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Springs, maximum extension.

  1. Jan 1, 2013 #1
    1. The problem statement, all variables and given/known data
    A perfectly elastic spring is attached to the ceiling and a mass m is hanging from the spring. he mass is in equilibrium when the spring is stretched a distance x(o). The mass is carefully lifted and held at rest in the position where the string is nether stretched nor compressed at all. The mass is then allowed to drop from rest. What is the maximum extension of the spring from its unstretched length?


    2. Relevant equations
    1/2 kx^2 perhaps...


    3. The attempt at a solution
    I know that the spring stretches more than x(o)... but cannot quite get myself together to prove it. 0.5kx^2=mgX, but that's by the time the spring stopped oscillating.. what's the length at 'MAX'imum? help me guys
     
  2. jcsd
  3. Jan 1, 2013 #2

    gneill

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    Staff: Mentor

    Maximum extension occurs at the instant the mass stops descending and is poised to return upwards, pulled up by the spring. What can you say about all the various forms of energy involved (KE, PEspring PEgrav) at the top point and the bottom point of the trajectory? Choose a suitable location to place your zero references and write the energy balance equation.

    Hint: You'd do well to first determine xo in terms of m,g,and k.
     
  4. Jan 1, 2013 #3
    gneill- i took your advice and first saw that the only difference between the motions of the situation when the mass is in equilibrium and the situation of the maximum extension is that for the situation where there is maximum extension, there exists an overall acceleration due upside.
    thus I got for the situation A which is in equilibrium: mg-kx(0)=0 thus x(0)=mg/k
    and for situation B - the maximum extension: ma=kx-mg thus x=(ma+mg)/k...
    but perhaps a is g directed up because of action reaction (or perhaps not action reaction. correct me if i'm wrong). <<is this process right? I think it's wrong!?!
    thus mg=kx-mg... so x=2mg/k thus 2x(0)?
     
  5. Jan 1, 2013 #4
    ohhh nvm i got it :D
    since ma(X)=0.5kx^2 so a=kx/2m,
    ma=kx-mg=kx/2 thus kx/2=mg
    thus x=2mg/k thus 2x(0). But would I be able to think of all this proof in like a minute or so, which this competition yields per question? =.=
     
  6. Jan 1, 2013 #5

    gneill

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    Staff: Mentor

    Your expression for xo looks fine. Keep it handy for later.

    For the bottom position, you don't know what the net force (or acceleration) is at this point, thus writing a force equation there is problematical. You'd be better off working with conservation of energy. What's the KE at the top and bottom? What are the PEs?

    Other hints: If you set your zero reference for gravitational PE at the top, it happily coincides with where the stored energy in the spring is zero. Write the spring extension at the bottom as xo + x, where x is the further offset from xo that the mass reaches.
     
  7. Jan 1, 2013 #6

    gneill

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    Staff: Mentor

    Okay, that works too :smile:

    Regarding time to solution, the only answer is to practice until the methods become automatic and you can quickly recognize profitable strategies.
     
  8. Jan 1, 2013 #7
    yeahy i did the energy conservation way as you told me and i still got it right :)
    thanks gneill!!
     
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