Question

Same amount of force (10 N) is acting on two bodies in two different ways.

In case of body 1: Force is acting in the direction of displacement of the body.
In case of body 2: Force is making an angle of 60° with the direction of displacement.
If the displacement is 'd' meters in both the cases, then what is the relation between the work done in both the cases?

• 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.

As the work done is defined as product of displacement and component of force in the direction of displacement. Work done can be calculated by multiplying the force with cosine of the angle between force and displacement. So,
$W_1=F\times d=10dJ$
$W_2=F\times d\times cos{60}^{\circ }$ $=5dJ$
$\Rightarrow$ $\frac{W_1}{W_1}=\frac{10d}{5d}=2$