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Math Help - Work problem

  1. #1
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    Work problem

    find how much work it takes to empty a full hemispherical water reservoir of radius 5m by pumping the water to a height of 4 m above the top of the reservoir. water weighs 9800 N/m^3.

    I need to find the volume first so i think it's v=integral (pi*r^2)

    so V = pi * integral (25-y^2) dy from 0 to 5
    which is 250*pi/3
    mass = density * volume = 9800n/m^3 * 250*pi/3 = 2.45*10^6*pi / 3
    force = mass * 9.8m/s^2 (gravity) = 2.401*pi*10^7 /3
    then work = force*x = [2.401*pi*10^7 /3] *4m

    am i doing this right?
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  2. #2
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    Quote Originally Posted by synnexster View Post
    find how much work it takes to empty a full hemispherical water reservoir of radius 5m by pumping the water to a height of 4 m above the top of the reservoir. water weighs 9800 N/m^3.

    I need to find the volume first so i think it's v=integral (pi*r^2)

    so V = pi * integral (25-y^2) dy from 0 to 5
    which is 250*pi/3
    mass = density * volume = 9800n/m^3 * 250*pi/3 = 2.45*10^6*pi / 3
    force = mass * 9.8m/s^2 (gravity) = 2.401*pi*10^7 /3
    then work = force*x = [2.401*pi*10^7 /3] *4m

    am i doing this right?
    Consider a disk of thickness ds s metres below the brim of the tank. this has volume:

    <br />
dV = \pi (5^2-s^2) ds<br />

    which has mass:

    <br />
dM=\rho dV=1000 dV<br />

    the 1000 is the density of water which is 1000 kg/m^3.

    The work done to lift this to a height of 4m above the brim of the tank is:

    <br />
dW=dM\ g\ (s+4)<br />

    that is the mass times g times the height lifted; as this is the change in potential energy of the water.

    So the total work to empty the tank is:

    <br />
W=\int_0^5 (s+4)\ g\ 1000\ \pi\ (5^2-s^2) ds=\int_0^5 (s+4)\ 9800 \ \pi\ (5^2-s^2) ds<br />

    (where we take g\approx 9.8 m/s^2.)

    RonL
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