A body of mass ' m ' dropped from a height ' h ' reaches the ground with a speed of 0.8 √ gh . The value of workdone by the air-friction is:

Question

A body of mass ' m ' dropped from a height ' h ' reaches the ground with a speed of 0.8 √ gh . The value of workdone by the air-friction is: (A) -0.68 mgh (B) mgh (C) 1.64 mgh (D) 0.64 mgh

Tardigrade Admin · Accepted Answer

Work done = Change in kinetic energy W mg + W text air-fiction =(1/2) m (.8 √ gh )2-(1/2) m (0)2 W text air - text fiction =(.64/2) mgh - mgh =-0.68 mgh

Q. A body of mass ' $m$ ' dropped from a height ' $h$ ' reaches the ground with a speed of $0.8 g h$ . The value of workdone by the air-friction is:

-0.68 mgh

mgh

1.64 mgh

0.64 mgh

Work done $=$ Change in kinetic energy
$W_{m g} + W_{air-fiction} = \frac{1}{2} m (.8 g h)^{2} - \frac{1}{2} m (0)^{2}$
$W_{air - fiction} = \frac{.64}{2} m g h - m g h = - 0.68 m g h$

Q. A body of mass ' m ' dropped from a height ' h ' reaches the ground with a speed of 0.8gh​. The value of workdone by the air-friction is:

-0.68 mgh

mgh

1.64 mgh

0.64 mgh

Work done = Change in kinetic energy Wmg​+Wair-fiction ​=21​m(.8gh​)2−21​m(0)2 Wair − fiction ​=2.64​mgh−mgh=−0.68mgh

Q. A body of mass ' $m$ ' dropped from a height ' $h$ ' reaches the ground with a speed of $0.8 g h$ . The value of workdone by the air-friction is:

Work done $=$ Change in kinetic energy
$W_{m g} + W_{air-fiction} = \frac{1}{2} m (.8 g h)^{2} - \frac{1}{2} m (0)^{2}$
$W_{air - fiction} = \frac{.64}{2} m g h - m g h = - 0.68 m g h$