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Springy stuff

  1. Apr 20, 2006 #1
    g'day, any chance of some help with this spring question... can't seem to get my head around it.

    an ideal spring is hung vertically from the ceiling. when a 2kg mass hangs at rest from the spring it extends by 6cm. a downward external force is now applied to the mass to extend the spring an additional 10cm. while the spring is being extended by the force what is the work done by the spring?

    i think the original force is simply calculated form F=ma.... right?

    but do i then find the springs constant from F=kx ?

    and add the extension distances together for U=(1/2)kx^2 ??

    not to sure....

    thanks in advance :smile:
  2. jcsd
  3. Apr 20, 2006 #2
    In a lot of cases (nearly all...), work = energy.
  4. Apr 21, 2006 #3
    The work done on the spring equals the increase in its elastic potential energy. So obviously the work done by the spring (in this case on everything external to the spring, which is the mass + the external agent that exerts the additional force on the mass) equals the decrease in its elastic potential energy.
  5. Apr 21, 2006 #4


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    In which cases does work [itex]\neq[/itex] energy?

    Last edited: Apr 21, 2006
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