A person is holding a bucket by applying a force of $10 N$ . He moves a horizontal distance of $5 m$ and then climbs up a vertical distance of $10 m$ . The total work done by him is equal to

50 J

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150 J

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100 J

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200 J

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Solution

The correct option is C $100 J$

Let $θ$ be the angle between force and displacement vectors. For horizontal motion,

$F = 10 N, s = 5 m, θ = 90 °$

Work done, W =
$\to F . \to s$
Work done,
$W_{1} = F s c o s θ = 10 \times 5 \times c o s 90 ° = 0$

For vertical motion, the angle between force and displacement is 0°.

Here, $F = 10 N, s = 10 m, θ = 0 °$

Work done, $W_{2} = 10 \times 10 \times cos (0) = 100 J$

Total work done $=$ $W_{1} + W_{2} = 100 J$

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A person is holding a bucket by applying a force of 10 N. He moves a horizontal distance of 5 m and then climbs up a vertical distance of 10 m. The total work done by him is equal to

A person is holding a bucket by applying a force of $10 N$ . He moves a horizontal distance of $5 m$ and then climbs up a vertical distance of $10 m$ . The total work done by him is equal to