A block of mass $1 k g$ is pushed up a surface inclined to horizontal at an angle of $30^{\circ}$ by a force of $10 N$ parallel to the inclined surface as shown in the figure.
The coefficient of friction between block and the inclines is $0.1$ . If the block is pushed up by $10 m$ along the incline, then the work against gravity is (Take $g = 10 m s^{- 2})$ .

10 J

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50 J

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100 J

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150 J

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Solution

The correct option is D $50 J$
The various forces acting on the block are as shown in the figure.
Here, $m = 1 k g, θ = 30^{\circ}, F = 10 N$
$μ = 0.1, d = 10 m$
Work done against gravity is
$W_{g} = m g d sin θ$
$= (1 k g) \times (10 m s^{- 2}) \times 10 \times sin 30^{\circ} m = 50 J$ .

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