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Question: conservation of E and Newton's second law

  1. Nov 28, 2005 #1
    I was wondering if anyone had some input on how, via calculus, to show how the law of conservation of mechanical energy for an object in free fall (ideal) is a direct consequence of the of newtons second law.
     
  2. jcsd
  3. Nov 29, 2005 #2

    Stingray

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    The energy of a particle moving in 1 dimension through a potential field V(x) is
    [tex]E=\frac{1}{2} m ( dx/dt )^{2} + V(x)[/tex]
    Then the rate of change of the energy is
    [tex]dE/dt = m ( dx/dt ) ( d^{2} x/dt^{2}) + dV/dt[/tex]

    Use Newton's second law on the first term to give
    [tex]dE/dt = F ( dx/dt ) + dV/dt [/tex]

    But by definition,
    [tex] F= - dV/dx = - (dV/dt) / (dx/dt) [/tex]

    Substituting this into the previous expression gives dE/dt=0, as expected.
     
  4. Nov 29, 2005 #3
    Thanks! but what about 2-dimensional case?
     
  5. Nov 29, 2005 #4
    Just substitute the correct kinetic energy into the problem. If you have rotation you will have to look the torque. And remember that:

    [tex] \mathbf{F}=-\mathbf{\nabla}V(\mathbf{r}) [/tex]
     
  6. Nov 29, 2005 #5
    Sorry but I am still confused, is there a way to express the relationship algebraically?
     
  7. Nov 29, 2005 #6

    Stingray

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    What do you mean?
     
  8. Nov 29, 2005 #7
    I am having a probelm understanding the formulas
     
  9. Nov 29, 2005 #8
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