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Beavers water and work

  1. Mar 2, 2010 #1
    1. The problem statement, all variables and given/known data
    A beaver is building a damn and fills a 5kg bucket with 25kg of water. Then he begins to climb a vertical ladder, but unfortunately there is a whole in the bucket where water leaks out at a constant rate. At the top of the ladder, only 10kg of water is left in the bucket. Find the work done by the beaver in this problem.

    Hint: Speed up the ladder is constant and writing an expression for mass of the bucket plus water as a function of height will help.


    2. Relevant equations
    (I believe):
    PE=mgh
    KE=(1/2)mv^2
    Integrals


    3. The attempt at a solution

    Well, I want to start with writing the expression and then integrating that, but I don't know what the expression should be... h=Mb+Mw seems too easy, or is this what they meant at first and then to do something with the PE equation and height?? Ack!
     
  2. jcsd
  3. Mar 3, 2010 #2

    ehild

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    First you need an expression for the mass of water as function of time.
    The mass is decreasing at a constant rate. Initially, it is 25kg. The final mass is 10 kg. The total time needed to climb up is T. t is the running time. Draw a plot first, and then write the expression for m(t).

    ehild
     
  4. Mar 3, 2010 #3
    Well, we don't know the time and the hint says to write mass as a function of height, so I went ahead and used the old school skills to get mass as a function of height:
    m(h)=-1.25h+30

    ...sooo, now should I be using integrals? Or am I taking the wrong approach?
     
  5. Mar 3, 2010 #4
    Okay, so integrating I get:
    12(30-0.655(12)) - 0 = 270 J

    Is this correct?
     
  6. Mar 3, 2010 #5

    ehild

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    I do not understand your old school skills. What is h at the top of the ladder? Will be the mass of water equal to 10 kg there?
     
  7. Mar 3, 2010 #6
    h at the top (at 12m) is 15 (10kg water plus 5kg bucket)
    h at the bottom (at 0m) is 30 (25kg water plus 5kg bucket)

    So I used these as coordinates (0,30) and (12,15) and have the equation
    m(h)=-1.25h+30
     
  8. Mar 3, 2010 #7

    ehild

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    So the heigh was given? You did not write it in the first post. Then the mass is all right. But you should integral the force with respect to h. As I see, you have integrated the mass. And I do not know either, where that 0.655 comes from.

    ehild
     
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