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The slope of a kinetic energy vs. potential energy graph

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data

    I have plotted a graph of Gravitational potential energy (y-axis) vs. kinetic energy (x-axis), and I have found the value of the slope to equal -1.4. The thing is I dont know what the -1.4 equals to? Since I have to find the percentage error, I need to know the theorectical value of the experiment, to put into the equation of % error. I have tried several ways to determine what the value would be like but I don't think they make sense:

    3. The attempt at a solution

    (kg(m/s^2)m)/((kg(m^2/s^2))/2) which equals finally 2 (theo value)

    and when I put into percentage error formula:

    (|theo-exp|/|theo|)x100→(2-(-1.4)/2)x100→1.7x100=170% (which is impossible)

    or something I found on the internet:

    For a freely falling body,
    P.E. + K.E. = cosntant
    => x + y = constant, where x = P.E. and y = K.E.
    => P.E. vs. K.E. graph is a line with slope = - 1

    I don't understand how -1 should be the theo value of the slope... But it works perfectly in my equation for percentage error!

    (|theo-exp|/|theo|)x100→(-1-(-1.4)/-1)x100→0.4x100=40% (which makes more sense)

    Could someone help me understand this? And tell me my mistakes? It doesn't make any sense to me! Thanks!

    2. Relevant equations

    em=ek+pe
     
  2. jcsd
  3. Apr 13, 2012 #2
    Hint:

    Write the equation as

    PE + KE = C, where C is a constant.

    Then put it in the form of y = mx + b

    What is m?
     
  4. Apr 14, 2012 #3
    Uh well "m" equals -1 right?
     
  5. Apr 15, 2012 #4
    That's correct.
     
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