A body dropped from a height H reaches the ground with a speed of 1.2 $\sqrt{gH}$. Calculate the work done by air-friction.

\(-0.28 mgh\)

The forces acting on the body are the force of gravity and the air friction. By work energy theorem, the total work done on the body is : 

\(W = \frac{1}{2}.m.(1.2 \sqrt{gH})^2 - 0\)

  \(= 0.72 mg \)

The work done by the force of gravity is mgh. Hence, the work done by the air friction is :

  \(= 0.72 mgh - mgh = -0.28 mgh\)