Question

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

• A. - 0.28 mgH
• B. -0.72 mgH
• C. -0.72 mgH
• D. 1.2 mgH

Answer 1

The correct option is A

- 0.28 mgH

The forces acting on the body are the forces of gravity and the air-friction.

By work-energy theorem, the total work done on the body is

W=$\frac{1}{2}m(1.2\times 1.2\times gH)-0=0.72mgH$

The work done by the force of gravity is mgH.

Hence, the work done by air-friction is

=0.72 mgH - mgH = - 0.28 mgH