Question

A body is sliding down a rough inclined plane of inclination ${60}^0$. If the angle of repose is ${30}^0$, find the acceleration of the body ($g=10m{s}^{-2}$) in $m{s}^{-2}$.

• A. $2.33$
• B. $3.46$
• C. $4.5$
• D. $5.77$

Answer 1

Answer: (D)

$5.77$

Angle of repose is the maximum angle of inclination when the downward force on the block is exactly balance by the frictional force.

i.e. $F=\mu R=\mu mgcos\theta =mgsin\theta$, where, $\theta$ is the angle of repose.

Solving this we will get:

$\mu =tan{30}^0=1/\sqrt{3}$.

Now when it will start moving then the acceleration of the box will be:

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

where $\alpha ={60}^o$

$\therefore a=10(0.866-\frac{1}{2\sqrt{3}})=5.77m/s^2$