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Springs, energy conservation

  1. Nov 11, 2011 #1
    This is just a concept question.

    Say you have a spring arranged vertically with an object on the end of it. The spring is compressed.. At some point you, the system is released from compression.

    When calculating the velocity at which the object loses contact, I understand that you use the conservation of energy.


    Eo =Ef ---> But do you have a final PE? Or is it all kinetic energy? Since it is returning to equilibrium position (where the compression x is zero), I would think that there is no mgh, but then again it has gained displacement in the vertical direction.

    0.5KX2 = 0.5mv2

    OR

    0.5KX2 = 0.5mv2 + mgh?
     
  2. jcsd
  3. Nov 11, 2011 #2

    BruceW

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    Homework Helper

    This isn't a homework question! wrong forum, surely?
     
  4. Nov 11, 2011 #3

    gneill

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    Staff: Mentor

    Any change in height in the gravitational field implies a change in potential energy. If the spring was compressed below its equilibrium point, and you've set your origin for PE at the equilibrium level, then it had a negative potential energy to 'recover' as it travels back to the equilibrium point (assuming your taking PE = mgh, with Δh being positive with increasing height).

    The salient fact about the equilibrium point, with regards to losing contact with the projectile, is that it is where the acceleration of the spring changes sign. If the projectile is not affixed to the spring end, then it is not compelled to decelerate along with the spring end and it continues with its current velocity (inertial motion).
     
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