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Simple harmonic motion problem

  1. Apr 19, 2005 #1
    I have been looking at this problem forever. It is probably easy, but I am making it harder. States: In preparation for shooting a ball in a pinball machine, a spring (k=675 N/m) is compressed by 0.0650m relative to its unstrained length. The ball ( m=0.0585kg) is at rest against the spring at point A. When the spring is released, the ball slides (w/o rolling) to point B, which is 0.300m higher then point A. How fast is the ball moving in B?

    you want to find the vf (velocity final), I think so I set up the equation as 1/2vf2 + 1/2 kxf2= 1/2 kxo2 (o=initial, f=final)
    vf=sqrt km(xo2-xf2) which I got an outrageous number for. Please help I am soo lost
     
  2. jcsd
  3. Apr 19, 2005 #2

    quasar987

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    Do it using conservation of energy.

    1) Calculate the potential energy stored in the spring when strechted. Once released, all that energy becomes kinetic.

    2) Calculate what energy the ball has converted in gravitational potential once it reaches B.

    3) The remains of the energy is still kinetic. Find the velocity associated to that amount of kinetic energy.
     
  4. Apr 19, 2005 #3
    So you would set up the equation as E= 1/2MV2 + mgh + 1/2kx2 right
     
  5. Apr 19, 2005 #4

    quasar987

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    Yeah, you rock! Then use the fact that [itex]E_f = E_0[/itex] to recover [itex]v_f[/itex].
     
  6. Apr 19, 2005 #5
    what do u mean to recover vf
     
  7. Apr 19, 2005 #6
    so 1/2(.058kg)(0 m/s) + 1/2(.058kg)(9.80 m/s2)(h?) + 1/2(675 N/m)(x?)2
    so xo=.0650 and then xf is .300 m higher so xf=.3650m right
     
  8. Apr 19, 2005 #7

    quasar987

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    lol, what? Be consistent with your notation. You got 2 values of xf there.
    What you gotta do is set Ef = Eo, plug all the values, solve for vf.
     
  9. Apr 19, 2005 #8
    no the problem says that point B is .300m higher then a so I thinmk that would mean that xf would = .3650m bc u would add the beginning xo value plus the .300m value. Also I dont know what h would be and xfor the equation
     
  10. Apr 19, 2005 #9
    so we would use 1/2mvf2 + 1/2 kxf2= 1/2mvo2 +1/2 kvo2 from the equation Ef =Eo
     
  11. Apr 19, 2005 #10

    quasar987

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    h is 0.300 and the Xo of the spring's pot.energy is 0.0650m.

    The problem is 2 dimensional. When they say point B is 0.300m higher, it means higher in the sense of 'h'.
     
  12. Apr 19, 2005 #11

    quasar987

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    No you're missing the gravitationa potential energy now. You had it right in your earlier post.
     
  13. Apr 19, 2005 #12
    /2mvf2 + mgho +1/2 kxf2= 1/2mvo2 + mghf +1/2 kvo2 right
     
  14. Apr 20, 2005 #13

    quasar987

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    Ef = Eo means (all final values) = (all initial values). Not a mix of both. Keep it to

    1/2mvf2 + mghf +1/2 kxf2= 1/2mvo2 + mgho +1/2 kxo2
     
  15. Apr 20, 2005 #14
    opps had it right on paper but put it in the wrong area on here. So would xf=.650 +.300, that is the only variable Im trying to find, other then the final speed
     
  16. Apr 20, 2005 #15

    quasar987

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    Xf is 0. It represent the lenght by which the spring is stretched! As soon as the ball leaves the spring x is 0.
     
  17. Apr 20, 2005 #16
    so hf=.300, ho= .0650 ??
     
  18. Apr 20, 2005 #17

    quasar987

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    ho = 0

    xo = 0.0650
     
  19. Apr 20, 2005 #18
    so we plug in the numbers and solve for vf right
     
  20. Apr 20, 2005 #19

    quasar987

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    right........
     
  21. Apr 20, 2005 #20
    that will then give us the speed at B, god you have been such a help thank you
     
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