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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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Application +4 | -1
A body of mass 2 kg2 \mathrm{~kg} is projected vertically upwards with velocity 20 m/s20 \mathrm{~m} / \mathrm{s}. If the body rises to a height of 15 m15 \mathrm{~m}, the work done by air drag is (g=10 m/s2)\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)
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Correct option: D
−100 J-100 \mathrm{~J}
Option D (−100 J-100 \mathrm{~J}) is correct because it represents the work done by air drag, calculated using the energy conservation principle.

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