Question

A block of mass $m$ is sliding down on a fixed inclined plane with velocity of $2\text{m/s}$ for a distance of $2\text{m}$. Angle of inclined plane is ${35}^{\circ }$. Calculate the work done by normal reaction on the block.

• A. $0.5mg\sin {35}^{\circ }$
• B. $0.5mg\cos {35}^{\circ }$
• C. $2mg$
• D. $0$

Answer 1

The correct option is D $0$

FBD of block:-



Now, since the normal reaction on the block is acting in perpendicular direction of displacement of block.
$\therefore$ Work done by normal reaction $=0$ (zero).