Question 148 of 210

A block of mass $m$ slides down the plane inclined at angle $30^{\circ}$ with an acceleration $\dfrac{g}{4}$ . The value of coefficient of kinetic friction will be :

A$\dfrac{2\sqrt{3}+1}{2}$B$\dfrac{1}{2\sqrt{3}}$C$\dfrac{\sqrt{3}}{2}$D$\dfrac{2\sqrt{3}-1}{2}$

Show answer and solution▾

Answer: (B) $\dfrac{1}{2\sqrt{3}}$

Applying Newton's second law along the direction perpendicular to inlined plane:
$N-mg\cos {30}^{^{\circ}}=0$ (As $\theta =30^{\circ}$)
(since the acceleration of block in the direction perpendicular to the inclined plane is $0$)
$∴N=mg\cos {30}^{^{\circ}}$ .....(1)
Let $\mu$ be the coefficient of kinetic friction between the block and the inclined plane. Applying Newton's second law along inclined plane:
$mg\sin {30}^{^{\circ}}-\mu N=ma=m\dfrac{g}{4}$ ....(2)
(since the acceleration along the inclined plane, $a=\dfrac{g}{4}$)
$∴mg\sin {30}^{^{\circ}}-\mu mg\cos {30}^{^{\circ}}=m\dfrac{g}{4}$ (from equation 1 and 2)
$\Rightarrow \dfrac{mg}{2}-\dfrac{\sqrt{3}}{2}\mu mg=m\dfrac{g}{4}$
$\Rightarrow$$\mu =\dfrac{1}{2\sqrt{3}}$

JEE Main Physics practice question on Laws of Motion (Frictional force), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.