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Potential energy

  1. Jul 14, 2004 #1
    This maybe an easy question for you, but for me im having trouble. Im working on 2D projectile motion and i need an equation to work out the inital velocity when i have the PE.
     
  2. jcsd
  3. Jul 14, 2004 #2
    you have the PE what other info do you have?
    jamie
     
  4. Jul 14, 2004 #3

    Gza

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    This question isn't really answerable with the limited information you've given, as jamie mentioned. You'll need the initial angle the object was fired at, since I worked out the problem using [tex]W = \Delta E [/tex] (work energy theorem) and came up with

    [tex]v_i = {\sqrt{\frac{2gh}{Sin^2\theta}}[/tex]

    where theta is the initial angle the object is fired at, and h is the max height of the trajectory.
     
    Last edited: Jul 14, 2004
  5. Jul 14, 2004 #4
    yea sorry i should have said, i have the projectiles weight, PE, initial angle. the equation you gave looks good, but surely i can only work out the maximum height after knowing the initial velocity?
    By the way im over here in England :biggrin: im not sure what levels of education are the same in england and america, but ill be going to university in 1 year and im 17 :surprise:
     
    Last edited: Jul 14, 2004
  6. Jul 14, 2004 #5
    Im in england too what uni are you at and where do you come from

    jamie
     
  7. Jul 14, 2004 #6

    NSX

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    PE at what point in time?

    i.e. MAX PE, 0 PE, etc...
     
  8. Jul 14, 2004 #7

    Gza

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    I'm sorry, I should have been more thourough in my explanation. The max height is contained within the PE. Just divide it by the mass of the object. And as NSX alluded to, you should be more specific about where relative to zero potential, your final PE is (I just assumed you were talking about the PE at the highest part of the path).
     
  9. Jul 15, 2004 #8
    sorry i should explain, its using a mangonel with a spring, i used the spring constant to work out the potential energy. So as you can see i was confused with the maximum trajectory height but thats my fault for not explaining. and im in Bath :P
     
  10. Jul 15, 2004 #9

    Gza

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    Not mentioning the spring makes this a completely different problem. You had said PE, which I assumed to be simply gravitational PE. But you neglected to mention the spring's PE. This is actually a much easier problem than the one I solved for you before, since a straight application of [tex]W = \Delta E [/tex] is required without any trigonometry. And from now on, please post all details of the problem to help avoid this kind of confusion.
     
    Last edited: Jul 15, 2004
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