{"id":1229,"date":"2015-08-26T00:24:23","date_gmt":"2015-08-26T00:24:23","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1229"},"modified":"2024-08-23T13:09:43","modified_gmt":"2024-08-23T13:09:43","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-lei-de-hooke-e-associacao-de-molas","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/forca-elastica-lei-de-hooke\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-lei-de-hooke-e-associacao-de-molas\/","title":{"rendered":"Lei de Hooke e Associa\u00e7\u00e3o de molas – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Lei de Hooke e Associa\u00e7\u00e3o de molas<\/b><\/span><\/span><\/span><\/p>\n

\u00a001<\/b><\/span><\/span>– A mola fica deformada de x=(22 \u2013 10)=12cm\u00a0 —\u00a0 x=12cm\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Numa deforma\u00e7\u00e3o de 12cm\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= P = 4N\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= K.x \u00a0—\u00a0 4 =K.12\u00a0 —\u00a0 K=1\/3 N\/cm<\/b><\/span><\/span><\/p>\n

K \u00e9 constante para qualquer deforma\u00e7\u00e3o (lei de Hooke)\u00a0 —\u00a0 para F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=P=6N\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=K.x\u00a0 —\u00a0 6=1\/3.x\u00a0 —\u00a0 x=18cm\u00a0 —\u00a0 fica deformada de 18cm e seu comprimento ser\u00e1 L=18 + 10 =\u00a0<\/b><\/span><\/span>28cm<\/b><\/span><\/span><\/p>\n

02<\/b><\/span><\/span>– R- D \u2013 vide teoria<\/b><\/span><\/span><\/p>\n

03<\/b><\/span><\/span>– K=300N\/m e \u00e9 constante (obedece \u00e0 lei de Hooke). A m\u00e1xima deforma\u00e7\u00e3o da tira de borracha \u00e9 de 28cm (Dx=28cm=0,28m)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

F=K.Dx\u00a0 —\u00a0 F=300.0,28\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=84N\u00a0 R-E<\/b><\/span><\/span><\/p>\n

04<\/b><\/span><\/span>– a) As for\u00e7as que atuam sobre a caixa s\u00e3o o peso,\u00a0 vertical e para baixo, a for\u00e7a normal, exercida pelo plano e perpendicular a ele, e a for\u00e7a el\u00e1stica, exercida pela mola.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) Como a caixa est\u00e1 em repouso, temos: F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0= F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/p>\n

\"\"<\/p>\n

m.g.sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0= K.x\u00a0 —\u00a0 5.10.1\/2 = 100.x\u00a0 —\u00a0 x=25\/100\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>x=0,25m<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span>\u00a0Em 2\u00a0 —\u00a0 F=Kx\u00a0 —\u00a0 10=K.5\u00a0 —\u00a0 K=2N\/cm\u00a0 —\u00a0 em 3\u00a0 —\u00a0 F=Kx\u00a0 —\u00a0 25=K.12,5\u00a0 —\u00a0 K=2N\/cm\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Sim, obedece, pois K \u00e9 constante<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span>\u00a0Ap\u00f3s o sistema entrar em movimento com acelera\u00e7\u00e3o\u00a0\"\", as molas j\u00e1 se encontram deformadas de x<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0e x<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0e a mola A sujeita \u00e0 for\u00e7a de tra\u00e7\u00e3o\u00a0\u00a0 —\u00a0\u00a0\"\"<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0\u00a0 bloco 2\u00a0\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.a\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 T=8a<\/b><\/span><\/span>\u00a0I\u00a0 —\u00a0 bloco 1\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F \u2013 T =12a\u00a0 II\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 resolvendo I com II\u00a0\u00a0<\/b><\/span><\/span>—\u00a0\u00a0 F=20a e T=8a\u00a0\u00a0 —<\/b><\/span><\/span>\u00a0 sendo as molas id\u00eanticas, elas possuem a mesma constante el\u00e1stica K\u00a0 — F=Kx<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 x<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=20a\/K\u00a0 —\u00a0 T=Kx<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 8a = Kx<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

x<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=8a\/K\u00a0 —\u00a0 x<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/x<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=8a\/K\/K\/20a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>x<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/x<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=2\/5\u00a0<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span>\u00a0a) balde e elevador em repouso\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 P=F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 mg=Kx<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=mg\/K<\/b><\/span><\/span><\/p>\n

b) balde e elevador subindo com acelera\u00e7\u00e3o a\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 — F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 P=ma\u00a0 —\u00a0 K.(x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ d) \u2013 mg=ma\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 a=(K.(x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ d) \u2013 mg)\/m<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span>\u00a0Quando o fio \u00e9 cortado, a esfera desce 1m e p\u00e1ra momentaneamente e, nesse instante temos o esquema abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

