Question

A block of mass $2\text{kg}$ 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. $9.8\text{N}$
• B. $0.7\times 9.8\times \sqrt{3}\text{N}$
• C. $9.8\times \sqrt{3}\text{N}$
• D. $0.7\times 9.8\text{N}$

Answer 1

The correct option is A $9.8\text{N}$


Since, block is in rest so applied force will be less than the limiting friction.

$\therefore \mu mg\cos \theta >mg\sin \theta$

$\therefore$ Force of friction is $f=mg\sin \theta$

$f=2\times 9.8\times \frac{1}{2}=9.8\text{N}$

Hence, option $\left(\text{A}\right)$ is correct.