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2 Things: Work-Energy theorem and parabolic trajectories.

  1. Oct 5, 2005 #1

    When reading through the work-energy thread, i just got reminded of something bugging me for quite a while. I don't think this is mentioned in the work-energy thread, firstly, how can we show that,

    [tex] \int f \bullet ds = \Delta E [/tex]


    I've heard that there's a way to show that when only one force acts on a particle or something, the particle follows a parabolic trajectory. Is there a way to show this?

  2. jcsd
  3. Oct 5, 2005 #2


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    If there is a constant (non-zero) force acting on something, it follows a parabolic trajectory. That's because, taking the x-axis perpendicular to the constant force, integrating a constant twice (from [itex]\frac{d^2y}{dt^2}= \frac{F}{m}[/itex] to [itex]\frac{dy}{dt}= \frac{F}{m}t+ v_0[/itex] to [itex]y= \fra{F}{2m}t^2+ v_0t+ y_0[/tex]) we get a quadratic in t while x, with 0 acceleration, is linear in t.

    (There's always "only one force" on something- the net force.)
  4. Oct 5, 2005 #3
    ohhhh ok, that helped alot, thanks..

    Anyone with the work question?
  5. Oct 5, 2005 #4

    Claude Bile

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    You can't derive the work formula, as it is a definition. You can, however justify why the definition is a good one with a little bit of logical reasoning.

  6. Oct 6, 2005 #5

    Doc Al

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    Staff: Mentor

    But you can certainly derive the Work-Energy theorem. Taking the simplest case, where the direction of the net force remains fixed:
    [tex] \int f \bullet ds = \int m a \bullet ds = m \int \frac{dv_s}{dt} ds = m \int v_s \ dv_s = \Delta (1/2 m v_s^2) [/tex]
  7. Oct 7, 2005 #6
    oh ok thanks, that's what i'm looking for.

    : )
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