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Question

A block of mass 10 kg is put gently on a belt-conveyor system of infinite length at t=0, which is moving with constant speed 20 m/s  towards right at all time. A constant force of magnitude 15 N is applied on the block continuously during its motion. Work done by the kinetic friction on the block of mass 10 kg is


A
2500 J
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B
2500 J 
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C
1250 J
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D
1250 J 
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Solution

The correct option is A 1250 J
Here by the FBD of 10 kg blockn conveyor belt will initially apply frictional force in the right direction
kinetic friction = μkN=0.25×100
fk=25N
Applying 2nd law of Newtons

15+25=ma

a=4010=4 m/s2
Now, distance travelled.
S=v22a=2022×4=4008=50 m

Work done by kinetic friction on the block of mass 10kg=fk.d
=fk.d
=25×50
Wk=1250J  Here the work done is positive as the frictional force and the displacement are in the same direction.

Hence option C is the correct answer

Physics

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