According to the question, acceleration of a block sliding down an inclined plane is shown in the following figure.

From third equation of the motion, velocity of disc when it leaves the inclined plane,


$$
v^2=u^2-2 a s
$$
or
$$
u=\sqrt{2 a s}
$$
$(\because v=0)$
$\because$ Acceleration of a block on a horizontal,
$$
\begin{gathered}
a=\mu g \\
\left.\Rightarrow \quad a=0.2 \times 10=2 \mathrm{~m} / \mathrm{s}^2 \quad \text { (Given, } \mu=0.2\right)
\end{gathered}
$$

After putting the value of $a$ in Eq. (i), we get
$$
\begin{array}{rlrl}
\therefore & u & =\sqrt{2 \times 2 \times 1} \\
& =2 \mathrm{~m} / \mathrm{s}
\end{array} \quad(\because s=1 \mathrm{~m}, \text { given })
$$

From above diagram,
the frictional force applied on the disc inclined plane $r$ is
$$
\begin{aligned}
F & =\mu N \\
& =\mu m g \cos 30^{\circ}
\end{aligned}
$$
and the net acceleration force down the inclined plane,
$$
m g \sin 30^{\circ}-f=m g \sin 30^{\circ}-\mu m g \cos 30^{\circ}=m a_1
$$

Hence, from the third equation of motion,
$$
v_1^2=u_1^2+2 a_1 s
$$

When $s$ is distance travelled by the disc,
$$
\begin{array}{rlrl}
& v_1=u=2 \mathrm{~m} / \mathrm{s}, u_1=0 \\
\therefore & (2)^2 & =2 a_1 s \text { or } s=\frac{2}{a_1}
\end{array}
$$

Putting the value of $a_1$ from Eqs. (i), we get
$$
\begin{aligned}
\Rightarrow \quad s & =\frac{2}{5(1-\sqrt{3} \mu)} \\
& =\frac{2}{5(1-\sqrt{3} \times 0.2)}=0.612 \mathrm{~m}
\end{aligned}
$$

Hence, the work done by the frictional force, $W=$ work done on inclined plane + work done on horizontal plane.
$$
=\left(\mu m g \cos 30^{\circ}\right) s+\frac{1}{2} m u^2
$$

Putting the given values, we get
$$
\begin{aligned}
& =0.2 \times \frac{100}{1000} \times 10 \times \cos 30^{\circ} \times 0.612+\frac{1}{2} \times \frac{100}{1000} \times(2)^2 \\
& =0.306 \mathrm{~J}
\end{aligned}
$$