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Homework Help: Grade 12 Physics-Finding the Spring Constant

  1. Jan 19, 2014 #1
    1. The problem statement, all variables and given/known data
    This was a mini-lab and the task was to fire a rubber band by stretching it a known distance, shooting it straight horizontally from a known height. Based on where the rubber band lands determine the elastic coefficient. Known data from experiment: mass=0.30g, height=0.412m, distance flung=1.362m, stretched length (Δx)=0.047m.


    2. Relevant equations
    d=vt
    Ee=1/2k(Δx)^2
    Eg=mgh
    Ek=1/2m(v)^2

    3. The attempt at a solution
    We tried to find the time with the vertical component
    d=Vit+1/2at^2, acceleration is 9.8m/s^2
    we got t=0.28997s

    Then we used the horizontal component to get the velocity
    d=vt
    v=4.697m/s (initial velocity)

    then we found the final velocity using a=(Vf-Vi)/(t)
    we got Vf=7.53m/s (right before the rubber band hits the ground)

    Then we thought that Eg+Ee (lost)=Ek (gained)
    We solved for k, but got 6.62, which is not the right value because it is too small. There must be a problem with our method. We think maybe it has something to do with the acceleration formula used to find the final velocity. We aren't sure what we did wrong but let us know if you have any corrections or any other methods to approach this question with.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 19, 2014 #2
    Good so far.

    Why would you need the final velocity? The contribution of the rubber band ends when the initial velocity is obtained.
     
  4. Jan 19, 2014 #3
    We thought we needed a final velocity to put into the Ek formula. Do we just use Vi for the v variable in Ek=1/2m(v)^2.
     
    Last edited: Jan 19, 2014
  5. Jan 19, 2014 #4
    That formula is generally correct. For the initial velocity, it gives you the initial kinetic energy.
     
  6. Jan 19, 2014 #5
    So what would you do after finding the initial velocity in the d=vt formula?
     
  7. Jan 19, 2014 #6
    Where does the initial kinetic energy come from?
     
  8. Jan 19, 2014 #7
    The elastic potential plus the gravitational potential energy is converted into kinetic energy.
     
  9. Jan 19, 2014 #8
    Gravitational potential energy is converted into the initial kinetic energy? How?
     
  10. Jan 19, 2014 #9
    We flung the rubber band from a height of 0.412m, so it initially had Eg.
     
  11. Jan 19, 2014 #10
    So at the beginning it had Eg and Ee and then when the rubber band landed 1.3m away, it had all converted to Ek, we thought.
     
  12. Jan 19, 2014 #11
    I thought we sorted out the (ir)relevance of "landing" in posts #1 - #3.
     
  13. Jan 19, 2014 #12
    So is it Ee=Ek then ?
     
  14. Jan 19, 2014 #13
    I am not sure what those symbols mean. Is Ee the elastic potential energy? Then it is correct.
     
  15. Jan 19, 2014 #14
    yes Ee is elastic potential
     
  16. Jan 19, 2014 #15
    Why dont you consider the fact that there is gravitational potential and that the elastic travels a certain disance?
     
  17. Jan 19, 2014 #16
    Again: what relevance does the gravity have on the work of the rubber band?
     
  18. Jan 19, 2014 #17
    Okay, I see what you're saying now but when I use Ee=Ek and sove for the spring constant, I only get 1.9.
     
  19. Jan 19, 2014 #18
    I get about 30 N/m from your data. Show your working in details.
     
  20. Jan 19, 2014 #19
    are you converting the mass correctly? Im still getting a different answer
     
  21. Jan 19, 2014 #20
    I can offer no further help unless I see exactly what you do.
     
  22. Jan 19, 2014 #21
    Ee=Ek
    (1/2)kx^2= (1/2)mv^2
    (1/2)k(0.047)^2=(1/2)(3.0 x 10^-4)(4.967)^2
    0.0011045k= 0.00309271
    k= 0.00309271/0.0011045
    k= 2.8
     
  23. Jan 19, 2014 #22
    (1/2)(3.0 x 10^-4)(4.967)^2 is not 0.00309271 but a little more.

    Anyway, the result is close to 3 N/m which I am getting now. I think I did make a mistake in mass conversion.

    Which brings me to this: are you sure it really was 0.3 g? That is very tiny. The rubber density is about 1000 kg per cubic meter, which makes that volume 0.0000003 cubic meter, or about a cube with 7 mm side; your reported extension is about 7 times that size.
     
  24. Jan 19, 2014 #23
    Im almost positive that is the correct mass and I googled masses of other rubber bands and they're all less than a gram. If the mass is larger it would make sense and that's the only place I can see that is the problem.
     
  25. Jan 19, 2014 #24
    The only realistic shape I can think that would have this mass is a thin and long strip of rubber; that should indeed have low stiffness. Can you describe the shape and its dimensions, even if approximate?
     
  26. Jan 19, 2014 #25
    It is just a few centimetres in diameter. Its a circular rubber band, kind of thick and pretty stiff.
     
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