A body of mass 'm' dropped from a height 'h' reaches the ground with a speed of $ 0.8 \sqrt {gh} $. The value of work done by the air-friction is:
(A) -0.68 mgh
(B) 0.64 mgh
(C) mgh
(D) 1.64 mgh
Solution
Short Method
Air friction force is resistive force whose work-done must be negative. The only option with negative value is (A).
Detailed Method
Work done by all forces $ W_{all} = \Delta K $
$ \therefore W_{mg} + W_{air friction} = \frac {1}{2} mv^2 - 0 $
$ \Rightarrow W_{fr} = \frac {1}{2} m \times (0.8 \sqrt {gh} )^2 - mgh $
$ \Rightarrow W_{fr} = 0.32 mgh - mgh = -0.68 mgh $
Answer: (A)
