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Homework Help: Spring compression help

  1. Sep 10, 2010 #1
    1. The problem statement, all variables and given/known data
    A 0.40kg object connected to a light spring with a force constant of 19.6N/m oscillates on a frictionless horizontal surface. If the spring is compressed 4.0 cm and released from rest, determine (a) the maximum speed of the object.


    2. Relevant equations

    F = -kx

    3. The attempt at a solution

    Does this involve kinetic energy?
     
  2. jcsd
  3. Sep 10, 2010 #2
    Re: Springs!!

    Yes, you need to use the conservation of mechanical energy principle.
     
  4. Sep 10, 2010 #3
    Re: Springs!!

    so KE + KEs = 0

    1/2mv^2 + 1/2kx^2 = 0

    solve for v?
     
  5. Sep 10, 2010 #4
    Re: Springs!!

    Close, but the spring has no kinetic energy only the attached mass. The spring can store what type of energy? Also set it up so its all the energy at state 1 equal to all the energy at state 2 (the point of interest).
     
  6. Sep 10, 2010 #5
    Re: Springs!!

    sorry not kinetic energy but potential energy.
     
  7. Sep 10, 2010 #6
    Re: Springs!!

    Yes so you should arrive at:
    PEs = KE
    which would give the same result as if you used the work kinetic energy theorm:
    Ws = KE2-KE1
    where Ws is the work done by the spring on the block and is equal to the integral of the spring force with respect to x or:
    Ws = integral of Fdx = (1/2)kx^2
     
  8. Sep 11, 2010 #7
    Re: Springs!!

    so 1/2mv^2 = 1/2kx^2
    v = square root (kx^2/m)
     
  9. Sep 11, 2010 #8
    Re: Springs!!

    correct.
     
  10. Sep 11, 2010 #9
    Re: Springs!!

    k. i got the answer, the max speed is 0.28m/s.

    The next part of the question is, for what value of x does the speed equal one-half the maximum speed.

    What i did was to solve for x with v = 1/2(max speed)

    I got the answer wrong.

    Can someone point out where i went wrong?
     
  11. Sep 11, 2010 #10
    Re: Springs!!

    Again use the conservation of energy. Is your energy at state 1 (the compressed state) any different from part a? Then what forms of energy are present at state 2?
     
  12. Sep 11, 2010 #11
    Re: Springs!!

    at state 2, is it half the compressed state?
     
  13. Sep 11, 2010 #12
    Re: Springs!!

    No, that distance x is the unknown you are asked to solve for. At state 1 the only energy is potential energy which you know. At state two you know the kinetic energy of the mass because you know v=vmax/2, but you don't know the potential energy of the spring=(1/2)kx^2...
     
  14. Sep 11, 2010 #13
    Re: Springs!!

    So in state 1 there is potential energy, no kinetic energy. State 2 there's kinetic energy, where v is half of the max speed and there's potential energy and that's when we solve for x.
     
  15. Sep 11, 2010 #14
    Re: Springs!!

    correct.
     
  16. Sep 11, 2010 #15
    Re: Springs!!

    Thanks for you help! =)
     
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