Question

A block of mass 2 kg rests on a rough inclined plane making an angle of ${30}^{\circ }$ with the horizontal. The coefficient of static friction between the block and the plane is 0;.7. The frictional force on the block is

• A. 9.8 N
• B. $0.7\times 9.8\sqrt{3}N$
• C. $9.8\times 7N$$
• D. $0.8\times 9.8N$

Answer 1

The correct option is A 9.8 N

Answer is A.

The frictional force is given by the formula, $F=mgsin\theta or\mu mgcos\theta$.
When the block is at rest on the plane, the frictional force acting on it will be $F=mgsin\theta$ (We have to take $F=\mu mgcos\theta$ only when the block is moving down the plane).
Let us take g= $9.8m/s^2$.
In this case, the block is ar rest. Therefore the frictional force on the block is $F=mgsin\theta =mgsin\left(30\right)=2\times 9.8\times 0.5=9.8N$.
Hence, the frictional force on the block is 9.8 N.