T \u2013 for\u00e7a de tra\u00e7\u00e3o em cada uma das molas e o peso da esfera – P=mg=5,1.10\u00a0 —\u00a0 P=51N\u00a0 — aplicando Pit\u00e1goras num dos tri\u00e2ngulos ret\u00e2ngulos\u00a0 —\u00a0 y<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>+ 1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 y=\u221a2=1,41m\u00a0 —\u00a0 observe que y \u00e9 o comprimento da mola na posi\u00e7\u00e3o normal (1m) e que \u0394x \u00e9 sua deforma\u00e7\u00e3o e que y=1 + \u0394x\u00a0 —\u00a0 1,41=1 + \u0394x\u00a0 —\u00a0 \u0394x=0,41m\u00a0 —\u00a0 observe tamb\u00e9m que sen\u04e8=1\/y=1\/\u221a2\u00a0 —\u00a0 sen\u04e8=\u221a2\/2=1,41\/2\u00a0 —\u00a0 sen\u04e8=0,7\u00a0 —\u00a0 T<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>=Tsen\u04e8=0,7T\u00a0 —\u00a0 como a esfera est\u00e1 em equil\u00edbrio, P=2T<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 51=2.0,7T\u00a0 —\u00a0 T\u224836N\u00a0 —\u00a0 T=F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=K.\u0394x\u00a0 —\u00a0 36=K.0,41\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>K=87,8N\/m<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span>\u00a0a) F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=Kx\u00a0 —\u00a0 0,40.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>=K.0,40.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 K=1,0N\/m<\/b><\/span><\/span><\/span><\/p>\n

b) F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=ma\u00a0 —\u00a0 a=25.10=250m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Kx=m.250\u00a0 —\u00a0 1.0,5.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>=m.250\u00a0 —\u00a0 m=5.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>\/250=0,02.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>m=9,0.10\u20119\u2011kg<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span>\u00a0Deforma\u00e7\u00e3o da teia quando em equil\u00edbrio \u0394x=0,01L\u00a0 —\u00a0 \u0394x=10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>L\u00a0 —\u00a0 no equil\u00edbrio P=F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 mg=K\u0394x\u00a0 —\u00a0 70.10=10<\/b><\/span><\/span>10<\/b><\/span><\/span><\/sup>.A\/L.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>L\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 A=10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

11<\/b><\/span><\/span>– Depois das oscila\u00e7\u00f5es iniciais terem sido amortecidas, as molas n\u00e3o se deformam mais e o conjunto se desloca coma acelera\u00e7\u00e3o constante a;<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Bloco 3\u00a0 —\u00a0 l<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0\u2013 l<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=ma I\u00a0 —\u00a0\u00a0 bloco 2\u00a0 —\u00a0 l<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0– l<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=ma II\u00a0\u00a0 —\u00a0\u00a0 bloco 1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>l<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=ma\u00a0<\/b><\/span><\/span>III\u00a0 —\u00a0 III em II\u00a0 —\u00a0 l<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 ma = ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>l<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2ma<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

Somando I, com II com III\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0I<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=3ma\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Associa\u00e7\u00e3o de molas<\/b><\/span><\/span><\/span><\/p>\n

12<\/b><\/span><\/span>– Per\u00edodo T da mola da figura 1\u00a0 —\u00a0 T = 2p\u221am\/k<\/b><\/span><\/span><\/p>\n

Como as molas est\u00e3o associadas em paralelo, a constante el\u00e1stica da mola equivalente, que, substituindo as duas produz o mesmo efeito ser\u00e1 k<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= k + k\u00a0 —\u00a0 k<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0=2k e seu per\u00edodo ser\u00e1 T\u2019 = 2pm\/2k\u00a0 —\u00a0 T\u2019 = 2p\u221am\/k.1\/\u221a2.<\/b><\/span><\/span><\/p>\n

