Question

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 system, the two blocks always move together. (a) Find the coefficient of kinetic friction between the bigger block and the horizontal surface. (b) Find the frictional force acting on the smaller block. (c) Find the work done by the force of friction on the smaller block by the bigger block during a displacement d of the system.
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Given that acceleration, a= F/2 (M +m )

Now for (a) :- Find the coefficient of Kinetic Friction .

Weight of the system = (M + m)g = N {Where N is the normal force}
The force of Friction = μN
Also given, a = F/2 (m +M)
Now for equation of force, along its surface
F-μN = (m + M)a
F-μ(m + M)g =(m +M) F/2 (M +m)
F-μ(m + M)g = F/2
2F-2μg(m + M) = F
μ = $\frac{F}{2g (m + M) }$ 
Hence the coefficient of Kinetic frcition is $\frac{F}{2g (M +m )}$ 


Now for (b) :- Frictional force :-

Let the frictional force be f acting on the small block
so the acceleration must be , a = F/2 (m + M) 
So the force = m × a
f = m [ F/2 (m + M )
f = mF/2 (m + M)
Hence the frictional force is mF/2 (m + M )


Now for (c) :- Work done :-

The velocity of the block during displacement d,
v² = u² + 2ad
v² = 2ad    { since u =0}
v² = 2Fd/2 (m + M )
v² = Fd (m + M )
Now the final Kinetic energy K.E = 1/2 (mv²)
K.E = mFd / 2 (m + M)

So the work done on smaller block is larger than bigger block by the frictional force , so change in K.E is mFd /2(m + M )


Hope it Helps.