Work Energy Power

JEE Main level, single correct, medium

94 of 183

Question 94 of 183

A body of mass mm dropped from a height hh reaches the ground with a speed of 0.8gh.0.8\sqrt{gh}. The value of work done by the air-friction is:

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Answer: (A) −0.68mgh-0.68mgh

K1+U1=K2+U2+Wf{K}_{1}+{U}_{1}={K}_{2}+{U}_{2}+{W}_{f}
Wf:{W}_{f}: work done against friction
0+mgh=12mv2+0+Wf0+mgh=\dfrac{1}{2}m{v}^{2}+0+{W}_{f}
mgh=12m(0.64)gh+Wfmgh=\dfrac{1}{2}m\left(0.64\right)gh+{W}_{f}
Wf=0.68mgh{W}_{f}=0.68mgh
Work done by friction Wf′=−Wf=−0.68mgh{W}_{f}^{'}=-{W}_{f}=-0.68mgh

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