T\/T\u2019 =\u00a0\u221a2\u00a0 —\u00a0 T\u2019 = T\/\u221a2\u00a0 — racionalizando\u00a0 —\u00a0 T\u2019= T\u221a2\/2\u00a0 Resposta C<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span>\u00a0 Como as duas molas de constantes k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0est\u00e3o em para, a mola equivalente ter\u00e1 constante k<\/b><\/span><\/span>e1<\/b><\/span><\/span><\/sub>\u00a0=30 + 30 = 60N\/m. Ent\u00e3o teremos:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

As duas molas acima est\u00e3o em s\u00e9rie, ent\u00e3o a mola equivalente ter\u00e1 constante k<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>, dada por: 1\/k<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= 1\/60 + 1\/30\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>k<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= 20N\/m,<\/b><\/span><\/span><\/p>\n

que \u00e9 a constante el\u00e1stica total\u00a0 equivalente do conjunto.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

14-<\/b><\/span><\/span>\u00a0As 3 molas de constantes k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0est\u00e3o em paralelo e ser\u00e3o substitu\u00eddas por uma \u00fanica mola de constante k<\/b><\/span><\/span>e1<\/b><\/span><\/span><\/sub>=3k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

As duas molas de constantes k<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0tamb\u00e9m est\u00e3o em paralelo e ser\u00e3o substitu\u00eddas por um \u00fanica mola de constante k<\/b><\/span><\/span>e2<\/b><\/span><\/span><\/sub>=2k<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/p>\n

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

\"\"<\/p>\n

A mola resultante das duas acima, que est\u00e3o em s\u00e9rie, ter\u00e1 k<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>, tal que: 1\/k<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= 1\/3k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ 1\/2k<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1\/k<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= 2k<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ 3k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\/ 6k<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

K<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= 6k<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\/ 2k<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ 3k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/p>\n

O per\u00edodo desse sistema vale\u00a0 —\u00a0 T = 2p\u221am\/6k<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\/ 2k<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ 3k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T = 2p\u221am(2k<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ 3k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)\/6k<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/p>\n

F = 1\/T = 1\/2p<\/b><\/span><\/span>\u221a<\/b><\/span><\/span>6k<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\/ m(2k<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ 3k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span><\/p>\n

 <\/p>\n

15-<\/b><\/span><\/span>\u00a0a) A mola inteira (mola equivalente) tem constante el\u00e1stica k\u2019=10N\/m sendo que 1\/k\u2019= 1\/k + 1\/k +1\/k, onde k \u00e9 a constante el\u00e1stica de cada parte.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

1\/k\u2019=3\/k\u00a0 —\u00a0 1\/12 = 3\/k\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>k =36N\/m<\/b><\/span><\/span><\/p>\n

b) Paralelo\u00a0 —\u00a0 k\u00ade=36 + 36 +36\u00a0 —\u00a0 k<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=108N\/m\u00a0 —\u00a0 T=2p\u221am\/k<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T=2p\u221a0,1\/108\u00a0 —\u00a0<\/b><\/span><\/span>T\u00a0@\u00a06..10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.p\u00a0s<\/b><\/span><\/span><\/p>\n

c) S\u00e9rie\u00a0 —\u00a0 k<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=12N\/m\u00a0 —\u00a0 T=2p\u221am\/k<\/b><\/span><\/span>e\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0 T=2p\u221a0,1\/12\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T@\u00a018.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.p\u00a0s<\/b><\/span><\/span><\/p>\n

 <\/p>\n

16<\/b><\/span><\/span>– a) Peso de cada massa\u00a0 —\u00a0 P=mg\u00a0 —\u00a0 P=0,01.10\u00a0 —\u00a0 P=0,1N. Como as molas s\u00e3o ideais, suas massas s\u00e3o desprez\u00edveis.<\/b><\/span><\/span><\/p>\n

Observe que a mola 1 est\u00e1 sujeita \u00e0 for\u00e7a F=0,3N (s\u00e3o as 3 massas que est\u00e3o deformando-a)<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=k<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0,3=0,1.x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= 3cm<\/b><\/span><\/span><\/p>\n

A mola 2 est\u00e1 sujeita \u00e0 F=0,2N (apenas duas massas est\u00e3o deformando-a)<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=k<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0,2=0,1.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= 2cm<\/b><\/span><\/span><\/p>\n

