Question

A bucket that weighs 3 lb and a rope of negligible weight are used to draw water from a well that is 50 ft deep. The bucket is filled with 42 lb of water and is pulled up at a rate of 2.5 ft/s, but water leaks out of a hole in the bucket at a rate of 0.25 lb/s. Find the work done in pulling the bucket to the top of the well. Show how to approximate the required work by a Riemann sum. (Let x be the height in feet above the bottom of the well. Enter xi* as xi.)

Answer 1

the work done in pulling the bucket to the top of the well is 2125 ft-lb

Explanation:

Given that;

Weight of the bucket is 3 lb

weight of water which can be filled in the bucket is 42 lb

Total weight of bucket and water = 3 + 42 = 45 lb

distance, the bucket filled with water is to be pulled 50 ft

now, let at any time t be the bucket at distance x ft from the bottom of the well

then, t = x/2 × S

where S is the rate at which water is leaking from the bucket

so at this time t, the amount pf water which leaked from the bucket is;

⇒ x ft / 2.5 ft/s × 0.25 lb/s

= 0.25 lb.s⁻¹ / 2.5 ft.s⁻¹ × x ft

= 0.1 lb/ft × x ft

= 0.1x lb

now, as x represent the distance that the bucket has been raised, we the force F applied to the bucket x to be;

F = ( 45 - 0.1x ) lb

so, the required worked by Riemann sum as;

`\lim_(n \to \infty)`ⁿ∑_
`_(i=1)` ( 45 - 0.1x
`_i`* ) Δx

so, the work done, pulling the bucket up will be;

`\lim_(n \to \infty)`ⁿ∑_
`_(i=1)` ( 45 - 0.1x
`_i`* ) Δx =
`\int\limits^(50)_0 {(45-0.1x)} \, dx`

= [ 45x - 0.1
`\frac{x^2}2}` ]⁵⁰₀

= 45 × 50 - 0.1/2 × (50)²

= 2250 - 125

= 2125 ft-lb

Therefore, the work done in pulling the bucket to the top of the well is 2125 ft-lb