Question

A brick of mass $2kg$ 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 $10m/s^2$)

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