Relation between stopping distance and Kinetic Energy

[fishnic]

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 $\frac{g\times cm^{2}}{s^{2}}$

Three points (From lab work):

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

Related Equations

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

After doing a basic slope calculation of $\frac{(y-y)}{(x-x)}$, 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 $\frac{s^{2}}{g\times cm}$

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.

 

[ehild]

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

 

[fishnic]

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 = $\frac{\frac{1}{2}Mv_{i}^{2}}{f}$

d = $\frac{KE}{f}$

y = mx + b

d = m(KE) + 0

d = $\frac{1}{f}\times KE$

 

[ehild]

Wait, I think I got it.

d = $\frac{\frac{1}{2}Mv_{i}^{2}}{f}$

d = $\frac{KE}{f}$

y = mx + b

d = m(KE) + 0

d = $\frac{1}{f}\times KE$

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

ehild