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/s²]

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(2) 5 J. Given: m=1 kg, θ=60°, μ=0.1, d=10 m along the incline. Normal reaction: N=mg cos 60°=1×10×1/2=5 N. Kinetic friction: f=μ N=0.1×5=0.5 N. Work against friction =fd=0.5×10=5 J.

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/s²]

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