Question

A block is lying on an inclined plane which makes an angle of $60^{\circ}$ with the horizontal. If coefficient of friction between block and plane is $0.25 $ and $g=10\, ms ^{-2},$ then the acceleration of block when it moves along the plane will be [Take $\sqrt{3}=1.7]$

• A. $2.50\, ms ^{-2}$
• B. $5.00\, ms ^{-2}$
• C. $7.25\, ms ^{-2}$
• D. $8.66\, ms ^{-2}$.

Answer 1

Answer: (C)

Along the inclined plane,
$m g \sin \theta-\mu m g \cos \theta=m a$
$a=g(\sin \theta-\mu \cos \theta)$

$\Rightarrow a=10 \frac{ m }{ s ^{2}}\left(\sin 60^{\circ}-0.25 \cos 60^{\circ}\right)$
$=10\left[\frac{\sqrt{3}}{2}-(0.25) \frac{1}{2}\right]=\frac{10(1.7-0.25)}{2}=7.25 \frac{ m }{ s ^{2}}$