The figure shows two blocks of mass `m_(1) and m_(2)` placed one above the other. These is no friction between the lower block and ground. The lower block is being pushed by a constant horizontal force F. There is sufficient friction between the blocks so that they do not slip over each other. Draw the free body diagram of the upper block in the frame of groung and in the frame of lower block. Find the work done by various forces on the upper block in the two frames as the arrangement moves through 'S'

Explanation

As the two blocks move together, their common acceleration is given by `F=(m_(1)+m_(2))a or a=(F)/(m_(1)+m_(2))`
(i) FBD of `m_(2)` with respect to ground.

`N=m_(2)g` (equilibrium along vertical)
`f_(r)=m_(2)a` (by newton's `2^(nd)` law).
`W_(N)=0,N_(m_(2)g))=0.`
`W_(f)=m_(2)axx S xx cos0^(@)`
`=(m_(2)FS)/((m_(1)+m_(2)))`
(ii) FBD of `m_(2)` in the frame of `m_(1)`

In the frame of `m_(1)` th mas `m_(2)` does not move. Therefore, work done by all the forces is zero.