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Relation between stopping distance and Kinetic Energy

  1. Dec 14, 2011 #1
    1. What is the meaning, when a graph of Stopping Distance vs. Kinetic Energy is made, of the slope of the line? Justify this answer by showing how y=mx+b corresponds to the quantities you actually plotted.


    D is given in CM, and KE given in [itex]\frac{g\times cm^{2}}{s^{2}}[/itex]

    Three points (From lab work):

    D = 14.0cm, KE= 344,450[itex]\frac{g\times cm^{2}}{s^{2}}[/itex]
    D = 40.0cm, KE= 911,250[itex]\frac{g\times cm^{2}}{s^{2}}[/itex]
    D = 111.0cm, KE= 2,553,800[itex]\frac{g\times cm^{2}}{s^{2}}[/itex]

    Related Equations

    W = Fd*cosθ
    KE = (.5)mv^2
    KE(net) = W(net)*d

    After doing a basic slope calculation of [itex]\frac{(y-y)}{(x-x)}[/itex], the numbers were inconsistent, and when dividing the units, I couldn't figure out what it represented.

    I worked the slope's units out to be [itex]\frac{s^{2}}{g\times cm}[/itex]


    I don't know where to go from here. Our physics teacher's answer to questions is that "[we] are accelerated students and should be able to find it out on our own" or that "[we] should see him after school" (which, he is never there). We JUST started working with these problems yesterday, and I want to understand this unit.
     
    Last edited: Dec 14, 2011
  2. jcsd
  3. Dec 14, 2011 #2

    ehild

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    Homework Helper
    Gold Member

    Recall the Work-Energy Theorem: Change of KE equal to the work done.
    The object stops, so its KE becomes zero if some force acts along the stopping distance D. So and ΔKE=-KE=W, and W=FDcosθ,so KE=-FDcosθ. The stopping distance vs KE graph corresponds to the function D=KE/(Fcosθ). How does it look like? What is the tangent of the slope? What physical quantity is it?
    As the points scatter a bit you need to fit a straight line to your graph which is as close to all points as possible, and find the slope of that line.


    ehild
     
  4. Dec 14, 2011 #3
    Given we are working horizontally with no angle, the function would then be D=KE/F, meaning that the slope represents the amount of force needed to stop the object with the given force at a certain distance?

    But I am confused as to where the "/(Fcosθ)" came from?

    Wait, I think I got it.

    d = [itex]\frac{\frac{1}{2}Mv_{i}^{2}}{f}[/itex]

    d = [itex]\frac{KE}{f}[/itex]

    y = mx + b

    d = m(KE) + 0

    d = [itex]\frac{1}{f}\times KE[/itex]
     
    Last edited: Dec 14, 2011
  5. Dec 14, 2011 #4

    ehild

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    Homework Helper
    Gold Member

    That is!!
    So the tangent of the slope is the reciprocal of force.

    ehild
     
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