Question

A block of mass m is released from top of a smooth fixed inclined plane of inclination $\theta$. Find out work done by normal force.

• A. 0
• B. $mgh$
• C. $2mgh$
• D. $4mgh$

Answer 1

The correct option is A 0

The normal force will act in the direction normal to the inclined plane whereas the displacement of the block is along the inclined plane.

Angle made b/w normal force and displacement of the block$=\varphi ={90}^o$

Now, Work done$=W=Fs\cos \varphi =N\times s\times \cos {90}^o=0$

Hence, correct answer is option $A$