Question

A plane surface is inclined at an angle of ${60}^0$. A body of mass 10 kg is uniformly accelerating along the inclined plane surface. If the value of coefficient of friction ${\mu }_K$, between the body and the inclined surface is 0.2, calculate the resultant force on the body. (Take $g=10m{s}^{-2}$)

• A. 56.6 N
• B. 66.6 N
• C. 76.7 N
• D. 86.6 N

Answer 1

Answer: (C)

76.7 N

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 in a direction perpendicular to inclined plane and equals $mgcos\theta$.

The net force acting on the body in a direction along the plane is:
$F_x=mgsin\theta -{\mu }_{K}N=ma$

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:
$a=10(\frac{\sqrt{3}}{2}-0.2\times \frac{1}{2})$
$a=7.67m{s}^{-2}$
On substituting this value in the equation:
$F_x=ma$
We obtain:
$F_x=10\times 7.67=76.7N$