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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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Application +4 | -1
A block of mass m=0.5 kg{m}=0.5 \mathrm{~kg} gently placed on a conveyor belt moving with velocity of 4 m/s4 \mathrm{~m} / \mathrm{s}. The coefficient of friction between the conveyor belt and bag is 0.2.0.2. Initially the block slips on the conveyor belt before it stops due to friction. The work done by kinetic friction on the block is [g=10 m/s2]\left[{g}=10 \mathrm{~m} / \mathrm{s}^2\right]
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Correct option: A
4 J4 \mathrm{~J}
Option A (4 J4 \mathrm{~J}) is correct because it is the work done by the kinetic friction on the block as it comes to rest relative to the belt.

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