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

Answer:

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

Step-by-step 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

$I = \frac{r}{v}$

=> $I = \frac{0.15 }{1.5 }$

=> $I = 0.1$

So

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

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

So

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

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

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

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