Question

A bucket that weighs 5 lb and a rope of negligible weight are used to draw water from a well that is 60 ft deep. The bucket is filled with 42 lb of water and is pulled up at a rate of 1.5 ft/s, but water leaks out of a hole in the bucket at a rate of 0.15 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.) lim n→[infinity] n Δx i = 1 Express the work as an integral. 0 dx Evaluate the integral. ft-lb

Answer 1

The value is
`W= 2640 \ ft \cdot lb`

Explanation:

From the question we are told that

The weight of the bucket is
`F = 5 lb`

The depth of the well is
`x_1 = 60 \ ft`

The weight of the water is
`W_w = 42 lb`

The rate at which the bucket with water is pulled is
`v = 1.5 \ ft/s`

The rate of the leak is
`r = 0.15 lb/s`

Generally the workdone is mathematically represented as

`W = \int\limits^(x_1)_(x_o) {G(x)} \, dx`]

Here G(x) is a function defining the weight of the system (water and bucket ) and it is mathematically represented as

`G(x) = F + (W_w- Ix)`

Here I is the rate of water loss in lb/ft mathematically represented as

`I = (r)/(v)`

=>
`I = (0.15 )/(1.5 )`

=>
`I = 0.1`

So

`G(x) = 5 + (42- 0.1x)`

=>
`G(x) = 47- 0.1x)`

So

`W = \int\limits^(60)_(0) {47- 0.1x} \, dx`]

=>
`W = [47x - (0.1x^2)/(2) ]|\left 60} \atop {0}} \right.`

=>
`W= [47(60) - 0.05(60)^2]`

=>
`W= 2640 \ ft \cdot lb`