A block of mass m is kept over another block of mass M and the system rests on a horizontal surface. A constant horizontal force F acting on the lower block produces an acceleration `(F)/(2(m+M))` in the sytem. The two blocks always move together. Consider displacement d of the system.
a. Find the work done by friction on bigger block.
b. Find the coefficient of kinetic friction between the bigger block and the horizontal surface.
c. Find the frictional acting on the smaller block.
d. Find the work done by the force of friction on the smaller block by the bigger block.
e. Find the work done by static friction on bigger block.

Explanation

As both blocks move together, there will be static friction betweem m and M. The friction between M and ground will be kinetic in nature. Common acceleration, `a=(F)/(2(m+M))`
a. As both block moves together, we can take m and M as a system. Let friction between M and ground is `f_M`.

From free body diagram of `m+M`
Equation of motion: `F-f_M=(m+M)a`
or `f_M=(m+M)(F)/(2(m+M))`
`impliesf_M=F/2`
b. Work done by friction on bigger block,
`(W_f)_M=vecf_M*vecd=-f_Md=-F/2*d`
As friction between M and ground is kinetic, then
`f_M=muN`
`F/2=mu(m+M)gimpliesmu=(F)/(2(m+M)g)`
c. Considering the free body diagram of m
`f_s=ma`
`f_s=m(F)/(2(m+M))=(Fm)/(2(m+M))`

d. Work done by friction force on smaller block by bigger block
`(W_(fs))_m=vecf_s*d`
`=f_s*d`
`=((Fm)/(2(m+M)))d`
`=(Fdm)/(2(m+M))`
e. Net work done by static fricition on a system is zero.
`(W_(fs))_m=(W_(fs))M=0`
`[(Fdm)/(2(m+M))]+(W_(fs))_M=0implies(W_(fs))_M=(-Fdm)/(2(m+M))`