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Question

A long conveyor belt is designed to transport packages of various weights. Each 20 kg package has a coefficient of kinetic friction 0.25. If the speed of the conveyor belt is 10 m/s and then it suddenly stops, the distance the package will slide before coming to rest is


A
6 m
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B
10 m
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C
15 m
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D
20 m
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Solution

The correct option is D 20 m
When the conveyor belt is suddenly stopped, the block is moving with velocity 10 m/s. Let a be the deceleration of the block (due to kinetic friiction between the block and the belt)

The FBD of the package can be given by


From the FBD, we have
N=20g=200 N
fk=μkN=0.25×200=50 N
a=fkm=5020=2.5 m/s2

Hence, using 3rd equation of motion,
Distance travelled before sliding stops is
s=v22a=(10)22×2.5=1005=20 m

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