Question

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. Initially 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.)

Answer 1

The work done shall be 14715 Joules

Step-by-step explanation:

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

`W=m(y)g* 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-(y)/(10))+0.9y`

Thus applying values we get

`W=\int m(y)g* dy\\\\W=\int_(0)^(10)(30(1-(y)/(10)+0.9y)* g)dy\\\\Solving\\W=294.3\int_(0)^(10)(1-(y)/(10)+0.9y)dy\\\\W=14715J`