Dear Souvik ,

PLease see the above free body diagram.

writing equations--

for mass 4m : Fr  =  4ma₂   ----------- (1)

for mass m placed above 4m : T - Fr = ma₁ ---------------(2)

For hanging mass m : mg -T = ma1 ( acceleration is same as string has to be tight )-------------(3)

Fr = µ_kmg ---------------------(4)

solving we get , a₂ = µ_kg /4

                        a₁ =g -µ_kg /2

When only half of the block is still on the 4m mass block , distance covered  = 3l/2

time taken  = (2s/a)^1/2  =(3l/a₁)^1/2

in that time distance travelled by m mass = 1/2 (a₂)t²  = 1/2 *(µ_kg /4)*(6l/(1-µ_k) g) =  3µ_kl/4(1 - µ_k)   m

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Respected Sir,

I have also proceeded like this way and arrived to the same answer provided by you, but the ans. given in the material is something like this: x= 7μl/[8(2-3μ)]

I hope you will clear this confusion.

how didi u come to the conclusion that distance travelled by the m block is 3L/2?