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. `4.44 J`
D. `10 J`

Answer

B

Explanation

Maximum frictional force between the slab and the block
`f_(max) = mu N = mu mg` , m = 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`.