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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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A block of mass 5 kg5 \mathrm{~kg} is moving along xx-axis under the action of a variable force F=10x NF=10 x \mathrm{~N}. The work done by the force in moving the block from x=4 mx=4 \mathrm{~m} to x=6 mx=6 \mathrm{~m} will be
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Correct option: D
100 J100 \mathrm{~J}
Option D (100 J100 \mathrm{~J}) is correct because it is the work done calculated by integrating the force over the given displacement.

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