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Q. A body of mass ' $m$ ' dropped from a height ' $h$ ' reaches the ground with a speed of $0.8 \sqrt{ gh }$. The value of workdone by the air-friction is:

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Solution:

Work done $=$ Change in kinetic energy
$W _{ mg }+ W _{\text {air-fiction }}=\frac{1}{2} m (.8 \sqrt{ gh })^{2}-\frac{1}{2} m (0)^{2}$
$W _{\text {air }-\text { fiction }}=\frac{.64}{2} \,mgh - mgh =-0.68\, mgh$