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Work with spring force

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data
    A spring with spring constant of 29 N/m is
    stretched 0.21mfromits equilibrium position.
    How much work must be done to stretch it
    an additional 0.14 m? Answer in units of J.


    2. Relevant equations
    W=-kx(x)/2


    3. The attempt at a solution
    I solved it and got 0.4263 and it was wrong. does it need to be negative?
     
  2. jcsd
  3. Oct 20, 2008 #2

    LowlyPion

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    How did you arrive at that number?
     
  4. Oct 20, 2008 #3
    i did ((29)(0.21)(0.14))/2 =0.4263. does it need to be negative?
     
  5. Oct 20, 2008 #4

    LowlyPion

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    That's not quite right.

    Additional work = Difference in work = 29*(.21+14)2/2 - 29*(.21)2/2
     
  6. Oct 20, 2008 #5
    Ok so that means it would equal
    29*(.21+14)2/2 - 29*(.21)2/2 =1.1368

    Thanks
     
  7. Oct 20, 2008 #6

    LowlyPion

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    That's what I got anyway. What you need to remember is that work is the area under the force and distance curve over the range of the curve.

    Cheers
     
  8. Oct 20, 2008 #7
    i will remember thanks
     
  9. Oct 21, 2008 #8
    wait...

    wait wait... why is it 14 instead of .14?
     
  10. Oct 21, 2008 #9

    LowlyPion

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    He got the right answer and the 14 was apparently a typo. Calculate it out and you will see the inner sum needs to sum to .35.
     
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