A block of mass 1kg 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 1m, then work done against friction is (Take g=9.8m/s2)
A
9.82J
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B
4.94J
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C
2.45J
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D
1.96J
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Solution
The correct option is C2.45J
From figure, normal reaction is N=mgcosθ =1×9.8×cos60∘ =4.9N
Force due to friction ;f=μN =μmgcos60∘ =0.5×4.9 N =2.45N
length of the plane is S=1m. ∴ Work done against friction(Wf) =f×S=2.45N×1m =2.45J