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Determing Spring Constant by Graphing

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data
    In class, we did a lab that consisted of putting a small wooden cylinder in a pipe that contained a spring, pushing it down on the spring and locking it in, then unlocking it which allowed it to launch straight up. We then would add fishing weights to the block to increase the mass and did 3 trials with each mass. This is my collected data:

    Constant Values: h0 = .258 m, x0 = .022 m, xf = .000 m, V0 = 0 m/s, Vf = 0 m/s

    Height of uncompressed projectile
    Trials 1-3: m = .0228 kg, hf (avg) = 1.367 m
    Trials 4-6: m = .0280 kg, hf (avg) = 1.304 m
    Trials 7-9: m = .0329 kg, hf (avg) = 1.076 m
    Trials 10-12: m = .0380 kg, hf (avg) = 0.951 m
    Trials 13-15: m = .0424 kg, hf (avg) = 0.860 m

    It now asks me to derive an equation for the spring constant of the projectile launcher based on the data I have. I came up with this: K = 2Us/X02 = 2(.5m2ahfavg - mgh0) / X02

    Then it says Your goal for the lab is to determine the spring constant. If you graph your data with x02/m on the x-axis and hf-h0 on the y-axis you should get a straight line with a slope of _____ and a y-intercept of _____.

    2. Relevant equations

    V2 = V02 + 2aΔy
    Us = .5Kx2

    3. The attempt at a solution

    As I said above, I derived K = 2Us/X02 = 2(.5m2ahfavg - mgh0) / X02

    I tested it with some old data in my notes (that I knew K for) and it worked... but now that I see it wants me to graph x02/m on the x-axis and hf-h0 on the y-axis, I feel like I did it wrong because I can't see where those values became relevant.

    So my question is did I derive that equation correctly? And how does it relate to what it wants me to graph? Thank you for your help!
     
  2. jcsd
  3. Nov 4, 2013 #2
    I would think that all the energy stored in the spring

    Us = 1/2 K X[itex]^{2}_{o}[/itex]

    would be converted to the potential enegy of the block at the top

    Ub = mg(hf - ho)

    so that

    hf - ho = [itex] ( \frac{K}{2g} ) \frac{X^{2}_{o}}{m}[/itex]
     
    Last edited: Nov 4, 2013
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