A particle of mass 5 kg is taken from point A to point B slowly with help of external force on a rough inclined surface as shown in figure. If coefficient of friction force between block & inclined plane is $μ = 0.5$ . Then find the work done by external force on block :-

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Since the block moves slowly, at every point $N = m g cos θ$ and friction work over a small element is $d W_{f} = μ m g cos θ d s = μ m g d x$ . Over the whole path:

$W_{f} = μ m g \times 5 = 0.5 \times 5 \times 10 \times 5 = 125 J$

$Δ U = m g h = 5 \times 10 \times 2 = 100 J$

$W_{e x t} = W_{f} + Δ U = 125 + 100 = + 225 J$

Ans. (C)

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