Solution (.pdf)

Solution (.pdf)

Solution (.pdf)

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)