Laws Of Motion Question 4
Question 4 - 29 January - Shift 1
A block of mass $\mathrm{m}$ slides down the plane inclined at angle $30^{\circ}$ with an acceleration $\frac{\mathrm{g}}{4}$. 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
$Mg \sin 30^{\circ}-\mu mg \cos 30^{\circ}=ma$
$\frac{g}{2}-\frac{\sqrt{3}}{2} \mu g=\frac{g}{4}$
$\frac{\sqrt{3}}{2} \mu=\frac{1}{4}$
$\mu=\frac{1}{2 \sqrt{3}}$
