Three identical uniform planks A, B and C of mass m = 1 kg each and length L= 2 m are placed on a smooth fixed horizontal surface as shown in the figure. There is friction between A and B (friction coefficient being  $\mu$) while there is no friction between A and C. At the instant shown, that is at t = 0 the block A has horizontal velocity of magnitude v = 6 m/s towards right, whereas speed of B and C is Zero. At the instant, block A has covered a distance L relative to block B velocity of all blocks are same. If heat dissipated due to friction in the system is H, find value of $\frac{H}{2\mu }$ in joule.

Motion of the system is a shown,
 
As final velocities of all blocks are equal (say  $v^{\text{'}}$ ) by conservation of momentum, 
$3m{v}^{\text{'}}=mv\Rightarrow v^{\text{'}}=\frac{v}{3}\mathrm{...}\text{(i)}$ 
By work- energy theorem,
 ${\text{W}}_{\text{friction}}=\text{change in kinetic energy}$
          $=\frac{1}{2}\times 3\text{m}\times {\left(\frac{v}{3}\right)}^2-\frac{1}{2}\text{m}{v}^2$  = -12 $\text{J}$       …..(ii)
$\Rightarrow$ Heat dissipated, H =12$\text{J}$         …(iii)
Also, as shown at general instant, assuming x length of A is in contact with B, normal force at contact
 $N=\frac{m}{L}\cdot \text{x}g$
 $\Rightarrow Friction=\text{μN=}\frac{\text{μmg}}{\text{L}}\text{x}$
$\Rightarrow$  Work done by friction =  $\int \text{f }dx$
 $=\underset{0}{\overset{L}{\int }}-\frac{\text{μmgx}}{L}dx=\frac{\text{-μ mg L}}{2}\mathrm{...}\left(iv\right)$
Using Eqs. (ii) and (iv), we get 
 $\frac{\text{μmgL}}{2}=12$
 $\Rightarrow \frac{\text{μ}\times \text{1}\times \text{10}\times \text{2}}{2}=12$                 …(v)
From Eqs. (iii) and (v),  $\frac{H}{\mu }=\frac{12}{1.2}=10$