A bucket filled with 20 pounds of water is sitting at the bottom of a 100 ft. well. A taut chain weighing 2 pounds per foot is attached to the bucket, and you are holding the other end. How much work would you do if you were to use the chain to lift the bucket of water from out of the well?

Question

A bucket filled with 20 pounds of water is sitting at the bottom of a 100 ft. well.  A taut chain weighing 2 pounds per foot is attached to the bucket, and you are holding the other end.  How much work would you do if you were to use the chain to lift the bucket of water from out of the well?

amistre64 · Answer

how heavy is the bucket itself?

anonymous · Answer

I assume that's negligible in comparison to the 20 pounds of water.

amistre64 · Answer

perhaps :)

amistre64 · Answer

so we integrate with respect to distance right?

anonymous · Answer

Yeah, that's the method, my issue just comes with figuring out the expression for force to use in the integrand.

amistre64 · Answer

Work = How heavy * how far if i recall correctly ...

anonymous · Answer

Yeah, in a negative y direction.  So it begins as 220 lbs * 100 ft,  both lbs and ft change as work is done.

amistre64 · Answer

20 + 2(100-d) is the weight at any given point; you see why?

anonymous · Answer

I think so, I understand the 20+ 200, but not the -2d.

anonymous · Answer

nevermind.. lol  I do understand

amistre64 · Answer

|dw:1314887910579:dw|

amistre64 · Answer

$$\sum(20+2(100-d))\ \Delta d$$
$$\int_{0}^{100} (20+200-2d)\ dd$$
is my thought

anonymous · Answer

Fantastic, thank you so very much!

amistre64 · Answer

another thought would be to integrate the bucket and add the integration of the chain ..

$$\int_{0}^{100}20dd+\int_{0}^{100}(2(100-d))dd$$
$$\int_{0}^{100}20dd+\int_{0}^{100}(200-2d)dd$$
$$20d+(200d-d^2)$$
$$20(100)+(200(100)-(100)^2)$$
$$2000+(20000-10000)$$
$$2000+10000=12000$$

amistre64 · Answer

yep, youre welcome :)