The figure shows two blocks of mass 2M and M respectively. The coefficient of friction between the block of mass 2M and the horizontal plane is $\mathrm{\mu }$. The system is released from rest. Find the velocity of the block of mass M when the block of mass 2M has moved a distance s towards right.

 

 As the block of mass M, descends down, it loses potential energy, which appears in the form of kinetic energy of the two blocks, besides doing work against friction. 

Now, if the block of mass 2M moves by a distance s, and its velocity be v, then the distance descended by block of mass M would be s/2 and so its velocity would be v/2.

Now, loss in potential energy of the system $=\mathrm{Mg}\left(\frac{\mathrm{s}}{2}\right)$  …..(i)

Work done against friction $=2\mathrm{M\mu gs}$ ………(ii)

Gain in kinetic energy of the two blocks 

$=\frac{1}{2}\left(2\mathrm{M}\right){\mathrm{v}}^2+\frac{1}{2}\mathrm{M}{\left(\frac{\mathrm{v}}{2}\right)}^2$ ………(iii)

Law of conservation of energy demands that:

$\begin{array}{l}\mathrm{Mg}\left(\frac{\mathrm{s}}{2}\right)=2\mathrm{M\mu gs}+{\mathrm{Mv}}^2+\frac{{\mathrm{Mv}}^2}{8}\\ \frac{\mathrm{gs}}{2}=2\mathrm{\mu gs}+\frac{9}{8}{\mathrm{v}}^2\\ \frac{9}{8}{\mathrm{v}}^2=\frac{\mathrm{gs}}{2}(1-4\mathrm{\mu })\Rightarrow \mathrm{v}=\sqrt{\frac{4}{9}\mathrm{gs}(1-4\mathrm{\mu })}\end{array}$

Velocity of mass, $\mathrm{M}=\frac{\mathrm{v}}{2}$ 

Required velocity $=\frac{1}{3}\sqrt{\mathrm{gs}(1-4\mathrm{\mu })}$