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Math Help - work by pumping

  1. #1
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    work by pumping

    A circular swimming pool has a diameter of 16 m, the sides are 4 m high, and the depth of the water is 3.5 m. How much work (in Joules) is required to pump all of the water over the side? (The acceleration due to gravity is 9.8 and the density of water is 1000 .)

    so i tried
    volume=64pi dx

    weight= vol *density

    64000pidx

    integral .5 to 4 64000pidx
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  2. #2
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    Quote Originally Posted by dat1611 View Post
    A circular swimming pool has a diameter of 16 m, the sides are 4 m high, and the depth of the water is 3.5 m. How much work (in Joules) is required to pump all of the water over the side? (The acceleration due to gravity is 9.8 and the density of water is 1000 .)

    so i tried
    volume=64pi dx

    weight= vol *density

    64000pidx

    integral .5 to 4 64000pidx
    did you learn anything from this post?

    http://www.mathhelpforum.com/math-he...k-pumping.html
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  3. #3
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    ya but its different i followed the exact way my teacher did it but its not working
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  4. #4
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    Won't the work required to be done just be the gain in in gravitational PE of the entire water?

    The total mass of the water= Volume of the water(density)
    = 1000(pi)(64)(3.5)=m(say.)

    Now when the water is pumped over the side, the gain in gravitational PE is mg(4-3.5) = mg(0.5)
    =3448220.16 J ??????
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  5. #5
    MHF Contributor Calculus26's Avatar
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    If interested in general problems which involve work to empty a tank

    See Work on my web site
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  6. #6
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    Quote Originally Posted by bandedkrait View Post
    Won't the work required to be done just be the gain in in gravitational PE of the entire water?

    The total mass of the water= Volume of the water(density)
    = 1000(pi)(64)(3.5)=m(say.)

    Now when the water is pumped over the side, the gain in gravitational PE is mg(4-3.5) = mg(0.5)
    =3448220.16 J ??????
    that is not right though
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  7. #7
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    Quote Originally Posted by dat1611 View Post
    that is not right though
    Oh god that was a monumental blunder!!!!

    Ignore my first post, please. Or preserve it as an example of easily avoidable human stupidity!!!!!!
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  8. #8
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    can anyone else try to solve it
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  9. #9
    MHF Contributor Calculus26's Avatar
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    Since the density is constant and the cross sections are constant

    we don't need an integral.

    So we think of the mass as being concentrated at the center of gravity

    here y= 1.75 and the distance 4 - 1.75 = 2.25 m

    weight is pi * 64*1000*9.8*3.5

    Calculate weight * dist = 64*1000*9.8*3.5*pi*2.25

    If you do integrate the limits are 0 to 3.5

    you have 64*1000*9.8*pi integral[(4-y)dy]
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