Figure shows a small block of mass m kept at the left end of a larger block of mass M and length $l$. The system can slide on a horizontal road. The system is started towards right with an initial velocity $\nu .$
The friction coefficient between the road and the bigger block is $\mu$ and that between the blocks is $\raisebox{1ex}{$\mu $}\!\left/ \!\raisebox{-1ex}{$2$}\right..$ If the time elapsed before the smallest block separates from the bigger block is $\sqrt{\frac{k\times Ml}{\mu g(m+M)}}$ find the value of k.  

Assuming both blocks move together, their acceleration is: $a=\frac{\mu (m+M)g}{(m+M)}=\mu g$  
The required friction between the blocks will become: $f_{m_1}=m\left(\mu g\right)but{f}_{\mathrm{m}}=\left(\frac{\mu }{2}\right)mg$       
Thus, the limiting friction between two blocks is not sufficient to keep the two blocks together. So, we establish that relative slipping will occur between the blocks and kinetic friction will act on each of them.
Retardation of upper block: $\frac{\left(\raisebox{1ex}{$\mu $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)mg}{m}=\frac{\mu g}{2}$  
Retardation of lower block: $\frac{\mu (m+M)g-\left(\raisebox{1ex}{$\mu $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)mg}{M}=(\frac{m}{2M}+1)\mu g$  
Relative acceleration: $a=\frac{\mu g}{2}(\frac{m}{M}+1)$  
Thus, time taken can be evaluated by using second equation of motion relative to the lower block 
considering initial relative velocity of the upper block to be zero. So, $\frac{1}{2}\left[\frac{\mu g}{2}(\frac{m}{M}+1)\right]{\left(\Delta t\right)}^2=1$
We get, $\sqrt{\frac{4MI}{\mu g(m+M)}}$