Math problem?

Question

A bucket that weighs 4 lb and a rope of negligible weight are used to draw water from a well that is 80 ft deep. The bucket is filled with 40 lb of water and is pulled up at a rate of 2 ft/s, but water leaks out of a hole in the bucket at a rate of 0.2 lb/s. Find the work done in pulling the bucket to the top of the well

Answer 1

The key to this one is finding the weight as a function of time.

Weight:

The initial weight will be 40 lbs of water + 4 lbs of bucket = 44 lbs.

Since the rate of water loss is linear, and since you're given the "slope", i.e. the rate of water loss, (-0.2 lb/s), you can find an equation for the weight w in terms of the time t:

(w - 44)/(t - 0) = -0.2
w - 44 = -0.2t
w(t) = 44 - 0.2t

Distance:

The initial distance is 80 feet, and it decreases linearly by 2 feet per second, so again the distance in terms of time can be found using the point-slope formula:

(d - 80)/(t - 0) = -2
d - 80 = -2t
d(t) = 80 - 2t

And then, since we're trying to figure out how much work it takes to get the bucket to the top of the well, we need to solve for when the distance is 0:

0 = 80 - 2t
2t = 80
t = 40

So it will take 40 seconds to raise the bucket.

Work:

Now it's just a matter of using the formula for work:

W = ∫ F(t) · ds(t)

where F(t) is the weight, and ds(t) is the derivative of the distance with respect to time. The distance is DECREASING by 2 feet per second, but it's also decreasing in the opposite direction of the force of the weight, so the negatives cancel:

40
∫ (44 - 0.2t)(2) dt =
0

40
∫ 88 - 0.4t dt =
0

40
[88t - 0.4/2 t²] =
0

[88(40) - 0.2(40)²] = [3,520 - 320] = 3,200 foot-pounds

I think this one is right.