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Distance Spring Stretches

  1. Feb 12, 2009 #1

    TG3

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    1. The problem statement, all variables and given/known data
    A block of mass 2.5 kg rests on a frictionless floor. It is attached to a spring with a relaxed length L = 3 m. The spring has spring constant k = 16 N/m and is relaxed when hanging vertically. The block is in a relaxed position and is given a push and has an initial speed of 16.884 m/s. (Determined from previous problem.) How far does the block travel before it stops? (For this problem, the block never leaves the floor.)
    2. Relevant equations
    Work done by a spring = 1/2 KX^2
    K = 1/2 MV^2

    3. The attempt at a solution
    Initial Kinetic energy = 1/2 2.5 16.884^2
    Initial K =356.337
    356.337 = 1/2 16 X^2
    356.337 = 8 X^2
    44.54 = X^2
    6.674 = X
    Wrong.

    Is there some way to incorporate sine into this equation? I'm not sure how to do that, since the angle between the spring and the table is changing...
     
  2. jcsd
  3. Feb 12, 2009 #2

    Delphi51

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    I can't seem to picture this. Is the spring hanging vertically, the block moving horizontally?
    I'm surprised it ever stops since there is no friction.
     
  4. Feb 12, 2009 #3
    For Delphi51, the block will stop even without friction because of the force the spring will push back on the block. I am a bit confused myself with other parts of this problem, but maybe me telling you this first part will help you solve it.
     
  5. Feb 12, 2009 #4

    TG3

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    The spring is originally hanging vertically, the block moves horizontally. As the block moves horizontally, the spring is stretched and becomes a diagonal, and pulls against the block.

    --Edit--
    Wow, that was wierd. We posted at the same time. And I still don't get it...
     
  6. Feb 12, 2009 #5
    I have a feeling that this is a problem of energy conservation. Since you say the table is frictionless, energy will be conserved. Therefore, it is something along the lines of setting the potential energy of the system equal to the kinetic energy of the system and solving for the delta x.
     
  7. Feb 12, 2009 #6

    Delphi51

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    Oh - I think I understand. It will stop momentarily, then go backwards as the spring releases its energy. It is the momentary stop that interests us! Of course.

    Aren't we missing the angle? The bottom of the spring will be pushed sideways while the top is fixed, so the spring is at an angle. Its stretch will be different from the distance the block moves. A sine or cosine is needed.
     
  8. Feb 12, 2009 #7

    TG3

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    Perhaps it's just because I'm dense, but that doesn't help me at all.... or rather, I'm not sure what you're trying to tell me.
    Yes, it is an energy conservation problem. The original kinetic energy is 356.337 (shown above) and the final kinetic energy is 0 since the velocity is zero, but I don't know where to go from there...
    ---Edit---
    This is really wierd. As soon as I post I can see the previous post. Sorry bout that...
     
  9. Feb 12, 2009 #8

    TG3

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    Gah. I still don't know how to do this, because the angle is changing. I'm going to go do other homework and come back to this. If you know how to do this one please help...
     
  10. Feb 12, 2009 #9
    I think you can assume the spring will not be pushed at an angle. That would make this problem a lot harder than it should be for an intro physics course.
     
  11. Feb 12, 2009 #10

    TG3

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    Still not sure I follow... the spring definitely forms an angle with the table, there's a picture of that with the problem. (Sadly, no angle is provided.)
     
  12. Feb 12, 2009 #11
    Where is there a picture?
     
  13. Feb 12, 2009 #12

    TG3

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    https://tycho-s.phys.washington.edu/cgi/courses/shell/common/showme.pl?courses/phys114/winter09/homework/05c/mass_on_spring_WKE/4.gif [Broken]
     
    Last edited by a moderator: May 4, 2017
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