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A large slab of mass `5kg` lies on a smooth horizontal surface, with a block of mass `4kg` lying on the top of it. The coefficient of friction between the block and the slab is `0.25`. If the block is pulled horizontally by a force of F `=6N`, the work done by the force of friction on the slab, between the instants `t=2s` and `t=3s`, is `(g=10ms^-2)` A. 2.4 JB. 5.55 JC. 4.44 JD. 10 J |
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Answer» Correct Answer - A `s=(a+bt)^(6)" or "v=(ds)/(dt)=(d)/(dt){(a+bt)^(6)}=(d{(a+bt)^(6)})/(d(a+bt))(d(a+bt))/(dt)=6(a+bt)^(5)(d(a+bt))/(dt)=6b(a+bt)^(5)` The acceleration of the particle is `a_(0)=(dv)/(dt)=6b(d)/(dt)(a+bt)^(5)=6b{5(a+bt)^(5-1)b}=30b^(2)(a+bt)^(4)` |
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