A block of mass m slides down the plane inclined at angle 30 ° with an acceleration g 4 . The value of…

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

$N-mg\cos {30}^{\circ }=0$ (As $\theta =30°$)

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

$\therefore N=mg\cos {30}^{\circ }$   .....(1)

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

$mg\sin {30}^{\circ }-\mu N=ma=m\frac{g}{4}$    ....(2)

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

$\therefore mg\sin {30}^{\circ }-\mu mg\cos {30}^{\circ }=m\frac{g}{4}$   (from equation 1 and 2)

$\Rightarrow \frac{mg}{2}-\frac{\sqrt{3}}{2}\mu mg=m\frac{g}{4}$

$\Rightarrow$$\mu =\frac{1}{2\sqrt{3}}$