JEE Main 2021 · 1 September · evening · Single correct · +4 / −1
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
Answer and solution
Answer (a) -0.68 mgh
Solution
Physics sir · 4 steps
Gravity puts in mgh, but the body lands with less kinetic energy; air friction took the difference.
- Work-energy theorem. Two forces act during the fall: gravity and air friction. Their total work equals the change in kinetic energy, and the body starts from rest.
W_(gravity) + Wₐᵢᵣ = 1/2mv² - 0
- Each term. Gravity does W_(gravity) = mgh. The final speed is v = 0.8√gh, so v² = 0.64 gh and 1/2mv² = 0.32 mgh.
- Solve. Wₐᵢᵣ = 0.32 mgh - mgh = -0.68 mgh. It is negative because air friction acts opposite to the motion.
- So. The work done by air friction is -0.68 mgh, option (a).
The idea to keep. When a non-conservative force acts, total work equals the change in KE. Put the known work done by gravity on one side and solve for the unknown force's work. Resistive forces always give a negative answer.
Answer from the NTA final answer key. Solution by PrepWiser.