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 $20m{s}^{-1}.$ The work done by the viscous air drag is

$\left(\mathrm{Take} \mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2\right)$

From work-energy theorem, 

net work done by all the forces (internal and external ) = change in kinetic energy
 $\Rightarrow$Work done by gravity + Work done by air drag = change in kinetic energy.

Kinetic energy gained $= \frac{1}{2} \times 70({20}^2-0^2)$=14000

Work done by gravity = mgh = 70(10)50 = 35000
$\Rightarrow$Work done by air drag = KE -  PE = 14000 - 35000 = - 21000J