300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Work done against friction is (take g=10 m s⁻²)

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(3) 100 J. Given: W=300 J supplied, m=2 kg, h=10 m, g=10 m s⁻², and the block is slid up (no kinetic energy left over). Idea: the work supplied goes partly into lifting the block and the rest into friction. Potential energy gained =mgh=2×10×10=200 J. W_supplied=Δ U+W_friction. W_friction=300-200=100 J. Notice the length of the incline never enters; only the height matters for the lift.

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Work done against friction is (take g=10 m s⁻²)

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