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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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A force F=(40+0.5x)F=(40+0.5 x) acts on a particle in the xx direction where FF is in NN and xx in m{m}. The work done by this force during a displacement from x=0x=0 to x=4 mx=4 {~m} is
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Correct option: D
164 J164 {~J}
Option D (164 J164 {~J}) is correct because it's the result of integrating the force function over the given displacement.

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