Work Prob: A bucket that weighs 6 lb and a rope of negligibl

[flakine]

A bucket that weighs 6 lb and a rope of negligible weight are used to draw water from a well that is 90 ft deep. The bucket is filled with 50 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.

Can someone help me set this one up?

 

[galactus]

When the bucket is x feet above the bottom of the well.

total weight=weight of bucket + weight of water.

\(\displaystyle \L\\56-\frac{0.2}{2}y\)

\(\displaystyle \L\\\int_{0}^{90}(56-0.1y)dy=4635\)

or

\(\displaystyle \L\\6(90)+\int_{0}^{90}(50-0.1y)dy=4635\)

 

[flakine]

Thanks.

I understand most, but could you explain the physics behind the 0.2/2

 

[galactus]

That is the change in weight as the leaky bucket is hoisted up.

\(\displaystyle \L\\\frac{0.2\frac{lbs}{sec}}{2\frac{ft}{sec}}\)

\(\displaystyle \L\\=0.2\frac{lbs}{\sout{sec}}\cdot{2}\frac{\sout{sec}}{ft}=0.1\frac{lbs}{ft}\)

 

[skeeter]

let y = 0 be the top of the well

position of the bucket as a function of time ...
y = 2t - 90

t = (y + 90)/2

weight of the bucket as a function of time ...
F = 56 - 0.2t

weight of the bucket as a function of position ...

F = 56 - 0.2(y + 90)/2 = 47 - 0.1y

a small "bit" of work on the bucket = dW = (47 - 0.1y)dy

\(\displaystyle \L Work = \int_{-90}^0 (47 - 0.1y)dy = 4635 ft-lbs\)

edit: fixed egregious definite integral error

 

[galactus]

\(\displaystyle \L Work = \int_{-90}^0 (47 - 0.1y)dy = 4635 ft-lbs\)

Actually, skeet, your method works out to the same answer as my way.

4635.

 

[skeeter]

yessir ... i did the integral using the integrand 47 - .01y ... my eyeballs are going south ...