1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work Done (integral)2

  1. Dec 22, 2011 #1
    1. The problem statement, all variables and given/known data
    WorkDoneIntegral.png


    2. Relevant equations
    [itex]W = \int F dx[/itex]

    For density of water i used
    D = 62.4lb/ft^3
    g = 32.2 ft/s^2

    [itex]g(D = \frac{m}{V})[/itex]

    [itex]gDV = mg[/itex]

    [itex]gDV = F[/itex]

    3. The attempt at a solution
    i use limit from 0 to 6
    [itex]W = \int_{0}^{6} F dx[/itex]

    [itex]W = \int_{0}^{6} gDV dx[/itex]

    [itex]W = \int_{0}^{6} (32.2)(62.4)(\pi{6}^{2}{15}) dx[/itex]

    [itex]W = 20451979.29 [/itex]

    help check this pls
     
  2. jcsd
  3. Dec 22, 2011 #2

    Mark44

    Staff: Mentor

    This is incorrect. You are assuming that all of the water has to be lifted 15 feet, and this is not the case. Mentally divide the water into cylindrical slices, each of which is [itex]\pi[/itex]*36*[itex]\Delta y[/itex] in volume. From the volume, you can get to the weight of each slice, but the distance a slice of water has to be pumped depends on where the slice is in the tank. The layers (slices) at the top require almost no effort to pump out, since they don't need to be lifted very far. The ones at the bottom require a lot more work, since the have to be lifted much farther.
     
  4. Dec 23, 2011 #3
    found the answer thx for the advice

    slice of cylindrical element anywhere inside cylinder
    [itex]dV = 36{\pi}dy[/itex]

    [itex]dF = 32.2(62.4(\pi{36})dy)[/itex]

    work done from this differential element to the top: then (distance = 15 - y)
    [itex]dW = 32.2(62.4(\pi{36})dy)(15-y) [/itex]

    [itex]W = \int_{0}^{15} dW[/itex]
    etc.

    [itex]W = 25564974 [/itex]lb-ft

    [itex]W = 793943 [/itex]slugs-ft
     
  5. Dec 23, 2011 #4

    Mark44

    Staff: Mentor

    I didn't verify your numbers, but what you have looks to be set up correctly.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Work Done (integral)2
  1. Work done integration (Replies: 2)

Loading...