a0 —\u00a0 P=2.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>N (peso de uma formiga) \u00a0—\u00a0 a intensidade da for\u00e7a de compress\u00e3o entre o suporte B e o apoio (N=40.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>=4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>N) equivale a duas vezes o peso total (P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>) das formigas, pois, ao mesmo tempo que A ficou P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0mais leve, B ficou P\u00ad<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0mais pesado\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=peso total das<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

formigas = (n\u00famero de formigas).(peso de uma formiga)\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0= n.2.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 2.(n.2.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 n = 4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\/4.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 n = 100 formigas\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Calculando a massa de um corpo que pesa P=1.000N\u00a0 —\u00a0\u00a0 P=m.g\u00a0 —\u00a0 1.000=m.10\u00a0 —\u00a0 m=100kg=10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"O procedimento I \u00e9 correto, pois a massa de um corpo \u00e9 uma grandeza fundamental e que pode ser medida por compara\u00e7\u00e3o com massa padr\u00e3o utilizando uma balan\u00e7a de pratos, onde colocamos \"\"<\/b><\/span><\/span><\/span><\/p>\n

num dos pratos a massa que queremos medir, e no outro, \u00a0massas padr\u00e3o, at\u00e9 que equilibrem a balan\u00e7a. Observe que esse processo independe da intensidade do campo gravitacional (\u00e9 v\u00e1lido para qualquer planeta), pois sempre ser\u00e1 o mesmo para os dois objetos.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0\u00a0o procedimento II \u00e9 correto, pois um dinam\u00f4metro mede a for\u00e7a de rea\u00e7\u00e3o ao peso do corpo, que nas condi\u00e7\u00f5es indicadas \u00e9 igual ao peso do corpo\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0o procedimento III n\u00e3o \u00e9 correto, pois corpos de massa diferentes possuem o mesmo tempo de queda<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0o procedimento IV \u00e9 equivalente ao procedimento I e desta forma medir\u00e1 a massa do corpo e n\u00e3o seu peso.<\/b><\/span><\/span><\/span><\/p>\n

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

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Peso do carro\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=m.g\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=700.10\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=7000N\u00a0 —\u00a0 chamando de P o peso do <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

contrapeso, de m sua massa, e colocando todas as for\u00e7as\u00a0 —\u00a0 observe que sobre o carro agem as for\u00e7as 7P (para cima) o peso do carro P<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=7000N (para baixo)\u00a0 —\u00a0 como ele est\u00e1 em equil\u00edbrio\u00a0 —\u00a0 7P=P<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 7P=7000\u00a0 —\u00a0 P=1000N\u00a0 —\u00a0 \u00a0m=100kg (massa m\u00ednima do contrapeso)\u00a0 —\u00a0 \u00a0o cabo central exerce uma for\u00e7a de 2P (veja figura)\u00a0 —\u00a0 F=2.1000\u00a0 —\u00a0 F=2000N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Os \u00f3rg\u00e3os internos s\u00f3 se movem ou tendem a se mover, por in\u00e9rcia, quando houver varia\u00e7\u00e3o de velocidade, ou seja, surgir acelera\u00e7\u00e3o, o que s\u00f3 ocorre na alternativa D\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Colocando as for\u00e7as que agem sobre o bloco 1 que s\u00e3o\u00a0 —\u00a0 seu peso – P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=mg\u00a0 —\u00a0 for\u00e7a com que o bloco 2 age sobre o bloco 1 – N<\/b><\/span><\/span><\/span>21<\/b><\/span><\/span><\/sub><\/span>=N\u00a0 —\u00a0 como o bloco 1 sobe com movimento uniformemente <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

retardado, a resultante das for\u00e7as sobre ele \u00e9 para baixo\u00a0 —F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a\u00a0 — mg \u2013 N = ma (I).<\/b><\/span><\/span><\/span><\/p>\n

