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Spring Problem and average force

  1. Dec 3, 2007 #1
    1. The problem statement, all variables and given/known data
    [​IMG]


    2. Relevant equations
    [tex]E_p = \frac{Fx}{2}[/tex]
    [tex]E_k = \frac{mv^2}{2}[/tex]

    3. The attempt at a solution
    [​IMG]

    I get 2.22 m/s, but the answer key says 3.1 m/s. Where did I go wrong?
     
  2. jcsd
  3. Dec 3, 2007 #2

    PhanthomJay

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    The average force is already given as 5.7N. You mistakenly divided it again by 2. E_p = F_avg(x)
     
  4. Dec 3, 2007 #3
    Check your value for Force. The problem gives average force...this is not equal to the actual force when the spring is compressed 1.3 cm.
     
  5. Dec 4, 2007 #4
    So Phanthom, you're saying that [tex]E_p = F_{ave}x[/tex] is an actual formula?
     
  6. Dec 4, 2007 #5

    PhanthomJay

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    The elastic potential energy stored in a spring is equal in magnitude to the work done by it when moving from its compressed distance, x, to its original uncompresed length, that is, [tex]E_p = \int Fdx[/tex], from the definition of work, where F, from Hookes Law, is [tex] F =kx [/tex]. Performing the calculus, you get the well known equation for the elastic potential energy of the spring, [tex]E_p =1/2kx^2[/tex]. Noting that [tex]F=kx[/tex] at its max compressed point, this is identically equivalent to [tex]E_p = F/2 (x) [/tex], or [tex]E_p = F_{avg} x[/tex].
    In your problem, the average force is given as 5.7N. This means the force in the spring is 0 in its uncompressed length, and 11.4N when it is compressed 1.3cm.
     
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