Laws Of Motion Question 4

Question 4 - 29 January - Shift 1

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

(1) $\frac{2 \sqrt{3}+1}{2}$

(2) $\frac{1}{2 \sqrt{3}}$

(3) $\frac{\sqrt{3}}{2}$

(4) $\frac{2 \sqrt{3}-1}{2}$

Show Answer

Answer: (2)

Solution:

Formula: Second Law Of Motion

$Mg \sin 30^{\circ}-\mu mg \cos 30^{\circ}=ma$

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

$\frac{\sqrt{3}}{2} \mu=\frac{1}{4}$

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