Mola 3\u00a0 —\u00a0 F<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=k<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>.x<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0,1=0,1.x<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>x<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>= 1cm<\/b><\/span><\/span><\/p>\n

b) mola 1 \u2013\u00a0<\/b><\/span><\/span>L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= 23cm<\/b><\/span><\/span><\/p>\n

mola 2 \u2013\u00a0<\/b><\/span><\/span>L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= 22cm<\/b><\/span><\/span><\/p>\n

mola 3 \u2013\u00a0<\/b><\/span><\/span>L<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>= 21cm<\/b><\/span><\/span><\/p>\n

c)\u00a0<\/b><\/span><\/span>6 cm<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span>\u00a0Soma (04 + 08) =12\u00a0 Vide teoria<\/b><\/span><\/span><\/p>\n

18-\u00a0<\/b><\/span><\/span>No momento em que o fio \u00e9 cortado a massa mais pr\u00f3xima do teto ficar\u00e1, neste instante, sujeita a uma for\u00e7a resultante de intensidade\u00a0 igual a 2P= 2mg e, portanto, a uma acelera\u00e7\u00e3o descendente de 2g\u00a0 —\u00a0 o outro corpo neste instante estar\u00e1 sujeito ao seu pr\u00f3prio peso P=mg e logo a uma acelera\u00e7\u00e3o igual a g.<\/b><\/span><\/span><\/p>\n

R- (01 + 08) = 09<\/b><\/span><\/span><\/p>\n

19-\u00a0<\/b><\/span><\/span>A for\u00e7a de tra\u00e7\u00e3o, o empuxo da \u00e1gua e o peso do corpo est\u00e3o relacionados: E + T = P\u00a0 —\u00a0 d.g.V + T = P\u00a0 —\u00a0 T = P – d.g.V\u00a0 —\u00a0 T = 10 – 1000.10.0,0004 = 10 – 4 = 6 N\u00a0 —\u00a0 o trabalho \u00e9 o produto da for\u00e7a pelo deslocamento\u00a0 —\u00a0 W=6.0,1 = 0,6 J\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

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

20-\u00a0<\/b><\/span><\/span>a) Para instalar os 20 m de comprimento (L), o n\u00famero n de passos de mola necess\u00e1rio \u00e9 dado por\u00a0 —\u00a0 L = n \u00d7 a + (n + 1) e\u00a0 —\u00a0 a = separa\u00e7\u00e3o entre os an\u00e9is\u00a0 —\u00a0 e = di\u00e2metro do fio (bitola)\u00a0 —\u00a0 L = comprimento do muro\u00a0 —\u00a0 20 = n 0,1 + (n + 1) 0,008\u00a0 —\u00a0 20 = 0,1n + 0,008n + 0,008\u00a0 —\u00a0 20 = 0,108n + 0,008\u00a0 —\u00a0 n \u2248 20 \/ 0,108\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>n \u2248 185<\/b><\/span><\/span>\u00a0 —\u00a0 n\u00famero de voltas N\u00a0 —\u00a0 N = n + 1 = 186\u00a0 —\u00a0 com as voltas compactadas e superpostas, a altura total \u00e9 dada por\u00a0 —\u00a0 H = N e = 186 \u00d7 0,008 m \u2248 1,5 m\u00a0 — \u00a0quantidade C de caixas necess\u00e1rias\u00a0 —\u00a0 C \u2265 1,5\/0,4 = 3,75\u00a0 —\u00a0 como o n\u00famero de caixas deve ser inteiro\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>C<\/b><\/span><\/span>m\u00ednimo<\/b><\/span><\/span><\/sub>\u00a0= 4<\/b><\/span><\/span><\/p>\n

b) O comprimento inicial da mola vale L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= 1,5 m e o comprimento final dever\u00e1 ser L = 20 m\u00a0 —\u00a0 Lei de Hooke\u00a0 —\u00a0 F = k (L \u2013 L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 F = 5 (20 – 1,5) N = 92,5N\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0F = 92,5N<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

21-\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 x = 21 cm = 0,21 m\u00a0 —\u00a0 F = P = m g = 22,7(10) = 227 N\u00a0 —\u00a0 da lei de Hooke: F = k x\u00a0 —\u00a0 K=F\/x=227\/0,21\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

