1. Given data

• Mass of disc: $m = 0.5\,\text{kg}$
• Initial linear speed: $u = 18\,\text{m/s}$
• Coefficient of friction: $\mu = 0.3$
• Gravity: $g = 10\,\text{m/s}^2$
• Time to attain pure rolling: $t = 2\,\text{s}$

For a uniform disc, $I = \frac{1}{2}mr^2$

2. Friction force during sliding

Since the disc is initially sliding, kinetic friction acts opposite to motion: $f = \mu mg = 0.3 \times 0.5 \times 10 = 1.5\,\text{N}$

So linear deceleration is $a = \frac{f}{m} = \frac{1.5}{0.5} = 3\,\text{m/s}^2$

Hence speed after $2$ s: $v = u - at = 18 - 3\times 2 = 12\,\text{m/s}$

3. Condition at $t=2$ s

At $t=2$ s, the disc starts pure rolling, so $v = \omega r$ Thus, $\omega = \frac{v}{r} = \frac{12}{r}$

4. Translational kinetic energy after 2 s

$K_{\text{trans}} = \frac{1}{2}mv^2 = \frac{1}{2}\times 0.5 \times 12^2 = 36\,\text{J}$

5. Rotational kinetic energy after 2 s

$K_{\text{rot}} = \frac{1}{2}I\omega^2$ Using $I=\frac{1}{2}mr^2$, $K_{\text{rot}} = \frac{1}{2}\left(\frac{1}{2}mr^2\right)\omega^2 = \frac{1}{4}mr^2\omega^2$ Now since $\omega = \frac{12}{r}$, $K_{\text{rot}} = \frac{1}{4}m r^2 \left(\frac{12}{r}\right)^2 = \frac{1}{4}m\cdot 144$ $K_{\text{rot}} = 36m = 36\times 0.5 = 18\,\text{J}$

6. Total kinetic energy after 2 s

$K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = 36 + 18 = 54\,\text{J}$

7. Comparison with stored answer

Derived answer = $54$

Stored correct answer = $54$

They match.