Question:
A body of mass ' $m$ ' dropped from a height ' $h$ ' reaches the ground with a speed of $0.8 \sqrt{\mathrm{gh}}$. The value of workdone by the air-friction is :
Correct Option: 1
Solution:
Work done $=$ Change in kinetic energy
$\mathrm{W}_{\mathrm{mg}}+\mathrm{W}_{\text {air-friction }}=\frac{1}{2} \mathrm{~m}(.8 \sqrt{\mathrm{gh}})^{2}-\frac{1}{2} \mathrm{~m}(0)^{2}$
$\mathrm{~W}_{\text {air }-\text { friction }}=\frac{.64}{2} \mathrm{mgh}-\mathrm{mgh}=-0.68 \mathrm{mgh}$
Option (1)
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