28)

As the string is inextensible ,whenever the lower end of
the string (connected to M) displaces by x units to the
right,the other end of the string (connected to block m)
lowers down by "2x" units.

So acceleration of m (vertically downward) = 2 (accelration
of block M in horizontal right direction )

Let accelration of M = a
So acceleration of m = 2a.

Also,
The block m is always in contact with the larger block M
as far as motion in the horizontal direction is concerned.
So acceleration of block m in horizontal direction is also
"a" (same as that of M)

 

Now look at the figure:
Motion of m:

the forces are:
1) mg downwards
2) R ( contact force by M ) towards right
3) ₁R ( frictional force) upwards

4) T ( Tension) upward

 

For horizontal direction.

R = ma

 

In vertical direction

mg - T - ₁R = m(2a)

or T = mg - ma(2 + ₁)

 

Motion of M:

 

The forces on M are:

1) Mg downwards

2) R ( contact force by m ) towards left

3) ₁R ( reaction of frictional force on m ) downwards 

4) T ( Tension due to string on the lower end ) towards right

5) N ( contact force by ground ) upwards

6) ₂N ( frictional force due to ground ) towards left

7) T ( Tension due to string on pulley attached to M ) towards right

8) T ( Tension due to string on pulley attached to M ) downwards

 

for vertical equilibrium

N = Mg + T + ₁R

N = Mg + T + ₁ma             {since R = ma}

 

In horizontal direction

2T - R - ₂N = Ma

 

Putting values of R, T and N.......

2T - ma - ₂(Mg + T + ₁ma) = Ma

(2 - ₂)T - ma - ₂(Mg + ₁ma) = Ma

(2 - ₂)[mg - ma(2 + ₁)] - ma - ₂(Mg + ₁ma) = Ma

2mg  - ₂g(m + M) = 5ma + 2ma₁ - 2ma₂ + Ma

a[M + m{5 + 2(₁ - ₂)}] = g[2m - ₂(m + M)]

a = g[2m - ₂(m + M)] / [M + m{5 + 2(₁ - ₂)}]