Q. A small iron sphere is dropped from a great height. It attains its terminal velocity after having fallen 32 m. Then, it covers the rest of the path with terminal velocity only. The work done by air friction during the first 32 m of fall is $W_{1}$. The work done by air friction during the subsequent 32 m fall is$W_{2}$. Then
A) $W_{1} > W_{2}$
B) $W_{1} < W_{2}$
C) $W_{1} = W_{2}$
D) $W_{2} = 32\ W_{1}$
Explanation
(b) Work done against air friction is the average gain in KE before attaining the terminal velocity
$W_{1} = \frac{0 + \frac{1}{2}mv_{ter}^{2}}{2} = \frac{1}{4}mv_{ter}^{2}$
Work done against air friction after attaining terminal velocity, velocity is
$W_{2} = \frac{1}{2}mv_{\max}^{2}$
$\therefore W_{2} > W_{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.