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Math Help - Emptying a pool

  1. #1
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    Emptying a pool

    Hello Mathhelpforum!
    I have two questions. One with an answer I wanted to double check, and the second needing help on.

    The first one is if it takes 100J of work to stretch a spring 0.5m from it's equilibrium position, how much work is needed to stretch an additional .75m?

    So I set my limits at x = .5 and x = 1.25.
    100 = F(.5)
    F = 200.
    kx = F
    k(.5) = 200
    k = 100.
    So the spring constant is 100. Then the integral of 100x dx from .5 to 1.25 = 50x -> 5(1.25) - (50(.5)) = 65.625J

    And this problem I'm stuck on.
    A swimming pool has a shape of a box with a base of 25m by 15m and a depth of 2.5m. How much work is required to pump the water out of the pool?
    Would it be Integral from 0 to 2.5m (pg)dy * integral from 0 to 2.5m of (2.5-y)dy?
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  2. #2
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    Re: Emptying a pool

    Some error here I think.
    Extending from 0 to 0.5 work done= integral of kx from 0 to 0.5=k(0.5)^2/2 So 0.25k/2=100 k=800
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  3. #3
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    Re: Emptying a pool

    Hooke's law says f = -kx so k = f/x?
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  4. #4
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    Re: Emptying a pool

    Quote Originally Posted by Verleiren View Post
    Hooke's law says f = -kx so k = f/x?
    the force is variable ... you're trying to treat it as a constant.

    W = \frac{1}{2}kx^2

    100 = \frac{1}{2}k(0.5)^2 , solve for k, then calculate \int_{0.5}^{1.25} kx \, dx

    for the pool ...

    W = \rho \cdot g \int_0^{2.5} (25 \cdot 15)(2.5 - y) \, dy , where \rho is the density of water and g is the acceleration due to gravity
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