A particle of mass 2 kg is at rest at point A (2m, 1m). It is displaced to point B (3m, 5m) under the action of a constant force F→=8i^+2j^N. Then the rate at which the kinetic energy of the particle is increasing at point B is

Displacement d→=r→B−r→A=3−2i^+5−1j^ m⇒d→=i^+4j^mWork done on the particle, W=F→.d→=8i^+2j^.i^+4j^⇒ W=16 JouleBy work - energy theorem, 12mV2=W⇒12.2.V2=16⇒V=4 m/sSince, velocity vector must lie along the force vector, we can write, V→=V.V^=V.F^=4688i^+2j^m/s∴ Rate of increase of kinetic energy =F→.V→=8i^+2j^.4688i^+2j^J/s=468 Watt