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?
Velocity of body is given as v=ax3/2 where a=5 Initial velocity of body vi∣∣∣∣∣x=0=5×0=0 m/s Final velocity of body vf∣∣∣∣∣x=2=5×23/2=102 m/s From work-energy theorem, work done by the force is equal to change in kinetic energy of the body. ∴ W=21mvf2−21mvi2 Or W=21(0.5)(102)2−21(0.5)(0) Or W=21(0.5)(200)−0 ⟹ W=50 J
