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Applications of Integration - Work. Please Help

by JohnRV5.1
Tags: applications, integration, work
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Jun28-05, 12:16 AM
P: 8
I am currently taking CALC II in the summer. So far we have gone over Applications of Integration(area under the curve, solids of revolution) to Techniques of Integration (integration by parts, Trig substitution, Partial Fractions, IMproper Integrals, etc.) From all the problems I have encountered, only this one gives me trouble. It is the chapter on applications of Integration that pertains to work. Here is the problem and the solution I obtained, altough I am unsure I tackled the problem correctly:

A bucket that weights 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.

My Solution:
First I found the work required to lift the bucket byself to the top of the well.
I got Force = (4 lb)(80 ft) = 320 ft*lb

Then I obtained the work done in pulling the leaking water to the top of the well using integration.
I found the distance = x
The Force = (40 lb)/(80 ft) - (.2 lb/s)/(2 ft/s) = .5 lb/ft - .1lb/ft = .4 lb/ft or 2/5 lb/ft
therefor the force is = 2/5 dx or .4dx.
I set up the integral using the above info
= Integral from 0 to 80 of .4xdx
evaluating the integral I obtained 1280 ft*lb

So I summed up both the work required to lift the bucket to the top and the work required to lift the water to the top of the well: 1280 ft*lb + 320 ft*lb = Work = 1600 ft*lb!! Am I correct

p.s. My calc instructor informed us that no work questions will be on the exam but he did assign homework for it. The problem was due to time constraints, the instructor was not able to lecture on the section on work. Plus I have never taken a single Physics class so I did not feel too confident with my solution.
Thank you for lending your time and efforts to help me.
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Jun28-05, 01:16 AM
P: 8
I attempted the problem a second time with a different approach and got 3200 ft*lb of work instead. This time I used the integral from 0 to 80 of
(40 - .1x)dx

Can anyone tell me if I obtained the right solution. BTW, sorry if your having difficulties reading my notation above. Help!
Jun28-05, 01:40 AM
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siddharth's Avatar
P: 1,197
One way to do this question.
At time 't' what is the mass of water left in the bucket? It is 40 - 0.2t lb

The mass of the bucket is 4lb. So the total mass is 40 -0.2t + 4 = 44 - 0.2t lb.
now, by newton's second law,


Therefore, mdv/dt+vdm/dt=F(net)

Now, dv/dt=0

so, vdm/dt=(T-mg)
where T is the force you apply to pull the bucket as a function of time.
So T= vdm/dt+mg.
T=(2)(-0.2) + (44-0.2t)(g) {Use the value of g in appropriate units}
and time taken to reach the top is h/v=(80)/(2)=40

So work done is
[tex] \int T.dx [/tex]
Which is [tex] \int T.(vdt) [/tex]
= [tex] \int T.(2)(dt) [/tex] from t=0 to t=40

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