For\u00e7as que agem sobre o bloco 2\u00a0 —\u00a0 seu peso \u2013 P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=mg\u00a0 —\u00a0 for\u00e7a com que o bloco 1 age sobre o bloco 2 \u2013 N<\/b><\/span><\/span><\/span>12<\/b><\/span><\/span><\/sub><\/span>=N\u00a0 —\u00a0 for\u00e7a com que o bloco 3 age sobre o bloco 2 \u2013 N<\/b><\/span><\/span><\/span>32<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 como o bloco 2 sobe com movimento uniformemente retardado, a resultante das for\u00e7as sobre ele \u00e9 para baixo\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=ma\u00a0 —\u00a0 mg + N \u2013 N<\/b><\/span><\/span><\/span>32<\/b><\/span><\/span><\/sub><\/span>\u00a0= ma (II)\u00a0 —\u00a0 somando as equa\u00e7\u00f5es (I) com (II)\u00a0 —\u00a0 2mg \u2013 N<\/b><\/span><\/span><\/span>32<\/b><\/span><\/span><\/sub><\/span>\u00a0= 2ma\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>32<\/b><\/span><\/span><\/sub><\/span>\u00a0= 2mg \u2013 2ma\u00a0 —\u00a0 for\u00e7a que 3 exerce sobre 2\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>32<\/b><\/span><\/span><\/sub><\/span>=2m(g \u2013 a)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como o sistema (barco + pessoa + pacote) se movem juntos, eles possuem a mesma acelera\u00e7\u00e3o que \u00e9 fornecida pela segunda lei de Newton\u00a0 — F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0= m.a\u00a0 —\u00a0 \u00a0240 = (100 + 58 + 2).a\u00a0 —\u00a0 240 = 160.a ==> a = 240\/160 = 1,5 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 apenas sobre o pacote de<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

2 kg\u00a0 F = m.a = 2.1,5 =\u00a03,0 N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Colocando as for\u00e7as sobre cada bloco:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

bloco A \u2013 sobe\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=ma\u00a0 —\u00a0\u00a0T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0– 40= 4.a (I)\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

bloco B \u2013 direita\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=ma\u00a0 —\u00a0T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=11.a (II)\u00a0\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

bloco C \u2013 desce\u00a0 —F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=ma\u00a0 —\u00a0 50 \u2013 T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=ma\u00a0 —\u00a0\u00a050 \u2013 T\u00ad<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=5.a (III)<\/b><\/span><\/span><\/span><\/p>\n

somando (I), (II) e (III)\u00a0\u00a0 —\u00a0\u00a010=20.a\u00a0\u00a0 —\u00a0\u00a0a=0,5m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 o bloco A sobe com acelera\u00e7\u00e3o de 0,5m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>, percorrendo\u00a0\u2206S=25cm=0,25m e tendo V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0 em X e V em Y\u00a0 —\u00a0 aplicando a equa\u00e7\u00e3o de Torricelli\u00a0 —V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= Vo<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.a.\u2206S\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.0,5.0,25\u00a0 —\u00a0 V=\u221a(0,25)\u00a0 —\u00a0\u00a0V=0,5m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R – A<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Decompondo, na figura I a for\u00e7a de tra\u00e7\u00e3o T<\/b><\/span><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0em suas componentes vertical e horizontal, e colocando todas as for\u00e7as que agem sobre a tora nas duas figuras, voc\u00ea observa que:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

figura I\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>1x<\/b><\/span><\/span><\/sub><\/span>\u00a0+ N<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=P\u00a0 —\u00a0\u00a0N<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=P \u2013 T<\/b><\/span><\/span><\/span>1x<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0figura II\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=P\u00a0 —\u00a0 portanto N<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>>N<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

