Question

A body of mass m 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. $0.14mgH$
• B. $0.28mgH$
• C. $-0.14mgH$
• D. $-0.28mgH$

Answer 1

The correct option is D $-0.28mgH$

K(final)-U (initial)= loss

$=\frac{1}{2}m{v}^2-mgH$

$=\frac{1}{2}m(1.2\sqrt{gH}{)}^2-mgH$

$=\frac{1}{2}m\left(1.44gH\right)-mgH$

$=0.72mgH-mgH$

$=-0.28mgH$.