A person of mass 70 kg jumps from a stationary helicopter with the parachute open. As he falls through 50 m height, he gains a speed of 20 ms-1. The work done by the viscous air drag is

From work-energy theorem, net work done by all forces (internal and external) = change in kinetic energy. ⇒ Work done by gravity + work done by air drag=12×70(202−02) ⇒Work done by air drag =14000 - 35000 = -21000 J