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Question

A block of mass 4 kg is placed gently on a conveyor belt moving at a constant speed of 5 m/s at t = 0. Coefficient of friction between the belt and the block is μ = 0.2. Power of friction force on block at t = 3 sec from ground frame of reference is [g=10m/s2](Write until two digits after the decimal point.)

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Solution

fk=μN=μmg
a=fkm=μg
Friction acts on the block till the block achieves 5 m/s.
a=μg=0.2×10=2m/s
v=u+at5=0+2×tt=2.5 sec
Therefore, friction only acts until 2.5 sec.
So, at t = 3 sec, friction = 0.
Hence, the power delivered is zero.

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