A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60^ by a force of 10 N parallel to the inclined surface as shown in figure. When the block is pushed up by 10 m along inclined surface, the work done against frictional force is : [g=10 ~m / s ^2 ]

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Correct answer: B. 5 J

Mass of block, m=1 ~kg Force of parallel inclined surface, F =10 ~N Work done against frictional force = _2 ~N 10= Mg 60 10=0.1 5 10=5 ~J

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