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. (a) `2.4J`
B. (b) `5.55J`
C. (c) `4.44J`
D. (d) `10J`

Answer

B

Explanation

Maximum frictional force between the slab and the block
`f_(max)=muN=mumg`
`=1/4xx4xx10=10N`
Evidently, `fltf_(max)`

So, the two bodies will move together as a single unit. If a be their combined accelration, then
`a=(F)/(m+M)=(6)/(4+5)=2/3ms^-1`
Therefore, frictional force acting can be obtained as
`f=Ma=2/3xx5N=10/3N`
Using `s=1/2at^2`
`s(2)=1/2xx2/3(2)^2=4/3` and `s(3)=1/2xx2/3(3)^2=3`
Therefore, work done by friction `=f[s(3)-s(2)]`
`=10/3[3-4/3]`
`=50/9=5.55J`