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Work needed to remove Keel-Aid from water

  1. Mar 15, 2005 #1
    You are visiting your friend Fabio's house. You find that, as a joke, he filled his swimming pool with Kool-Aid, which dissolved perfectly into the water. However, now that you want to swim, you must remove all of the Kool-Aid contaminated water. The swimming pool is round, with a 14 foot radius. It is 12 feet tall and has 1 feet of water in it.
    How much work is required to remove all of the water by pumping it over the side? Use the physical definition of work, and the fact that the weight of the Kool-Aid contaminated water is 65.7lbs/ft^3


    first of all, i figure the bounds of integration to be from 1 to 12.

    volume = pi(14)^2

    =196pi*ft = 65.7lbs/ft^3

    so...
    196pi(65.7)

    my function:
    distance would just be equal to x right?
    [tex]\int_{1}^{12} 196*pi(65.7) * x dx [/tex]
     
  2. jcsd
  3. Mar 15, 2005 #2

    learningphysics

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  4. Mar 15, 2005 #3
     
  5. Mar 15, 2005 #4

    learningphysics

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    But there is only 1 foot of water in the pool. You are integrating over the water that needs to be pumped, not the distance that needs to be travelled.

    Anyway here's the integral I figured:
    [tex]\int_{0}^{1} 196*pi(65.7) * (12 - x) dx [/tex]
     
  6. Mar 15, 2005 #5
    I see. My mistake. you're right, that integrals is correct. I mixed up the distance traveled with the integrals bounds.

    Regards,

    Nenad
     
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