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Waterfall Question Work and Energy

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data

    As the water in a river approaches a 10.0m vertical drop, its average speed is x m/s. As that water falls, its kinetic energy increases. The speed of the water based on 1kg of mass is 14.9m/s just before it hits the ground. Find the average speed of the river water before it goes over the edge.




    3. The attempt at a solution

    i used Eg=Ek
    mgh=1/2mv^2
    2gh=v^2
    2(9.81)(10.0)=v^2
    196.2=v^2 (square root not to isolate for velocity)
    14.0m/s = v

    the answer was about 5.0m/s im just not sure how to get to that.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 13, 2013 #2

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    Hello Arythmatic,

    Welcome to Physics Forums!

    Your approach would be valid if the problem asked you to calculate the final speed of the water, and if the water was at rest when reaching the beginning of the vertical drop.

    But that's not the case.
    • The final velocity of the water, after the vertical drop, is given to you in the problem statement: 14.9m/s. You don't need to solve for this.
    • Just before the vertical drop, the water is not at rest. It is moving; it has kinetic energy.

    If P.E. is potential energy and K.E. is kinetic energy, use conservation of mechanical energy:
    [tex] P.E._{\mathrm{before}} \ + \ K.E._{\mathrm{before}} \ = \ P.E._{\mathrm{after}} \ + \ K.E._{\mathrm{after}} [/tex]

    (Hint: the K.E.before is the one you are looking for. :wink:)
     
  4. Nov 13, 2013 #3

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to PF Arythmatic

    You have to be a little bit more careful with your expression for the conservation of energy. It's:

    (total initial energy) = (total final energy)

    [1] Eki + Egi = Ekf + Egf

    (where i means initial and f means final). Your mistake was in assuming that the initial kinetic energy was zero. It wasn't. You can also rearrange this formula to get:

    [2] Egi - Egf = Ekf - Eki

    which says:

    -ΔEg = ΔEk

    (change in kinetic energy) + (change in potential energy) = 0

    This is just another way of expressing that energy is conserved. From equation [2], since the final potential energy is zero at the bottom, we have:

    mgh - 0 = (1/2)mvf2 - (1/2)mvi2

    In other words, what you have computed in your original post (mgh) is the change in kinetic energy, not the initial kinetic energy.
     
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