Question

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

• A. - 0.28 mgH
• B. -9.8mgH
• C. 9.8mgH
• D. Zero

Answer 1

The correct option is A - 0.28 mgH

The forces acting on the body are the force of gravity and air-friction. By work-energy theorem, the total work done on the body is equal to change in kinetic energy,
$W=\frac{1}{2}m(1.2\sqrt{gH}{)}^2-0=0.72mgH$
The work done by the force of gravity is mgH. Hence the work done by air-friction is
$0.72mgH-mgH=-0.28mgH$.