The disc is at rest at the top of the hill.
$\therefore$ Potential energy of the disc, $U=m g h$
Let the velocity of the sliding disc before the friction is $v$.
$\therefore$ The kinetic energy, $K=\frac{1}{2} m v^2$


From energy conservation,
$$
\begin{aligned}
\frac{1}{2} m v^2 & =m g h \\
v & =\sqrt{2 g h}
\end{aligned}
$$
$$
\Rightarrow \quad v=\sqrt{2 g h}
$$
The disc now gets onto the plank and slows down and moves as one piece with the plank.
Let the velocity of the plank is $v^{\prime}$.
From conservation of momentum,
$$
\begin{aligned}
m v & =(m+M) v^{\prime} \\
\Rightarrow \quad v^{\prime} & =\frac{m \sqrt{2 g h}}{(m+M)}
\end{aligned}
$$
$\therefore$ The kinetic energy of the plank,
$$
\begin{aligned}
K_2 & =\frac{1}{2}(m+M) v^{\prime 2}=\frac{m^2 \times 2 g h}{2(M+m)} \\
\Rightarrow \quad K_2 & =\frac{m^2 g h}{(m+M)}
\end{aligned}
$$
$\therefore$ The total work done by the frictional force is
$$
\begin{aligned}
W_f & =K_1-K_2=m g h-\frac{m^2 g h}{m+M} \\
& =\frac{M m g h}{m+M} \\
\Rightarrow \quad W_f & =\frac{100 \times 10^{-3} \times 1 \times 10^{-3} \times 10 \times 10}{(100+1) \times 10^{-3}} \\
\Rightarrow \quad W_f & =0.1 \mathrm{~J}
\end{aligned}
$$