A uniform disc of mass $$0.5$$ kg and radius $$r$$ is projected with a velocity $$18$$ m s$$^{-1}$$ at $$t = 0$$ s on a rough horizontal surface. It starts off with a purely sliding motion at $$t = 0$$ s. After $$2$$ s it acquires a purely rolling motion. The total kinetic energy of the disc after $$2$$ s will be ______ J.
(given, coefficient of friction is $$0.3$$ and $$g = 10$$ m s$$^{-2}$$).

$$\text{Friction force } (f_k) = \mu mg$$

$$\text{Deceleration } (a) = \frac{f_k}{m} = \mu g$$

$$v = u - at$$

$$v = 18 - (0.3 \times 10) \times 2$$

$$v = 18 - 6 = 12\text{ m/s}$$

$$KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

For a uniform disc, the moment of inertia is $$I = \frac{1}{2}mr^2$$. In pure rolling, $$v = \omega r \implies \omega = \frac{v}{r}$$.

$$KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2$$

$$KE_{total} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$$

$$KE_{total} = \frac{3}{4} \times 0.5 \times (12)^2$$

$$KE_{total} = 0.375 \times 144 = 54\text{ J}$$