A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction μk=0.1. Calculate the work done by kinetic friction in 10sec(g=9.8m/s2)
A
246.96J
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B
−246.96J
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C
882J
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D
−882J
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Solution
The correct option is B−246.96J by the FBD of the block,
N=mg=2×9.8=19.6N
∴ Kinetic friction =μkN =0.1×19.6 =1.96N
Now. Fnet=F−fk=ma ⇒7−1.96=2×a Retardation due to friction ⇒a=2.52m/s2 ∴ Distance moved, s=12at2=12×2.52×102=126m
∴ work done by kinetic friction =−F.s=−1.96×126 =−246.96J, Here the work done is negative as the force and displacement are in opposite direction