Observ. \u2013 A intensidade da for\u00e7a de atrito entre a tora e o solo n\u00e3o depende da \u00e1rea de contato entre os mesmos e, assim, F<\/b><\/span><\/span><\/span>at1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>at2<\/b><\/span><\/span><\/sub><\/span>\u00a0e T<\/b><\/span><\/span><\/span>1x<\/b><\/span><\/span><\/sub><\/span>=T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como o andaime se encontra parado (equil\u00edbrio est\u00e1tico) a resultante das for\u00e7as que agem sobre ele \u00ea nula\u00a0 —\u00a0 \u00a0observe que os pesos sobre o andaime encontram-se no meio (peso de um oper\u00e1rio + peso do andaime, supondo-o homog\u00eaneo) e na extremidade esquerda (peso do outro oper\u00e1rio)\u00a0 —\u00a0 assim, a tens\u00e3o na corda X \u00e9 maior que a na corda Y\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>For\u00e7as que agem sobre o balc\u00e3o na dire\u00e7\u00e3o do movimento\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 parcela do peso que tenta fazer o bloco descer o plano inclinado\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=m.g.sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0—\u00a0 P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=200.10.0,5\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=1.000N\u00a0 —\u00a0 F\u00a0 — intensidade da for\u00e7a aplicada pelo empregado sobre o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0F\u00a0 — intensidade da for\u00e7a aplicada pelo empregado sobre o balc\u00e3o fazendo-o descer com velocidade constante, ou seja, est\u00e1 em equil\u00edbrio est\u00e1tico (F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=0)\u00a0 —\u00a0 assim,\u00a0F=P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=1.000 N\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

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

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) A mola inteira (mola equivalente) tem constante el\u00e1stica k\u2019=12N\/m e, nela, as tr\u00eas partes est\u00e3o associadas em s\u00e9rie, sendo que 1\/k\u2019= 1\/k + 1\/k +1\/k, onde k \u00e9 a constante el\u00e1stica de cada parte.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

1\/k\u2019=3\/k\u00a0 —\u00a0 1\/12 = 3\/k\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0k =36N\/m<\/b><\/span><\/span><\/span>\u00a0(constante el\u00e1stica de cada parte da mola).<\/b><\/span><\/span><\/span><\/p>\n

b) Paralelo (figura 2)\u00a0 —\u00a0 k\u00ad<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=36 + 36 +36\u00a0 —\u00a0 k<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=108N\/m\u00a0 —\u00a0 per\u00edodo de oscila\u00e7\u00e3o – T=2\u03c0\u221a(m\/k<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 T=2\u03c0\u221a(0,1\/108)\u00a0 —\u00a0\u00a0T=2\u03c0\u221a9,26\u00a0 —\u00a0T=2\u03c0.0,03\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>T=6.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u03c0 s\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

c) S\u00e9rie\u00a0 —\u00a0 k<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=12N\/m\u00a0 —\u00a0 T=2\u03c0\u221a(m\/k<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 T=2\u03c0\u221a(0,1\/12)\u00a0 —\u00a0 T=2\u03c0\u221a(0,008)\u00a0 —\u00a0 T=2\u03c0.0,09\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>T=18.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u03c0 s\u00a0<\/b><\/span><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/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 \u00a0 Aplica\u00e7\u00f5es das leis de Newton sem Atrito 01-\u00a0As leis de Kepler n\u00e3o justificam a afirma\u00e7\u00e3o do astronauta porque elas versam sobre forma da \u00f3rbita, per\u00edodo da \u00f3rbita e \u00e1rea varrida na \u00f3rbita\u00a0 —\u00a0 a afirma\u00e7\u00e3o explica-se pelo Princ\u00edpio Fundamental da Din\u00e2mica, pois o que est\u00e1 em quest\u00e3o s\u00e3o a massa e o peso do telesc\u00f3pio\u00a0 —\u00a0 como<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3621,"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-3623","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3623","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=3623"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3623\/revisions"}],"predecessor-version":[{"id":10629,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3623\/revisions\/10629"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3621"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3623"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}