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Math Help - amount of work

  1. #1
    Junior Member
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    amount of work

    If water is pumped in through the bottom of the tank in the picture, how much work is done to fill the tank:
    a. To a depth of 2 feet?
    b. From a depth of 4 feet to a depth of 6 feet?
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  2. #2
    Junior Member
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    Here's what I make of it.

    - - -

    The tank is conical, clearly 6 units high, and it looks like it's roughly 4 units in 'radius' at the top.

    Call the radius at height y R(y). If R(6)=4 and it runs linearly (as you'd expect), then in general

    R(y) = \frac{2y}{3}.

    Using conservation of energy, the amount of work done to fill the tank to a height y (call it W(y)) is equal to the total gravitational potential energy of the water in the container.

    The gravitational potential energy is found by integrating (calculus) over circular "slices" of water, at height y and with thickness dy. (Let me know if you don't get this part; it's explained better with a diagram.)

    The slice will have a volume

    \pi \left(R(y)\right)^2 dy.

    Its mass comes just from multiplying by the density of water, \rho:

    m = \rho \pi \left(R(y)\right)^2 dy.

    Finally, the potential energy of the slice is

    mgy = \rho \pi \left(R(y)\right)^2 dy \cdot gy = \rho \pi gy \left( \frac{2y}{3} \right)^2 dy = \frac{4}{9} \rho \pi g y^3

    (where I have substituted for R(y).)

    Then just integrate with respect to y.

    W(y) = \int_0^y \frac{4}{9} \rho \pi g y^3= \frac{4}{9} \rho \pi g \left[ \frac{y^4}{4}\right]_0^y = \frac{\rho \pi g y^4}{9}.

    The answer to part a) is W(2).
    The answer to part b) is W(6) - W(4) (as the tank was already filled to 4 units.)
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