Work Energy Power

JEE Main level, single correct, medium

11 of 183

Question 11 of 183

A block of mass 1kg1kg is pushed up a surface inclined to horizontal at an angle of 60∘60^{\circ} by a force of 10N10N parallel to the inclined surface as shown in figure. When the block is pushed up by 10m10m along inclined surface, the work done against frictional force is : [g=10ms−2]\left[g=10m{s}^{-2}\right]

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Answer: (B) 5J5J

Work done against frictional force
=μN×10=μ(mgcos⁡60∘)×10=\mu N\times 10 =\mu \left(mg\cos 60^{\circ}\right)\times 10
=0.1×5×10=5J=0.1\times 5\times 10=5J

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