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JEE Main 2021 (1 Sep, Shift 2) · Work-Energy Theorem

Work-Energy Theorem question from JEE Main 2021 (1 Sep, Shift 2)

Q15JEE Main 2021 · 1 SEP · SHIFT 2Easy

Work-Energy Theorem

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:

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Answer: (a)

Work-energy theorem: $W_g + W_f = \frac{1}{2}mv^2$ .
$\frac{1}{2}m(0.8\sqrt{gh})^2 = 0.32\,mgh$
$W_f = 0.32\,mgh - mgh = -0.68\,mgh$ , so the answer is (a).

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