A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of $6 0^{\circ}$ 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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Mass of block, m = 1 kg Force of parallel inclined surface, F = 10 N Work done against frictional force = μ 2 ​ N × 10 = μ Mg cos 60 × 10 = 0.1 × 5 × 10 = 5 J

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