Relation between stopping distance and Kinetic Energy

  • #1
fishnic
2
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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.
 
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  • #2
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
 
  • #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]
 
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  • #4
fishnic said:
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]

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

ehild
 
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