Forces on the block along the incline

The block slides down a plane inclined at $\theta = 60^\circ$.

Along the plane:

• Component of gravity downward: $mg\sin 60^\circ$
• Friction upward: $f_k = \mu_k N$

Normal reaction is $N = mg\cos 60^\circ$

Apply Newton’s second law along the incline

Since the block accelerates downward with $a = \frac{g}{2},$ we write $mg\sin 60^\circ - \mu_k mg\cos 60^\circ = m\frac{g}{2}$

Substitute trigonometric values

$\sin 60^\circ = \frac{\sqrt{3}}{2}, \qquad \cos 60^\circ = \frac{1}{2}$

So, $mg\left(\frac{\sqrt{3}}{2}\right) - \mu_k mg\left(\frac{1}{2}\right) = m\frac{g}{2}$

Cancel $mg$

$\frac{\sqrt{3}}{2} - \frac{\mu_k}{2} = \frac{1}{2}$

Multiply by $2$: $\sqrt{3} - \mu_k = 1$

Therefore, $\mu_k = \sqrt{3} - 1$

Match with options

$\mu_k = \sqrt{3} - 1$

This corresponds to Option D.