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poizon [28]
3 years ago
14

A leaky 10-kg bucket is lifted from the ground to a height of 10 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

ly the bucket contains 30 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 10-m level. Find the work done. (Use 9.8 m/s2 for g.) Show how to approximate the required work by a Riemann sum. (Let x be the height in meters above the ground. Enter xi* as xi.)

Physics
1 answer:
sladkih [1.3K]3 years ago
8 0

Answer:

The work done shall be 14715 Joules

Explanation:

The work done by a force 'F' in a displacement 'dy' is given by

W=m(y)g\times dy

At any position 'y' the weight shall be sum of weft of water and weight of string

\therefore m(y)=m_{water}(y)+m_{string}(y)\\\\m(y)=30(1-\frac{y}{10})+0.9y

Thus applying values we get

W=\int m(y)g\times dy\\\\W=\int_{0}^{10}(30(1-\frac{y}{10}+0.9y)\times g)dy\\\\Solving\\W=294.3\int_{0}^{10}(1-\frac{y}{10}+0.9y)dy\\\\W=14715J

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