Question

A brick of mass $2\, kg$ slides down an incline of height 5m and angle $30^{\circ}$. If the coefficient of friction of the incline is $\frac{1}{ 2 \sqrt{3}}$ , the velocity of the block at the bottom of the incline is (Assume the acceleration due to gravity is $10\, m/s^2$)

• A. 5 m/s
• B. 50 m/s
• C. 7 m/s
• D. 0
• E. 10 m/s

Answer 1

Answer: (C)

$a=g(\sin \theta-\mu \cos \theta)$

Given, mass of brick $m=2\, kg$,
height of plane $H=5\, m , \theta=30^{\circ}$ and
coefficient of friction $\mu=\frac{1}{2 \sqrt{3}}$.
$\therefore a=g\left(\sin 30^{\circ}-\frac{1}{2 \sqrt{3}} \cos 30^{\circ}\right) $
$\Rightarrow a=10\left[\frac{1}{2}-\frac{1}{4}\right]=2.5\, m / s ^{2}$
Length of inclined plane
$x=H \sin \theta=5 \times \sin 30^{\circ}=10\,m$
From 3rd equation of motion
$v^{2}-u^{2}=2\, a s$
$\Rightarrow v=\sqrt{2 a s}$
$\Rightarrow v=\sqrt{2 \times 25 \times 10}=\sqrt{50}=7.07$