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Question

A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction μk=0.1.
Calculate the work done by kinetic friction in 10 sec (g=9.8 m/s2)

A
246.96 J
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B
246.96 J
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C
882 J
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D
882 J
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Solution

The correct option is B 246.96 J
By the FBD of the block,


N=mg=2×9.8=19.6 N

Kinetic friction =μkN
=0.1×19.6
=1.96 N

Now.
Fnet=Ffk=ma
71.96=2×a
Retardation due to friction a=2.52 m/s2
Distance moved, s=12at2=12×2.52×102=126 m

work done by kinetic friction =F.s=1.96×126
=246.96 J,
Here the work done is negative as the force and displacement are in opposite direction

Hence option B is the correct answer

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