## The Exchange of Potential and Kinetic Energies
Let’s get a sense of the scale of Joule units of energy. With fruit.
```{admonition} Please study Example 4:
:class: warning
dropping an apple...halfway
```
```{admonition} Please answer Question 4 for points:
:class: danger
the mother off all apples
```
### Keeping Track of Energies Geometrically
By now you won't be surprised that I want to bring this energy conservation message home by recreating our thermometer graphs for before, in-between, and after. Get comfortable with this and our next energtic steps will be a lot easier!
```{admonition} Pens out!
:class: warning
Let's re-cast the apple problem with our thermometer graphs. Look at this figure:
Three stages of the falling apple. (a) is just before the apple is released, (c) is just before the apple hits the floor, and (b) is when the apple is 0.3 m above the floor.
> **Wait.** What’s with the negative scale on the energy axis? Hard to imagine a negative kinetic energy.
> **Glad you asked.** I put that here in order to get your attention, so thanks for that. A negative potential energy will be more clear when we talk about gravitation, atoms, and, well, the birth of the universe! (BTW, yes. Hard to envision a negative kinetic energy...you walk slower, and slower so your kinetic energy gets smaller and smaller...and then you stop walking and somehow acquire a negative *speed*? Nope. Can't happen.
In Example 4 we established the energy scale of our 0.1 kg apple, 1 m above the floor --- using the approximation that $g=10$ m/s$^2$.
**Step (a)** The potential energy is $U=1$ J which reflects the fact that $U=0$ at the floor. The red "thermometer" over the $U$ position is 1 J "long." Since the apple is dropped, it has no kinetic energy at that point, and so $K=0$ and that's shown as the blue circle above the $K$ position. Now you see how to read these plots.
* The only energy that the apple has is all potential and so the total energy of that system is always going to be equal to 1 J and so the total energy is indicated on the right of (a) as $T=1$ J (for $E_T$, "total").
>Any subsequent plot of the energetics --- the $U$ and $K$ thermometers --- of this system must total to that gray, $T$ thermometer.
**Step (c)** The apple has reached the floor making $U=0$. Since $T$ is still, and always 1 J, since the combination of the two thermometers must equal the right-hand gray one, the $K$ "thermometer" can be constructed to be equal the $T$ length. Just a long way of saying that at this point: $K=T=1$ J.
>Now we could construct an answer to a question like, "What's the kinetic energy of the apple when it's fallen to 30 cm above the floor?"
**Step (b)** That's the situation in (b), where $U=0.3$ J and since $T$ is still and always 1 J, we can construct the kinetic energy thermometer so that when it is end to end to the potential energy thermometer, their combined length would be 1 J. That's show in (b) and is the obvious $K=0.7$ J.
Now you could answer a slightly more complicated question like,"What's the velocity of the apple when it's 30 cm above the floor?" Since you know $K$ you can easily calculate
$$
\begin{align*}K &=1/2mv^2 = 0.7J \\
v&= \sqrt{2K/m} = \sqrt{(2\times 0.7)/0.1} = \sqrt{20} = 3.7 \text{ m/s}\end{align*}
$$
I'll leave it to you to convince yourself that this is what we could have gotten from our Galileo discussion in Lesson [note: MOTION] So what good is an energy discussion of this?
```
Falling in a straight line is one thing. Falling but through a curved path is something else. Let's go to the beach.
```{figure} ./../_images/energy/parkboth.png
---
width: 450px
name: parkboth
align: center
---
(a) An aerial view of Jolly Roger Amusement Park in Ocean City, Maryland. (b) A photo of me on the giant water slide.
```
I love water parks and this is one of my favorites. I always feel safe because I respect the conservation of energy. This next figure labels a variety of points along my trip on the slide.
```{figure} ./../_images/energy/mepark.png
---
width: 450px
name: mepark
align: center
---
I gingerly start myself from rest at point A, which is 10 m above the ground. I pass point B and start up the other side passing point D on the way to point C, which is at the same height as A.
```
Let's analyze some of the energetics of this situation: A, B, and C. You do D.
```{admonition} Pens out!
:class: warning
First, let's remind ourselves of what energy conservation would say.
$$
K(A)+U(A) = K(B)+U(B)=K(C)+U(C) = K(D)+U(D) \label{slideE}
$$
Four important notes:
* I start from rest, so $K(A)=0$.
* We can define our "zero" of potential energy anywhere, but the most natural place is to say that $U=0$ on the ground. So, $U(A)=mgh_A$
* I weigh 200 pounds, so $m=90$ kg, approximately
* The total energy, $T$ is then equal to the potential energy at $A$: $T=mgh=(90 \text{ kg})(10 \text{ m/s}^2)(10 \text{ m})=9000$ J, or 9 kJ.
So, what's my kinetic energy at B?
$$
\begin{align*}
T&=K(B)+U(B) = K(B)+0 \\
K(B)=T &= 9000 \text{ J}\end{align*}
$$
How fast am I going at $B$?
$$
\begin{align*}
K(B)&=1/2mv^2 \\
v &= \sqrt{2K/m}=\sqrt{(2)(9000)/90} = \sqrt{200} = 14 \text{ m/s}
\end{align*}
$$
That's moving right along: 30 mph.
```
And, as you now know: we can do this with thermometers. So here is the algebra from above, reproduced as a geometrical "solution."
```{figure} ./../_images/energy/slidetherm.png
---
width: 450px
name: slidetherm
align: center
---
(a) represents the energetics at $A$; (b), at $B$; and (c) at $C$.
```
```{admonition} Please answer Question 6 for points:
:class: danger
A two dimensional collision without fruit...of billiard balls
```