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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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A force given by the relation F=8t\text{F}=8 \text{t} acts on a body of mass 2 kg2 \mathrm{~kg} initially at rest. The work done by this force on the body during first 2 s2 \mathrm{~s} of its motion is:
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Correct option: A
64 J64 \mathrm{~J}
Option A (64 J64 \mathrm{~J}) is correct because it calculates the work done using the work-energy theorem, finding the final velocity after 2 seconds.

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