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Work done by us on the spring

  1. May 25, 2012 #1
    If you apply a force F and compress a spring of spring constant k by x distance what is the work done by us?
    I know energy stored is 1/2kx2 but is the work done by us the same?
    We are applying a cosntant force F and dispalcement is x so work done by us should be Fx right? And F = kx so work done by us is kx2

    Where does the other 1/2kx2 go?
     
  2. jcsd
  3. May 25, 2012 #2

    Doc Al

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    Staff: Mentor

    Imagine there was a mass on the end of the spring. If you used the minimum force required to compress the spring, then that force would vary from 0 (at the start) to the maximum value of kX at full compression. The work you'd do would equal 1/2kx2. But if you use a constant force F = kX to compress the spring, then the extra work that you do goes into increasing the kinetic energy of the mass.
     
  4. May 25, 2012 #3
    I agree with DocAl
    When you start to compress(or stretch) the spring you start with zero force and increase to final force,F and F = kx. The work done = average force x extension.
    If the spring behaves linearly the average force = F/2 therefore work done = kx/2 . x
    = 0.5kx^2
     
  5. May 25, 2012 #4
    Okay thank you - but one more doubt
    Suppsoe a mass m attached to a spring with charge q. Suddenly electric field is switched on so it will exert a constant force right? So in this case the other 1/kx^2 goes into kinetic energy and the spring oscillates? Have i got it right?
     
  6. May 26, 2012 #5

    Doc Al

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    Staff: Mentor

    That's right. The mass will oscillate about the new equilibrium point (where the net force on the mass is zero).
     
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