A man is drawing water from a well of depth 10 m using a bucket with hole in it, such that only half of the water remains when bucket arrives at top of the well.
When bucket is full of water, its mass is 40 kg and pulling rate of bucket is constant. The work done by man in kilo joules to pull bucket up is ( g = 10 m/ $s^{2}$ ) [ Neglect weight of the bucket.]

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Solution

Work done $= m g h = \frac{40 + 20}{2} \times 10 \times 10 = 3000 J$

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Q. A man is drawing water from a well of depth 10 m using a bucket with hole in it, such that only half of the water remains when bucket arrives at top of the well.
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Work done =mgh=40+202×10×10=3000 J

Work done $= m g h = \frac{40 + 20}{2} \times 10 \times 10 = 3000 J$