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

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-09-24T05:59:48+02:00

Answer:

The value is [tex]W= 2640 \ ft \cdot lb[/tex]

Step-by-step explanation:

From the question we are told that

The weight of the bucket is [tex]F = 5 lb[/tex]

The depth of the well is [tex]x_1 = 60 \ ft[/tex]

The weight of the water is [tex]W_w = 42 lb[/tex]

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

The rate of the leak is [tex]r = 0.15 lb/s[/tex]

Generally the workdone is mathematically represented as

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

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

[tex]G(x) = F + (W_w- Ix)[/tex]

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

[tex]I = \frac{r}{v} [/tex]

=> [tex]I = \frac{0.15 }{1.5 }[/tex]

=> [tex]I = 0.1[/tex]

So

[tex]G(x) = 5 + (42- 0.1x)[/tex]

=> [tex]G(x) = 47- 0.1x)[/tex]

So

[tex]W = \int\limits^{60}_{0} {47- 0.1x} \, dx[/tex]]

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

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

=> [tex]W= 2640 \ ft \cdot lb[/tex]