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)`

Options

A. `2.4 J`
B. `5.55 J`
C. `4.444 J`
D. `10 J`

Answer

`(2)`

Explanation

Maximum frictional force between the slab and the block
`f_(max) = mu N = mumg` mass of upper block Evidently, F lt` f_(max)` so, the two bodies will move together as a single unit. If.a. be their combined acceleration, then `a = F/(m + M)` Therefore, frictional force acting can be obtained as `f= Ma`, Using s = `1/2 at^(2), find `S_(2)` and `S_(3)`,
`W = (S_(3)-S_(2))f`