JEE Main 1-Sep-2021 Shift 2 Solved Paper

Given, the mass of the body = m
The height from which the body dropped = h
The speed of the body when reached the ground, vf=0.8√gh
Initial velocity of the body, v = 0 m / s
Using the work-energy theorem,
Work done by gravity + Work done by air-friction = Final kinetic
energy - Initial kinetic energy
Wmg+Wair−friction=

1

2

mvf2−

1

2

mvi2
Here, work done by gravity = mgh

⇒ Wair−fricion=

0.64mgh

2

−mgh
= 0.32 mgh - mgh = -0.68 mgh
The value of the work done by the air friction is -0.68 mgh.