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Spring Constant

  1. Nov 18, 2007 #1
    A spring having constant of 2500N/m is compressed 6.0cm and then additional 2.cm. Then the compressed spring is used to propel a 0.25-kg ball vertically upward.

    a) The additional work done on the spring in compressing te final 2.0cm was ?
    b) the maximun heignt of the ball was



    part a) I know the formula for spring constant is Fs=-kx, and I know cm is not the right unit of measured it should be changed to m so it be .06m and .02m and thats a total of .08

    So for a it should be Fs= 2500N/M * .08m= 200

    now part b im a little confused
    I dont know what equationi should used,



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 18, 2007 #2

    Dick

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    Good job on putting the units in, so now use them. 2500(N/m)*0.08m=200N. N is not a unit of work. You've computed the force, not the work. With a non-constant force you can either integrate F*ds or in the case of spring use PE=(1/2)*k*x^2.
     
  4. Nov 18, 2007 #3
    Okay I see what you are saying I found the force instead of work?

    1/2*2500N/M*(.060)^2= 4.50J

    1/2*2500N/m*(.080)^2 -4.50J= 3.5J

    Now how do I find the max height, im confused about that part
     
  5. Nov 18, 2007 #4
    so for part b can I used the equation h=Kx^2/2mg

    2500N*M * (.080)^2 / 2*.25kg*9.87m/s = 19.74m
     
  6. Nov 18, 2007 #5

    Dick

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    Use conservation of energy. The potential energy that you put into the spring turns into the gravitational potential energy (do you know the formula for that?) of the ball at it's maximum height h.
     
  7. Nov 18, 2007 #6

    Dick

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    You beat me. Yes. The equation you've written down is actually PE(grav)=mgh=PE(spring)=k*x^2/2.
     
  8. Nov 18, 2007 #7
    ok, Yeah I just saw the equation in my notes.
     
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