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Work of spring

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data

    A spring has a relaxed length of 5 cm and a stiffness of 50 N/m. How much work must you do to change its length from 10 cm to 14 cm?

    2. Relevant equations

    W= -U = F(delta R) = (Ks)(s)

    3. The attempt at a solution

    -attempted solving this problem using the formula above
    -got a value of 0.08 J
    -value is apparently incorrect
    -not quite sure where I went wrong
  2. jcsd
  3. Oct 17, 2007 #2

    Doc Al

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    Staff: Mentor

    The spring force is not constant, so you can't just use W = Fx. You'll have to integrate the force over the distance.

    Alternatively, you can use the formula for the potential energy stored in a stretched spring.
  4. Oct 17, 2007 #3
    so is the formula (0.5) K s^2 ?

    K=stiffness = 50
    s= 14-10 = 4

    is this an accurate calculation?
  5. Oct 17, 2007 #4

    Doc Al

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    Staff: Mentor

    The formula is correct but your use of it is not. In the formula, s stands for the amount of stretch beyond the unstretched length.

    When the spring has length 10, how much is it stretched?
    When the spring has length 14, how much is it stretched?

    Compare the energy at each of those positions.
  6. Oct 17, 2007 #5
    oh ok... I think I understand

    so essentially, I should use that formula separately to find work at 10cm (s=5) and 14cm(s=9)

    and then find the difference of the too work values?
  7. Oct 18, 2007 #6

    Doc Al

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    Staff: Mentor

    That's correct.
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