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Rock on Vertical Spring

  1. Nov 9, 2006 #1
    An 8.00 kg stone rests on a spring. The spring is compressed 7.0 cm by the stone.

    (a) What is the spring constant?
    11.2 N/cm
    (b) The stone is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release?
    76.7 J
    (c) What is the change in the gravitational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height?
    76.7 J
    (d) What is that maximum height, measured from the release point?

    a, b, and c are correct. part d just confuses me.
    I said:
    [tex]mgh_f - mgh_i = \Delta U_G=76.7J[/tex] (as per part c)

    so [tex]mgh_f = 76.7J + mg(-0.37m)[/tex]

    so [tex]h_f=\frac{76.7-(8)(9.81)(0.37)}{(8)(9.81)}=0.608[/tex]

    Lovely WebAssign puts next to this a lovely red x. Somehow, that makes me wonder if perhaps my answer to part (d) is wrong...
  2. jcsd
  3. Nov 9, 2006 #2
    as you said
    (a) 11.2 N/cm or 1120 N/m correct
    (b) 76.7 J correct
    (c) 76.7 J correct
    (d) I get something different
    It says "measured from the release point", NOT from equilibrium point as you assumed. So you simply made a silly mistake, nothing big.
  4. Nov 9, 2006 #3


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    The gravitational potential energy change of 76.7 joules is in reference to the release point where the initial GPE is established as 0. So mgh_i = 0.
    h = 76.7/mg as measured from the release point
  5. Nov 10, 2006 #4
    But how do I calculate the release point? I have no idea how far the spring can stretch!
  6. Nov 10, 2006 #5
    Your equation implies that Ug = 0 @ 37 cm above the release point. Given the conditions of the problem, that is the equilibrium point of the spring. So the final height you get is the height above the equilibrium point, that's not what you want.

    If you want the height above the release point (the point where the spring is initial set into motion, after being compressed another 30 cm), then set Ug = 0 at that point. Then mgh_i = 0.

    This is basically what PhantomJay is telling you, and I think he is definitely correct.

  7. Nov 10, 2006 #6


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    The spring is not going to stretch. It is assumed massless, so it has no momentum once the stone is gone. The stone is not connected to the spring. The release point is where the stone is when the spring is at the maximum stated compression.
  8. Nov 10, 2006 #7
    By release point, the problem means the point at which you let go of the rock. So right where the stone is before it gets launched, when it is still stationary.
    Last edited: Nov 10, 2006
  9. Nov 10, 2006 #8
    Oh! I assumed that the release point was the point where the rock no longer touched the spring; ie where the spring was stretched to its maximum (so the spring would start moving toward earth while the rock continued moving upward).

    Thank you, all!
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