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Another spring question

  1. Mar 15, 2005 #1
    Work of 5 Joules is done in stretching a spring from its natural length to 13 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (13 cm)?


    here's what i done:

    W = F * D
    5 = F * 13
    F = 5/13
    F = kx
    5/13 = k(13)
    k = 5/39

    [tex] \int_{0}^{13} 5/39 *x dx [/tex]

    i'm really confuse about this problem
     
  2. jcsd
  3. Mar 15, 2005 #2

    learningphysics

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    Remember that the force depends on x. So using W=F*D does not work since the force is not constant. You need to use the integral.

    Use:
    [tex] W = \int_{0}^{13} k *x dx [/tex]
    to figure out k.

    Then get the force.
     
  4. Mar 15, 2005 #3
    F = kx

    from this the only thing i have is X, how would i found out k?
     
  5. Mar 15, 2005 #4

    learningphysics

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    To get k, use the integral:
    [tex] W = \int_{0}^{13} k *x dx [/tex]

    Solve the integral on the right side. What do you get?
    Then substitude W=5 on the left side.

    Now you should be able to get k.
     
  6. Mar 15, 2005 #5
    5=169/2*k
    k = 10/169

    would i have to setup another integral?

    [tex] \int_{0}^{13} 10/169 *x dx [/tex]

    i get 5 if i solve that integral
     
  7. Mar 15, 2005 #6

    learningphysics

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    I'm sorry. You need to convert to meters.

    So
    [tex] W = \int_{0}^{0.13} k *x dx [/tex]

    I get k=591.7 N/m

    Then F=kx=591.7* 0.13=76.92N
     
  8. Mar 15, 2005 #7
    thanks alot!
     
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