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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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Velocity of a particle of mass 4 kg4 \mathrm{~kg} moving along x\mathrm{x}-axis is given as v=2(x)3/2{v}=2({x})^{3 / 2} then work done by forces acting on the particle when it moves from x=0{x}=0 to x=2 m{x}=2 \mathrm{~m} is :-
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Correct option: B
64 J
Option B (64 J) is correct because the work done is equal to the change in kinetic energy, which is calculated using the given velocity function.

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