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Anti Newton equation

  1. Nov 14, 2006 #1
    Energy (E) = Work
    E = Fs
    E = mas
    E = msv/t --------(because a =v/t)
    E = mvs/t
    But s/t = v
    .: E = mv2 :surprised -----------(1)
    But according to Isaac Newton E = mv2/2
  2. jcsd
  3. Nov 14, 2006 #2


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    This is what happens when you simply plug in things without understanding what they mean.

    Notice that when you use "a=v/t", you have already made an assumption that this is a VARYING "v", and that this is really some average value of a uniformly varying v. Yet, you used "s/t=v", which would only work for a CONSTANT v. Just think about it, if v is changing and getting larger when there's an acceleration, what value did you just compute using s/t = v? Which "v" is this when it is a changing value?

  4. Nov 14, 2006 #3
    But s/t = v

    What happened to the acceleration?

    Edit: Late as usual. Opened this thread and went to browse other places in the web. :D
  5. Nov 14, 2006 #4


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    Why does everyone want to pick on good old Newton these days? :smile:
  6. Nov 14, 2006 #5
    There are several problems here, all of which stem from ignoring calculus.

    The work energy theorem, when stated correctly says:

    [tex]E = \int \vec{F} \cdot d\vec{s}[/tex]

    This only takes the form you stated when [tex]\vec{F}[/tex] is constant and the path traveled is in the same direction [tex]\vec{F}[/tex] points.

    Replacing [tex]\vec{F}[/tex] by [tex]m\vec{a}[/tex] is only valid if what you're considering is the net force on the object, not one force among many. If it is the net force, this step is fine.

    However, the next step is not. Acceleration is the rate of change of velocity:

    [tex]\vec{a} = \frac{d\vec{v}}{dt}[/tex]

    Even if you only care about the average acceleration, you need the change in velocity. The only situation where you can write [tex]\vec{a} = \frac{\vec{v}}{t}[/tex] is when [tex]\vec{v} = 0[/tex] at time 0.

    The same sort of considerations apply when talking about velocity:

    [tex]\vec{v} = \frac{d\vec{s}}{dt}[/tex]

    But, now we have the worse problem that you've already assumed that [tex]\vec{v}[/tex] isn't constant; so there's no situation where you can just write [tex]\vec{v} = \frac{\vec{s}}{t}[/tex].

    Finally, Newton never said anything about energy, much less that [tex]E = \frac{1}{2} mv^2[/tex]. He believed that momentum told you everything you needed to know. It was Leibniz who first introduced the idea of energy (although he called is "vis viva").

    Here is the quickest way to get the form for kinetic energy out of the work energy theorem algebraically. I will need to assume that the force discussed is the net force on the object, that the net force is constant, and that the force acts along the direction of motion.

    [tex]\Delta E = F\Delta x[/tex]

    [tex]\Delta E = ma\Delta x[/tex]

    [tex]\Delta E = m \frac{\Delta v}{\Delta t} \Delta x[/tex]

    [tex]\Delta E = m \Delta v \frac{\Delta x}{\Delta t}[/tex]

    [tex]\Delta E = m (v_f - v_i) \overline{v}[/tex]

    [tex]\Delta E = m (v_f - v_i) \frac{v_f + v_i}{2}[/tex]

    [tex]\Delta E = \frac{1}{2} m (v_f^2 - v_i^2)[/tex]

    [tex]\Delta E = \frac{1}{2} m \Delta v^2[/tex]

    [tex]\Delta E = \Delta \left (\frac{1}{2} mv^2\right )[/tex]
  7. Nov 15, 2006 #6
    excellent parlyne...good proof
  8. Nov 15, 2006 #7


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    I don't know, but it is getting quite annoying. :grumpy:.
  9. Nov 15, 2006 #8


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    Dearly Missed

    It's a homophobic reaction, I guess.
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