K=1.080,95\u00a0 —\u00a0 K=1,081.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N\/m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span>\u00a0N\/m\u00a0<\/b><\/span><\/span><\/p>\n

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

22-\u00a0<\/b><\/span><\/span>Pela tabela\u00a0 —\u00a0 K=F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\/X=160\/10=320\/20=480\/30=16N\/cm\u00a0 —\u00a0 K=1.600N\/m=1,6kN\/m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

23-\u00a0<\/b><\/span><\/span>As for\u00e7as que agem sobre a bola e sua soma vetorial est\u00e3o indicadas na figura\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=2F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>cos\u03b8\u00a0 —\u00a0 cos\u03b8=x\/\u2113\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=-K(\u2113 –<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u2113o<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=-2.K. (\u2113 – \u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>).x\/\u2113\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=-2K.(x –\u00a0 (\u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.x)\/\u2113)\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=-2Kx(1 – \u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/\u2113)\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=-2Kx(\u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/\u221a (\u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>))\u00a0 —\u00a0 para chegar na aproxima\u00e7\u00e3o do enunciado\u00a0 —\u00a0 \u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/\u221a (\u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>) = \u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/\u221a (\u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)= \u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/\u2113<\/b><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/sub>x \u00a0\u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/\u221a (\u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)=1\/\u221a(1 + x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/ \u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=1\/\u221a(1 + x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/ \u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>) \u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/\u221a (\u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)= (1 + \u00a0x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)=(1 + x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/ \u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>-1\/2<\/b><\/span><\/span><\/sup>=1 + (-1\/2.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/ \u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=1 + (-1\/2.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/ \u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>)=Ma\u00a0 —\u00a0 Ma= -2Kx{1 \u2013 (-1\/2.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/\u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>) 1 + (-1\/2.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/ \u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>)}\u00a0 —\u00a0 a= -Kx<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>)\/M\u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Como o sistema est\u00e1 em equil\u00edbrio as intensidades da for\u00e7a P=mg e das for\u00e7as el\u00e1sticas F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=k.d devem se igualar\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P=2F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 m.g=2.k.d\u00a0 —\u00a0 k=mg\/2d \u00a0—\u00a0\u00a0<\/b><\/span><\/span>R- C.<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

 <\/p>\n

I- Falsa\u00a0 —\u00a0 quando a esfera atinge a mola, sua velocidade vai aumentando, enquanto houver resultante para baixo (P>F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>)\u00a0 — a partir do instante em que as for\u00e7as peso e el\u00e1stica se anulam, a for\u00e7a el\u00e1stica continua aumentando ficando maior que a for\u00e7a peso, a resultante agora \u00e9 para cima, diminuindo a velocidade da esfera.<\/b><\/span><\/span><\/p>\n

II. Falsa\u00a0 —\u00a0 a mola atinge sua m\u00e1xima deforma\u00e7\u00e3o quando a velocidade da esfera \u00e9 nula (ela inverte o sentido de seu movimento e fica em repouso, para come\u00e7ar a voltar)\u00a0 —\u00a0 chamando esse ponto de C e colocando nele o n\u00edvel zero de<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

altura, a energia mec\u00e2nica nele ser\u00e1\u00a0 —\u00a0 E<\/b><\/span><\/span>mC<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + m.g.h + kx<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=m.0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 m.g.0 + kx<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span>mC<\/b><\/span><\/span><\/sub>=kx<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span>mC<\/b><\/span><\/span><\/sub>=50x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(I)\u00a0 —\u00a0 no ponto mais alto (A) de onde a esfera e\u00b4abandonada (V=0), sua energia mec\u00e2nica vale\u00a0 —\u00a0 E<\/b><\/span><\/span>mA<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + m.g.h\u2019=m.0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + m.g.h=1.10.(6 + x)\u00a0 —\u00a0 E<\/b><\/span><\/span>mA<\/b><\/span><\/span><\/sub>=60 + 10x (II)\u00a0 —\u00a0 igualando (I) com (II)\u00a0 —\u00a0 50x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 60 + 10x\u00a0 —\u00a0 5x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 x – 6=0\u00a0 —\u00a0 resolvendo essa equa\u00e7\u00e3o\u00a0 —\u00a0 x=1,2m.<\/b><\/span><\/span><\/p>\n

