Question

A force of 12 N is acting on a body and the same force 12 N is acting on the same body but in a different direction. If the body moves with a displacement of ${}^{\prime }{d}^{\prime }$ metres, then what is the work done?

• A. Work done in case 1 is 2 times the work done in case 2.
• B. Work done in case 2 is 2 times the work done in case 1.
• C. Work done in both the cases is equal because force applied and displacement are equal.
• D. Cannot be determined because work done will depend on the size and weight of the bodies.

Answer 1

The correct option is A

Work done in case 1 is 2 times the work done in case 2.

The work done, $W$ is defined as product of displacement and the net force in the direction of displacement.
$\therefore W=(F_1-F_2)\times d=(12-12)\times d$ J

Here, $F_1$ and $F_2$ are the equal forces acting on the body. Since equal forces are acting on the body in opposite directions, thus the net force on the body is zero

$\Rightarrow W=0J$