In the figure shown upper block is given a velocity of $6 \mathrm{m}/\mathrm{s}$ and lower block $3 \mathrm{m}/\mathrm{s}.$ When relative motion between them is stopped.

From conservation of linear momentum $(1+2)\mathrm{v}=(6\times 1)+(2-3)$

$\therefore v=4 \mathrm{m}/\mathrm{s}$ (of both the blocks)

From work energy therom

i.e., $W_{\text{total }}=\Delta \mathrm{KE}$ on $1 \mathrm{kg}$ block,

$W_f=\frac{1}{2}\times 1\times \left(4^2-6^2\right)=-10J$

on $2 \mathrm{kg}$block $W_f=\frac{1}{2}\times 2\left(4^2-3^2\right)=+7 \mathrm{J}$

$\therefore$ Net work done by friction is $-3 \mathrm{J}.$