III. Correta\u00a0 —\u00a0 a velocidade \u00e9 m\u00e1xima quando F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=P (veja I)\u00a0 —\u00a0 kx=mg\u00a0 —\u00a0 100.x=1.10\u00a0 —\u00a0 x=0,1m=10cm.<\/b><\/span><\/span><\/p>\n

IV. Correta\u00a0 — a velocidade \u00e9 m\u00e1xima quando x=0,1m (veja III)\u00a0 —\u00a0 chamando esse ponto de D e colocando nele o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0n\u00edvel zero de altura\u00a0 —\u00a0 E<\/b><\/span><\/span>mD<\/b><\/span><\/span><\/sub>=m(V<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + k.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + m.g.0=1.(V<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + 100.(0,1)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + 0\u00a0 —\u00a0 E<\/b><\/span><\/span>mD<\/b><\/span><\/span><\/sub>=0,5 + 0,5(V<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(I)\u00a0 —\u00a0 em A\u00a0 —\u00a0 E<\/b><\/span><\/span>mA<\/b><\/span><\/span><\/sub>=m.g.(h + x) + mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=1.10.(6 + 0,1) + 0\u00a0 —\u00a0 E<\/b><\/span><\/span>mA<\/b><\/span><\/span><\/sub>=61J (II)\u00a0 —\u00a0 igualando (I) com (II)\u00a0 —\u00a0 61=0,5 + 0,5(V<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=\u221a121\u00a0 —\u00a0 V<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=11m\/s.<\/b><\/span><\/span><\/p>\n

V- Correta\u00a0 —\u00a0 como o sistema \u00e9 conservativo, os atritos s\u00e3o desprezados e as for\u00e7as s\u00e3o conservativas, a velocidade com que a esfera atinge a mola \u00e9 a mesma com que ela deve retornar para atingir a mesma altura de h=6m\u00a0 —\u00a0 conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0 A\u00a0 —\u00a0 E<\/b><\/span><\/span>mA<\/b><\/span><\/span><\/sub>=m.g.h=1.10.6=60J\u00a0 —\u00a0 quando atinge a mola (ponto B)\u00a0 sua energia mec\u00e2nica ser\u00e1\u00a0 —\u00a0 E<\/b><\/span><\/span>mB<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=1.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span>mA<\/b><\/span><\/span><\/sub>\u00a0= bem\u00a0 —\u00a0 60 = V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 V=\u221a(120)=2.(30)<\/b><\/span><\/span>0,5<\/b><\/span><\/span><\/sup>m\/s.<\/b><\/span><\/span><\/p>\n

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

27-\u00a0Como cada um dos t\u00eanis tem 3 molas, a pessoa em p\u00e9 est\u00e1 apoiada sobre dois t\u00eanis, ent\u00e3o voc\u00ea ter\u00e1 6 molas associadas em paralelo\u00a0 —\u00a0 assim, cada mola suportar\u00e1 uma for\u00e7a de\u00a0 —\u00a0 P=m.g=84.10=840N\/6\u00a0 —\u00a0 F=140N (for\u00e7a suportada por cada mola, que \u00e9 a for\u00e7a el\u00e1stica Fe)\u00a0 —\u00a0 Fe=k.x\u00a0 —\u00a0 140=k.4.10-3\u00a0 —\u00a0 k=140\/4.10-3\u00a0 —\u00a0 k=35.103N\/m\u00a0 —\u00a0 k=35kN\/m\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/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 Lei de Hooke e Associa\u00e7\u00e3o de molas \u00a001– A mola fica deformada de x=(22 \u2013 10)=12cm\u00a0 —\u00a0 x=12cm\u00a0 Numa deforma\u00e7\u00e3o de 12cm\u00a0 —\u00a0 Fe\u00a0= P = 4N\u00a0 —\u00a0 Fe\u00a0= K.x \u00a0—\u00a0 4 =K.12\u00a0 —\u00a0 K=1\/3 N\/cm K \u00e9 constante para qualquer deforma\u00e7\u00e3o (lei de Hooke)\u00a0 —\u00a0 para Fe=P=6N\u00a0 —\u00a0 Fe=K.x\u00a0 —\u00a0 6=1\/3.x\u00a0 —\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1225,"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-1229","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1229","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=1229"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1229\/revisions"}],"predecessor-version":[{"id":10829,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1229\/revisions\/10829"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1225"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1229"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}