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Momentum is Potential Energy?

  1. Jan 3, 2005 #1
    the derivative of potential energy (inrespects to time) is Force right?

    F=-dPE/dt

    I also read that the derivative of momentum (inrespects to time) is force.

    F=dp/dt

    Can I conclude that momentum is Potential Energy?

    p=-PE
     
  2. jcsd
  3. Jan 3, 2005 #2

    Andrew Mason

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    No. This is incorrect. Force is the space derivative of Work or Energy.: F = dW/dx
    This is more often seen as Force x distance = work or: dW=Fdx

    A good example of this is gravity: PE = mgh. dPE/dh = mg which of couse is the force due to gravity.

    dPE/dt = dW/dt = Power or the time rate at which work is done


    AM
     
    Last edited: Jan 3, 2005
  4. Jan 3, 2005 #3
    Okay, let me restate then...

    the integral of force (inrespects to time) is Impulse right?

    ∫F=Impulse=p(t)

    the derivative of Potential Energy (inrespects to distance) is Force right?

    PE'(x)=F(x)


    is this correct?
     
    Last edited: Jan 3, 2005
  5. Jan 3, 2005 #4
    Is it true that work in respects to time is the same as the negative Potential Energy?

    w(t)=-PE(t)

    Is it true that work in respects to distance is the same as the negative Potential Energy?

    w(x)=-PE(x)
     
  6. Jan 3, 2005 #5

    dextercioby

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    I believe that Andrew was more than clear.The force is the gradient of the potential energy,which involves DERIVATIVES WRT TO SPACE COORDINATES AND NOT WRT TO TIME...So there is no simple connection between momentum and potential energy.To give a clear and hopefully comprehedable example:an apple in a tree has momentum zero (it's speed is zero,as long as the wind's not blowing),but has a potential energy 'mgh',where 'm' is its mass,'g' is the acceleration due to gravity and 'h' is the height above the ground assumed the level with 0 gravitational potential energy.

    Daniel.
     
  7. Jan 3, 2005 #6
    Yeah, I sorta realized that. It's changed
     
  8. Jan 3, 2005 #7

    dextercioby

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    What do you mean "work in respects to distance".Work is a physical quantity and mathematically it is function which can depend both on the coordinate and time.You might have meant:
    [tex] W(x,t)=-E_{pot} (x,t) [/tex] (1)
    Compare the definitions and tell me whether (1) is correct:
    [tex] W(x,t)=\int F(x,t) dx [/tex] (2)
    [tex] F(x,t)=-\frac{dE_{pot}(x,t)}{dx} [/tex](3)

    I guess it is.But it's not fair to break the functional dependence and state "work in respects to distance".

    Daniel.
     
    Last edited: Jan 3, 2005
  9. Jan 3, 2005 #8
    I know the integral of force in respects to distance is w(x).
    I know that the integral power in respects to time is w(t).

    I assumed that w(x)=delta KE(x) = -PE(x)
    and that w(t)=delta KE(t) = -PE(t)
     
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