A block of mass m slides down the plane inclined at angle 30 ° with...

Applying Newton's second law along the direction perpendicular to inlined plane:

$N - mgcos{30}^{∘} = 0$ (As $\theta = 30 °$)

(since the acceleration of block in the direction perpendicular to the inclined plane is $0$)

$∴ N = mgcos{30}^{∘}$ .....(1)

Let $\mu$ be the coefficient of kinetic friction between the block and the inclined plane. Applying Newton's second law along inclined plane:

$mgsin{30}^{∘} - \muN = ma = m\dfrac{g}{4}$ ....(2)

(since the acceleration along the inclined plane, $a = \dfrac{g}{4}$)

$∴ mgsin{30}^{∘} - \mu mgcos{30}^{∘} = m\dfrac{g}{4}$ (from equation 1 and 2)

$⇒ \dfrac{mg}{2} - \dfrac{\sqrt{3}}{2}\mumg = m\dfrac{g}{4}$

$⇒$$\mu = \dfrac{1}{2\sqrt{3}}$