### Q.20:- A body of mass 0.5 kg travels in a straight line with velocity v = ax^{3/2} where a = 5 m^{1/2} s^{-1}. What is the work done by the net force during its displacement from *x *= 0 to *x *= 2 m?

**Answer:-**

Mass of the body, *m* = 0.5 kg

Velocity of the body is governed by the equation, v = ax^{3/2} and a = 5 m^{1/2} s^{-1}

Initial velocity, *u *(at* x *= 0) = 0

Final velocity *v *(at* x *= 2 m) = 10√2 m/s

Work done, *W* = Change in kinetic energy

= (1/2) *m* (*v*^{2} – *u*^{2})

= (1/2) × 0.5 [ (10√2)^{2} – 0^{2}]

= (1/2) × 0.5 × 10 × 10 × 2

= 50 J