Question

The angle of inclination of an inclined plane is ${60}^{\circ }$. Coefficient of friction between $10kg$ body on it and its surface is $0.2,g=10m{s}^{-2}$. The acceleration of the body down the plane in $m{s}^{-2}$ is

• A. $5.667m{s}^{-2}$
• B. $6.66m{s}^{-2}$
• C. $7.66m{s}^{-2}$
• D. Zero $m{s}^{-2}$

Answer 1

The correct option is C $7.66m{s}^{-2}$

At angles greater than the the critical angle of inclination, the block slides down the incline with uniform acceleration $a$.
The frictional force is ${\mu }_{K}N$. Here $N$ is the normal reaction.

The net force acting on the body in a direction along the plane is:

$F=mgsin\theta -{\mu }_{K}mgcos\theta =ma$

Hence, the acceleration $a$ of the body is related to $\theta$, ${\mu }_K$ by the equation:
$a=g(sin\theta -{\mu }_{K}cos\theta )$

On substituting the respective values:
$a=10(\frac{\sqrt{3}}{2}-0.2\times \frac{1}{2})$
$a=7.66m{s}^{-2}$