Two long blocks $\text{m=1kg\hspace{0.17em}and\hspace{0.17em}M=3kg}$ are placed as shown. The coefficient of friction between the blocks is ${\mu }_2=0.8$  and between the floor and M is ${\mu }_1=0.1$. Initially M is at rest and m is given a speed. Mechanical energy dissipated out of the system of  $(m+M)$  due to friction between the blocks is……times the loss due to friction at ground $(g=10m/s^2)$

$f_2={\mu }_{2}mg=8N \& f_{1\max }=\mu (m+M)g=4N$

$\therefore f_2>f_{1\max }$
3kg will move and 1kg will have relative motion w.r.to 3kg will 1kg will show down and 3kg will speed  up till they attain same velocity.

Let ‘t’ be the time taken
$u-\left(\frac{8}{l}\right)t=0+\left(\frac{8-4}{3}\right)t\Rightarrow t=3\sec$
Displacement of  $1kg=28\left(3\right)-\frac{1}{2}\times 8\times 3^2=48m$
Displacement of  $3kg=0\left(3\right)+\frac{1}{2}\times \left(\frac{4}{3}\right)3^2=6m$
Work done by friction on M by m is  $f_2\times 6=48J$
At t=3s, the blocks will attain a common velocity
$v=28-8\times 3=4m{s}^{-1}$
From now onwards, they move together as one system with a common deceleration. If the friction between
$a=\frac{4}{4}=1m{s}^{-2}$
Friction between the blocks should be  $f_2=1\times 1=1N<f_l$
It takes 4 seconds more to come to rest and displacement  $=4\times 4-\frac{1}{2}\times 1{\left(4\right)}^2=8m$
Work done by friction on M by m is  $f_2\times 8=8J$
Total work by  $f_2=48+8=56J$
Total work by friction on m by $M=\Delta KE$  of  $M=(\frac{1}{2}\times {\left(28\right)}^2-0)=392J$
392 J is taken from m and only 56 J is given to M
Loss at contact = 392-56=336J

Loss of energy at ground is equal to all the energy given to M by ‘m’ which is 56 J.