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Physics problem involving work

  1. Mar 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A conical tank filled with kerosene is buried 4 feet underground. The density of kerosene is 51.2 lbs/ft3. The kerosene is pumped out until the level drops 5 feet. How much work is needed to pump the kerosene to the surface if the variable is given as:

    A. The distance between the vertex of the cone and the “slice”?
    B. The distance between the top of the cone and the “slice”?*
    C. The distance between the surface and the “slice”?*
    D. The distance between the final level of the kerosene and the “slice”?*
    2qsnzip.jpg
    2. Relevant equations
    W = Force*distance
    Force=vol*density

    3. The attempt at a solution
    My first question is- do I have the schematic set up right? Am I integrating the volume of the slice as indicated in A or B (I used A to set up my integral):
    2mcwoja.jpg

    *I am struggling with where to put the displacement (distance):
    A. The distance the slice must travel is the full length minus h (since it is only emptied to a certain depth) plus the 4 ft above ground; so d = (8-h) + 4
    B. I would say distance the slice must travel is up the top of the cone (so the origin is at the rim, and h goes down to the slice), plus the 4 ft above ground, so d = h+4
    C. I would say the distance the slice must travel is the full length plus the additional 4 ft, subtracted from the depth of h, so d = 12 - h [but why would it be the same as in part A?]
    D. The distance here is what made me think I initially set up my integral completely wrong, if so- would the distance then become (5-3)-h = 2-h?
     
  2. jcsd
  3. Mar 16, 2016 #2

    Mark44

    Staff: Mentor

    You're summing (by integrating) the work needed to bring a "slice" of kerosene to the surface. The limits of integration will be the bottom-most level that's pumped (lower limit) to the top of the tank (upper limit).

    It would be helpful in your drawing to explicitly label the axes, especially the vertical axis. That will help you get a handle on the various distances for the four parts of this problem.
     
  4. Mar 20, 2016 #3
    Thanks for your tip, Mark!
    So I corrected the limits to be from just the slab that is moving out 3 ft - 8 ft, and the distance as 4 + (8-h).
    Does this seem correct to you, or would you have any other comments?

    2m3n2nm.jpg


     
  5. Mar 20, 2016 #4

    Mark44

    Staff: Mentor

    Which part of the problem (a, b, c, or d) is this the work for?
     
  6. Mar 20, 2016 #5
    Sorry, just A
     
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