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Shooting an Arrow

  1. Oct 6, 2004 #1
    Help pls! Here is the question:

    "Assuming you have a bow that behaves like a spring with a spring constant of 80 N/m and you pull it to a draw of 51 cm, to the nearest tenth of a m/s, what is the speed of the 81-gram arrow when it is released?"

    What equations do you use? I'be been trying W=.5kx^2, but don't know where to go from there. Pls. help me think through this! Thanks!

    Hughey :smile:
     
  2. jcsd
  3. Oct 6, 2004 #2

    arildno

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    Mechanical energy is conserved. Does that help you?
     
  4. Oct 6, 2004 #3
    No, not really. If you found the Force of the "draw-back" bow, could you figure out the speed of the arrow? Because it says when the arrow is released, does that mean you don't need to take into account the weight of the arrow?
     
  5. Oct 6, 2004 #4

    Pyrrhus

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    In the Conservartion of Mechanical Energy

    [tex] \Delta K + \Delta \Omega = 0 [/tex]

    The potentials energy not necessarily has to be both, it could be one, like Spring-Potential Energy [tex] \frac{1}{2} kx^2 [/tex], and/or Gravitational Potential Energy [tex] mgh [/tex]
     
  6. Oct 6, 2004 #5

    arildno

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    You are quite correct in assuming that in the general, non-horizontal firing of the arrow, a change in potential energy due to gravity will be present.
    However, do you think a change of potential gravitational energy of a distance over maximally 51 cm is significant compared to the potential energy stored in the drawn-back string?
    Disregard the the effect of gravity, it shouldn't matter.
    (That's one of the reasons why they haven't provided you with information concerning the shooting angle..)
     
  7. Oct 6, 2004 #6

    arildno

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    Alternatively, solve the problem under the EXPRESS condition that you are aiming horizontally.
     
  8. Oct 6, 2004 #7
    I'm still a little confused. What value do you have to find first? If you use Spring-Potential Energy equation, how would you find the speed of the arrow?
     
  9. Oct 6, 2004 #8

    Pyrrhus

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    Conservation of Mechanical Energy

    [tex] \Delta K + \Delta \Omega = 0 [/tex]

    At any points of the system, the mechanical energy will be the same. You can find speed with Kinetic energy.
     
    Last edited: Oct 6, 2004
  10. Oct 6, 2004 #9

    arildno

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    Initially, all you've got is potential energy; finally, all you've got is kinetic energy.
     
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