A block of mass m is placed on the block of mass M as shown in figure. The horizontal force `vecF` acts on M during time interval t. If the horizontal surface is smooth, assuming no relative sliding between the blocks, find the
a. work done by friction on the blocks
b. work done by `vecF` on the lower block

Explanation

a. As there is no relative sliding between the blocks, they move with a common acceleration,
`a=(F)/(M+m)`
Then the static frictional force on the block m is
`f=ma`, where `a=(F)/(M+m)`. Then
`f=m(F)/(M+m)`
Work done by static frictional force on the block m
Since `(W_f)_m=vecf.vecs=fscos0^@=fs`
where `s=1/2at^2`
We have `(W_f)_m=f(1/2at^2)`
Substituting `f=(mF)/(M+m)` and `a=(F)/(M+m)`
We have `(W_f)_m=(mF^2t^2)/(2(M+m))`
As no relative sliding between m and M, hence, net work done by static friction is zero.
Since `(W_f)_m+(W_f)_M=0`
Hence, the work done by the static frictional force on the block m
We have `(W_f)_m=(mF^2t^2)/(2(M+m))`
b. The displacement of lower block during time t,
`impliess=1/2((F)/(M+m))t^2`
Hence, work done by F, `W_F=vecF.vecs`
`W_F=F*s=F[1/2((F)/(M+m))t^2]=(F^2t^2)/(2(M+m))`