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Question

A block of mass 1 kg slides down on a rough inclined plane of inclination 60 starting from its top. If the coefficient of kinetic friction is 0.5 and the length of the plane is 1 m, then work done against friction is (Take g=9.8 m/s2)

A
9.82 J
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B
4.94 J
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C
2.45 J
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D
1.96 J
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Solution

The correct option is C 2.45 J

From figure, normal reaction is
N=mg cos θ
=1×9.8×cos 60
=4.9 N
Force due to friction ;f=μN
=μ mg cos 60
=0.5×4.9 N
=2.45 N

length of the plane is S=1 m.
Work done against friction(Wf)
=f×S=2.45 N×1 m
=2.45 J

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