# Definite Integral Work

chukie
A bucket that weighs 3 lb and a rope of negligible weight are used to draw water from a
well that's 60 feet deep. Suppose the bucket starts with 37 lb of water and is pulled up by a rope at 2 ft/sec, while water leaks out of the bucket at a rate of 1/4 lb/sec.

A. How long does it take for the bucket to get to the top of the well? Write an equation that
expresses the total weight of the bucket as a function of time, as the time varies from 0 until the time the buckets gets to the top.

B. Recall that work equals force times distance. Calculate the work done in lifting the bucket
to the top of the well, keeping in mind that here force is equal to weight.

For A:I did 60ft/2ft/s=30 sec so it takes 30 secs for the bucket ot get to top of well
equation: Weight=40+ integral sign from 0 to 30 (-1/4)dx =32.5 lb

For B: i did total work= integral sign from 0 to 60 (32.5x)dx

i dunno if im doing this right. ive searched the internet and found a question similar to mine (question #6):
http://count.ucsc.edu/~bauerle/images/19BW07HWK/Homework 6.4 Math 19B Winter 2007, Bauerle.pdf"

and ive used the same steps that i did for this question but i got the wrong answer.

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Homework Helper
For A:I did 60ft/2ft/s=30 sec so it takes 30 secs for the bucket ot get to top of well
equation: Weight=40+ integral sign from 0 to 30 (-1/4)dx =32.5 lb
There isn't any integral involved here yet. You have to express your answer in terms of t, time. The weight would be wt of bucket + water remaining. Find an expression for the amount of water remaining at some given time t. No integration is required.

For B: i did total work= integral sign from 0 to 60 (32.5x)dx
One way you could do this would be to change the variables of your integration. Work done is given by $$\int^b_a \vec{F} \cdot d \vec{r} = \int^{b(t)}_{a(t)} \vec{F}(t) \cdot \frac{d\vec{r}}{dt} dt$$.

In this case since the problem is strictly one dimensional you can treat the vector functions as scalar functions. You have to do a change of variables from r to t. Firstly express r in terms of t. r is the distance from the bottom of the well to the top. Then you can continue from there.

chukie
There isn't any integral involved here yet. You have to express your answer in terms of t, time. The weight would be wt of bucket + water remaining. Find an expression for the amount of water remaining at some given time t. No integration is required.

One way you could do this would be to change the variables of your integration. Work done is given by $$\int^b_a \vec{F} \cdot d \vec{r} = \int^{b(t)}_{a(t)} \vec{F}(t) \cdot \frac{d\vec{r}}{dt} dt$$.

In this case since the problem is strictly one dimensional you can treat the vector functions as scalar functions. You have to do a change of variables from r to t. Firstly express r in terms of t. r is the distance from the bottom of the well to the top. Then you can continue from there.

so for A:the equation would be W(t)=40-(1/4)t

for B: i dun quite understand wut u mean.do u mean i do the integral from 0 to 30? if that's the case i dunno wut the equation that im trying to integrate is

Homework Helper
Do you know the difference between "dun" and "don't" or "wut" and "what"?

Homework Helper
so for A:the equation would be W(t)=40-(1/4)t

for B: i dun quite understand wut u mean.do u mean i do the integral from 0 to 30? if that's the case i dunno wut the equation that im trying to integrate is
You have to first do a change of variables from r to t, as I said earlier. The integral for work done in terms of r (displacement) goes from 0 to 60. But if you evaluate in terms of t, it goes from 0 to 30. Replace the limits of r with that of t and the corresponding integrand also changes.

The reason why the change of variables is made is because the earlier part of the question asked you express the weight of the bucket + water as a function of time, which makes its easier to evaluate the work done in t instead or r. You could of course evaluate it in terms of r as well, but you now have to find an expression for the total load weight as a function of r.

chukie
okay thanks. i finally got the answer =)