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Kinetic and potential energy

  1. Mar 28, 2004 #1
    These “types of energy” are pure result of Newtonian physics.
    Their common element is the definition for the work done by constant force:
    [tex]W = dE = Fdx[/tex]
    If the force were variable then the definition would have looked like this:
    [tex]W = dE = Fdx + xdF <=> E = Fx[/tex]

    The kinetic energy comes out as a combination of this definition with Newton’s second flaw:
    [tex]W = dE = Fdx = madx = m \frac{dV}{dt}dx = mVdV[/tex] from where
    [tex]E_{kinetic} = \frac {mV^2}{2}[/tex]

    The potential energy comes out as combination of the same definition with Newton’s gravity law:
    [tex]W = dE = Fdx = \frac {GM_1M_2}{x^2}dx[/tex] from where
    [tex]E_{potential} = - \frac {GM_1M_2}{x}[/tex]

    What if I tell you that these Newton’s laws are flaw?

    Just like every thing else in Archimedes’s physics energy also has its potential expressed in rather different counter parts than Jules (force is geometrical potential but we don’t measure it in meters).
  2. jcsd
  3. Mar 28, 2004 #2


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    But work is not a differential. Work is defined as

  4. Mar 28, 2004 #3


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    You missed a factor of 2 there...
  5. Mar 28, 2004 #4
    Oh, dear
    you kind a confuse me a little.

    [tex]\int_{x_1}^{x_2} Fdx = F (x_2 - x_1) = E_2 - E_1 = \delta E = W[/tex]
    [tex]\int_{E_1}^{E_2} dE = E_2 - E_1 = \delta E = W[/tex]
    [tex]Fdx = dE[/tex]

    dE is approximatelly E_2 - E_1 when small enough!?!
  6. Mar 28, 2004 #5
    I think I didn't

    [tex](x^{-1})' = -1 x^{-2} dx[/tex]
  7. Mar 28, 2004 #6


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    Yes, but W != dE, [itex]W = \int_{x1}^{x2}{dE} [/itex]

    You're right about the integration. S'wat I get for trying to do calculus at 5am... :frown:
  8. Mar 28, 2004 #7
    You misspelled joules ! But still, if v is an rms phasor e=mv^2 ? naw i guess the 1/2 is still there it is just that i always divide by the potential so it goes away. you know ... 1/2 kx^2. Doesn't everyone use eigenvalues ? i find it so easy to forget the basis when using stuff like FEA or BEM. i just try to remember f=ma and f=kx which solves any problem according to ohms law.
    Last edited: Mar 28, 2004
  9. Mar 29, 2004 #8
    what's the point of this definition?

    You know, I’m rightfully confused here because if [tex]W = \int Fdx[/tex] then differentiated [tex]dW = Fdx[/tex]
    [tex]dW = Fdx => limes_{dx -> 0} dW = \delta W = W_2 – W_1[/tex]
    What could be that difference of the work done?
    Sure it measures in Joules but... is it also some work done or is it energy alone?

    After all let it be whatever and don’t let it drag us away from the subject. That’s how Newtonian physics comes up with kinetic and potential energy. I’m not inventing it. If you have better way of showing it please do. If your result is different from the one presented here then I have little to worry because you are actually on my revolutionary side.

    What if I convince you that these Newtonian laws are flaws?
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