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Prove energy of a falling body is constant

  1. Jun 1, 2014 #1
    1. The problem statement, all variables and given/known data

    Prove that the enrgy of a falling body remains constat using the derivate of a function

    2. Relevant equations

    We need to prove that:

    [tex] \frac{mv^2}{2} + mgh [/tex] is constant, trhat is its derivate equals 0

    3. The attempt at a solution

    Let the positive direction be downward. Applying Newton's second law we get:

    [tex] mg = F = ma [/tex]

    Now I differentiated the expression and I got:

    [tex] Fv + \frac{dh}{dt}mg [/tex]

    If I replace mg with F, I get:

    [tex] F(v+\frac{dh}{dt}) [/tex]

    Now, how do I show that this equals 0? I don'y get it. I know that probably some elemntary fact is escaping me, but what? Could you please explain me? Thank you very much!
     
  2. jcsd
  3. Jun 1, 2014 #2

    TSny

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    Hello.

    How is dh/dt related to v?
     
  4. Jun 1, 2014 #3
    I think I've found the answer. If dx is the change in the position of the body, then dx=-dh, so dh/dt=-dx/dt=-v, right? Another way I thought about it would be. Let A be a point such that h=AB, where B it's at ground level. Let O an arbitrary point in space, that it's on the direction of AB. Then we have that OA+ AB is constat, that is x+h is constant, and this means that d(x+h)/dt=0 . I believe that both arguments are solid and correct, but could you please give me your opinion?
     
  5. Jun 1, 2014 #4

    TSny

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    Yes. That looks good.
     
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