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Vertically oriented spring and energy transfers

  1. Feb 20, 2007 #1
    I am posing this question due to a statement made by my TA:

    Suppose there is a spring pointed vertically upwards. Further, suppose that the spring is compressed a distance x so that it stores some potential energy. A mass is then placed on top of the spring. Assume all energy is conserved.

    Here's the question: upon releasing the spring, what is the kinetic energy of the ball as it passes the equilibrium position of the spring? Is it equal in magnitude to (1/2)kx^2 where x is the amount the spring is compressed, or is it (1/2)kx^2 - mgx?

    My TA said the former is correct, while I cannot figure out why it is not the latter. My reasoning: as the ball is rising from its initial height (the point at which the spring is fully compressed) some of that energy is being converted to gravitational potential energy. The rest goes to kinetic energy.

    Any insight will be greatly appreciated,
    -Zachary Lindsey
  2. jcsd
  3. Feb 20, 2007 #2

    Doc Al

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    Staff: Mentor

    Your reasoning is correct; you cannot ignore gravitational PE in calculating the KE of the mass.
  4. Feb 20, 2007 #3
    Thank you.
  5. Feb 20, 2007 #4
    Who is correct depends on whether x is the amount the spring is compressed from its free equilibrium length or that from its equilibrium length with the mass sitting on it. If it is the former, you are correct. If it's the later, your TA is correct.

    To see this, let's let y be the distance the mass is sitting above the spring's free equilibrium position. Then, the potential energy is:

    [tex]V(y) = \frac{1}{2} k y^2 + mgy[/tex]

    We can rearrange this:

    V(y) &= \frac{1}{2} k \left (y^2 + 2\frac{mg}{k} y\right ) \\
    &= \frac{1}{2} k \left (y^2 + 2\frac{mg}{k} y + \frac{m^2g^2}{k^2} \right ) - \frac{m^2g^2}{2k} \\
    &= \frac{1}{2} k \left (y + \frac{mg}{k} \right )^2 - \frac{m^2g^2}{2k}

    From this, we can see that the mass' height above the compressed equilibrium length is [tex]x = y + \frac{mg}{k}[/tex].

    So, [tex]V(x) = \frac{1}{2} k x^2 - \frac{m^2g^2}{2k}[/tex].

    The difference between the potential energy when the spring is compressed by a distance x from the compressed equilibrium position ([tex]y = \frac{-mg}{k}[/tex]) and at the compressed equilibrium is:

    [tex]\begin{align*} \Delta V &= V(x) - V(0) \\
    &= \frac{1}{2} k x^2 - \frac{m^2g^2}{2k} - \left (\frac{-m^2g^2}{2k} \right ) \\
    &= \frac{1}{2} k x^2
    \end{align*} [/tex]
  6. Feb 21, 2007 #5

    Doc Al

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    Staff: Mentor

    Parlyne is certainly correct. From the way the conditions were specified in the OP:
    I presumed that x measures the compression from the uncompressed position of the spring. Only later is a mass involved:
    The answer to that question depends on the meaning of "equilibrium postion" and where x is measured from. If, as I assumed from your description, x is measured from the uncompressed position of the spring (and equilibrium means x = 0), then 1/2kx^2 represents spring PE only and does not include gravitational PE.

    But, as Parlyne explained, if x is measured from the equilibrium position of the mass-spring system (which is not the uncompressed position of the spring), then 1/2kx^2 represents changes in both spring PE and gravitational PE together.

    CaptainZappo: Better tell us exactly what the TA stated, not your interpretation of it (if you can recall).

    Regardless of what the TA says, learn what Parlyne explained and you'll be ahead of the game. :wink:
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