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

 

[asteriatic]

The slope of a line on a graph of Stopping Distance vs. Kinetic Energy represents the relationship between the two quantities. In this case, the slope represents the change in stopping distance for a given change in kinetic energy.

To understand this further, let's look at the equation for kinetic energy: KE = 0.5mv^2. This equation shows that kinetic energy is directly proportional to the square of the velocity of an object. This means that as the velocity of an object increases, its kinetic energy increases at a faster rate.

Now, let's look at the equation for work: W = Fd*cosθ. This equation shows that work is equal to the force applied to an object multiplied by the distance it moves in the direction of the force. In the context of stopping distance, this means that the work done to stop an object is directly proportional to the force applied and the distance it travels while being stopped.

Combining these two equations, we can see that the kinetic energy of an object is directly related to the work done to stop it. This is because the kinetic energy of an object is converted into work when it is stopped.

Using the data points given, we can see that as the kinetic energy increases, the stopping distance also increases. This is because with a higher kinetic energy, the object is moving at a faster velocity and therefore requires a longer distance to be stopped.

Therefore, the slope of the line on the Stopping Distance vs. Kinetic Energy graph represents the relationship between these two quantities, showing how the stopping distance changes as the kinetic energy changes. In this case, the slope represents the ratio of stopping distance to kinetic energy, which is why the units are \frac{s^{2}}{g\times cm}. This means that for every unit increase in kinetic energy, the stopping distance increases by \frac{s^{2}}{g\times cm}.

In conclusion, the slope of the line on a graph of Stopping Distance vs. Kinetic Energy represents the relationship between these two quantities and can be used to understand how they are related and how one affects the other.