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

From work-energy theorem, net work done by all the forces (internal and external ) = change in kinetic energy ⇒Work done by gravity + Work done by air drag = change in kinetic energy.            Kinetic energy gained      = 12 × 70(202-02)=14000Potential energy lost = mgh = 70(10)50 = 35000⇒Work done by air drag = KE_PE=14000-35000 = - 21000J