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K in Hookes Law vs. Spring Potential Energy

  1. Mar 7, 2006 #1
    Today in physics class we were discussing the conservation of energy using a ball on a spring as the example. When the instructor completed the problem one of the students stated that the value of K (spring constant) that we found was different (by a factor of 2) than the value of K calculated using Hooke's Law. The instructor could not figure out why the values of K were not the same. Why are they different?

    Potential Energy(before)=Potential Energy (after)
    (1/2)kx^2+mgh = (1/2)kx^2+mgh
    ball on spring at rest = ball on spring extended

    ---------- = k is not equal to k= F/delta h
  2. jcsd
  3. Mar 7, 2006 #2


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    You can equate the sum of these Potential Energies only if KE = 0.
    This will occur at the bottom of the bounce and at the top of the bounce.
    [tex]\frac{1}{2} k x_{bottom}^2 + m g h_{bottom} = \frac{1}{2} k x_{top}^2 + m g h_{top}[/tex]
    suppose the spring was unstretched at the top, so x_top = 0 ;
    suppose we measure height from the bottom , so h_bottom = 0 .
    Then : (1/2) k (x_bottom)^2 = m g h_top
    => k = 2 m g h_top / (x_bottom)^2 = 2 m g / x_bottom .

    The adjectives "top" and "bottom" are important here. The ball was moving
    - in particular, the acceleration was NOT = 0 in either location.

    The Force by the spring when the ball was at the bottom was
    F = - k s = - [2 m g / x_bottom] x_bottom = - 2 m g (upward).

    The Force by the spring when the ball was at the top was zero.
    The average Force by spring during this motion was F_average = - m g .

    The form to remember is F = - k s , which will remind you that
    s is the stetch of the spring from its "relaxed" length.
    (not to be confused with horizontal coordinate "x" , nor with height)
  4. Mar 8, 2006 #3


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    There are two versions of Hooke's law:
    [tex]T = \frac{kx}{L}[/tex]

    [tex]F = kx[/tex]
    Each equation will yield a different value of k for a given tension or (after integration) potential energy.
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