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Compressed vertical spring

  1. Jun 21, 2014 #1
    1. First let me start by saying that there are similar posts about this, but I wasn't able to figure out what I need through those.
    A 3.45 kg mass vertically compresses a spring 67.0 cm before it starts to rebound. How high will the Mass move above the uncompressed if the mass is left to “bounce” back up?




    2. Relevant equations
    mg=-kx
    k=-mg/x
    U=1/2kx^2
    E(top) = E(bottom)




    3. The attempt at a solution:
    To figure this out (or at least try to) I set the spring's neutral position as 0m and the distance from the spring's neutral up to the top of the rebound as 'd'. I also set 0.670m as 'a' to help work my way through the problem.

    Here's how I tried to work my way through:

    Energy at the bottom:
    K=0
    U(spring)=1/2k(a^2)
    U(g)=mg(-a)

    Energy at the top:
    K=0
    U(spring)=1/2k(d^2)
    U(g)=mgd

    Then I set the energy at the bottom equal to the energy at the top:
    1/2k(d^2)+mgd=1/2k(a^2)-mg(-a)

    Here is how I broke that equation down:

    Get rid of the 1/2:
    k(d^2)+2mgd=k(a^2)-2mg(-a)

    Get rid of the k by using k=-mg/a:
    d^2-2ad=a^2-2a^2

    Get everything on one side:
    d^2-2ad+a^2=0
    (d-a)(d-a)=0
    d=0.67

    The spring has to work against gravity on the way up, does it make sense that it goes the same distance up as it was compressed?
     
    Last edited: Jun 21, 2014
  2. jcsd
  3. Jun 21, 2014 #2

    haruspex

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    Not enough information.
    The initial condition could be describing a static state of affairs: the mass is sitting on the spring and holding it compressed 67 cm wrt its relaxed length.
    Or maybe the mass is placed carefully on top of the spring, while it is at its relaxed length, and then released. 67 cm is the max compression in the subsequent motion.
    Or maybe the mass was dropped onto the spring from some height?
     
  4. Jun 21, 2014 #3

    tms

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    The problem states that the mass compresses the spring. It doesn't say that anything else compresses it, so your third case is ruled out. Your second case is ruled out because the problem states that the spring starts compressed.
     
  5. Jun 22, 2014 #4

    haruspex

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    Not really. It says it compresses the spring. That could be a statement of circumstance ("is pressing") or a statement of an action - like, Sally kicks a ball.
    Indeed, "before it starts to rebound" suggests the action interpretation.
    In each of my first two interpretations, the problem appears to be trivial.
    That is indeed the answer given by my second interpretation:
    It follows immediately from conservation of energy.
    danyork, Are you quite sure this is the whole question?
     
  6. Jun 22, 2014 #5
    Yes, I did a copy/paste. I have a couple of friends who also didn't like the question. Are there any assumptions that could be made to finish the problem? What if we assume it's a SMH? (I'm new, so forgive me if I'm not helping the situation).
     
  7. Jun 22, 2014 #6

    haruspex

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    No, I think what you did originally is the best bet. Assuming SMH won't change anything.
    (On dimensionality grounds, you can figure out that the mass and the value of g are irrelevant. The only useful datum you have is the 67cm.)
    Note, though, that it asks for the height "above the uncompressed". So the answer is not 67cm.
     
  8. Jun 23, 2014 #7

    ehild

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    The potential energy at the bottom is mg(-a).

    If the spring is compressed just by the weight of the body, the net force on the body is -mg+kx=0. There is zero force, zero acceleration, no 'rebound'.



    ehild
     
    Last edited: Jun 23, 2014
  9. Jun 23, 2014 #8
    If you had the value of the spring constant you could tell if the given initial position is one of equilibrium or not. This is the missing information.

    It cannot be that the answer will be the same for ANY spring in the world, for a given mass and initial compression. Do you know the spring constant?
     
  10. Jun 23, 2014 #9
    I could get the spring constant by k=-mg/x. I divided k out of the energy equation though. What else would I use if for?
     
  11. Jun 24, 2014 #10
    To get the spring constant from that equation you assume that the initial position (when the spring is compressed by 67 cm) is one of equilibrium. So there is no rebound and no problem to solve.
    This was already pointed in previous posts.

    You only have a problem (to solve) if the initial position is not of equilibrium but one where the elastic force is larger than the weight of the mass. Or if the mass has some speed at that point. But then to say that "it starts to rebound" is not compatible with such initial condition.

    How do you "divide out" k?
     
  12. Jun 24, 2014 #11
    I used k=-mg/x (but I used 'a' for the distance instead of 'x', see my original post for the value of 'a').

    After I multiplied the formula to get rid of the 1/2 I was left with:

    k(d^2)+2mgd=k(a^2)-2mg(-a)

    Then, I divided out k, but used -mg/a in place of k which gave me:

    d^2-2ad=a^2-2a^2

    After that I moved the a^2 values to the left side of the equation and got:

    d^2-2ad+2a^2=0

    Aside from the fact that I'm missing information to complete the initial problem, does the way I manipulated the equation seem legit/make sense?

    Thanks,
    Dan
     
  13. Jun 24, 2014 #12
    The algebra does not seem right. The second term on the right hand side should be positive, the way you wrote it.

    But this is quite irrelevant for the problem. You need to know the physics before you can write the right mathematical equations. And the physics is incomplete. So trying to manipulate equations is sterile.

    Another missing piece is: is the mass attached to the spring or is just sitting on it?
    The way you wrote the equation, you assume that the spring is stretched in the final position. This may be the case if the mass is attached. The solution will depend on which situation is assumed.
    The vague formulation of the problem, I would tend to assume that is not attached.

    If you want to practice a problem with masses and springs, you can try one that is clearly defined rather than guessing about this one.
     
  14. Jun 25, 2014 #13
    Thanks for all your help.
     
  15. Jun 25, 2014 #14
    After further reflection, I think that it may be a way to interpret the wording of the problem in a way that make it workable.
    If we start with the spring initially uncompressed, place the mass on top of the spring and let is go.
    The mass will go down compressing the spring and it will stop when the compression is the one given. Then it starts to move back upwards (it rebounds). So the word "compresses" in the text should be taken as an active meaning, an action of compressing the spring to the extreme position rather than a static one (is in equilibrium, keeping the spring compressed).
    This scenario seems to be compatible with the wording of the problem. It is not the only one, as harsupex has mentioned. But from the three that he proposed (post 2), this is the only one that makes the problem determined.
     
  16. Jun 25, 2014 #15

    NascentOxygen

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    I'm confident that's the only interpretation that would be countenanced, were this to appear on an assignment in an engineering course. :wink: I suggest that OP run with this.
     
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