How do we get to Energy?

There is a progression which will take us from Position to Energy. This progression goes through Velocity, Acceleration and Force. This progression is a “trip” that will give you sever

Velocity is needed to get from the location where you are to the location where you want to be.
- A position is a location.
- Distance is the difference between the values assigned to the two positions.
Acceleration is needed to change Velocity.
A Force is needed to give Acceleration to a Mass.
Energy is needed to maintain a Force as it moves over a Distance.

Bonus Features

Origin

Take some time to think about position and location. Inagine that you live on a road with mile markers and that there’s a mile marker by your house, “05”, and that there’s a mile marker by your friend’s house, “07”. Your mind is so fast, you knew your friend lives two miles away from you before I had a chance to finish saying this sentence.

Distance = position2 – position1

$d = x_2 - x_1$

2 m = 7 m – 5 m

The idea of an origin is born when you get to the position marked by “00” and you think “this place is important because this is where it started.”

Velocity

Velocity has units and the number may include a negative sign. This is how it is different from speed.

$Velocity = \dfrac { \Delta x} { \Delta t}$

For simplicity, we are not worrying about details such as you didn’t travel at the same velocity the whole time. Your physics textbook may talk about average velocity and define it the same as what we show above.

At a velocity of 15 mph $15 \dfrac {miles}{hour}$ you will get from your house to your friends house in 8 minutes.

To get from your friend’s house to your house in 8 minutes, you need to travel with a velocity of -15 mph. Does it make sense that the negative sign changes the direction of travel?

We made two choices, the location of the origin, and the direction for positive, and from that we got that your house is at mile marker 5 and your friend’s house is at mile marker 7.

You are granted certain powers in the realm of physics, and you can put the origin where you want it, and also choose which direction do the positive direction. Maybe you decide to put the origin at your house and have the positive direction point towards your friend’s so that your friend is now at mile marker “02”. Does it make sense that the location that was previously the origin is now a location we would designate by “-05”?

Acceleration

Acceleration is a change in velocity per unit time. This can become change in position per time squared if the units for time in the two places are the same. Just make sure if you have seconds and hours that you convert one to the other.

Force

For a given force, the I’m out to acceleration if we get will depend on the mass. It should make sense that increasing the mass will decrease the acceleration for a given amount of force.

$F = ma$

$Newtons = kg \cdot \dfrac {m}{s^2}$

Energy

Energy is a calculation of the requirenent to exert the Force that moves an object a known distance.

This careful wording is needed because energy and work share the same units. From one point of view it took so much energy for the whatever to happen. From another point of view so much work was done when that whatever was done.

$Work = Force \cdot Distance$

$Joule = Newton \: meter$

We could say that we are done at this point. Would you like to look at one more equation?

Two different equations for energy will be shown, and at the end we will show that in the end they both have the same units

$Energy = \dfrac {1}{2} mass velocity^2 = \dfrac {1}{2} kg \dfrac {m^2}{s^2}$

$Energy = mass \: acceleration \: distance = mass ( \dfrac {meter} {second^2}) meter$

We could say

Appendix A – Average Velocity

As a child, we took car trips in the summer to visit family, and even though we drove at near 60 mph, when we calculateod distance traveled divided by the amount of time we calculated 50 mph.

We lost time when the car needed gas and when we needed to stop for a meal.

Time-independent Velocity

The equations of Kinematics connect the dots between Position, Velocity and Acceleration.

Time is present in two equations and we can, using some mathematical trickery, put the two together in a way that we remove time.

We will do this after we make some initial conditions intended to make the math more simple.

Everything starts at time t=0.
The position of the object at the start of the experiment is $x_0=0$ .
The velocity of the object at the start of the experiment is $v_0=0$ .

$x = x_0 + v_0t + \dfrac {1}{2} a t^2$

$x = 0 + 0 \cdot t + \dfrac {1}{2} a t^2$

$x = \dfrac {1}{2} a t^2$

$\dfrac {2x}{a} = t^2$

$t^2 = \dfrac {2x}{a}$

$t = \sqrt{ \dfrac {2x}{a}}$

$v = v_0 + at$

$v = 0 + at$

$v = at$

$v = a \sqrt{ \dfrac {2x}{a}}$

$v = \sqrt{ \dfrac {2x a^2}{a}}$

$v = \sqrt{ 2x a}$

Speed of Light from two Other Variables

Mockingbird students found this:

speed of light in a vacuum = $\dfrac {1}{ \sqrt { \epsilon_0 \mu_0}}$

Pieces of their “chickenscratch” are provided below. Can you put the pieces together to get the speed of light?

$\mu_0$ – Vacuum Permeability
$\epsilon_0$ – Vacuum Permittivity

$\epsilon_0 = 8.8541878128 x 10^{-12} \dfrac {F}{m}$

$\mu_0 = 1.256637062 x 10^{-6}\dfrac {H}{m}$

$\mu_0 \epsilon_0 = 1.112650056 x 10^{-17} \dfrac {H F} {m^2}$

$\dfrac {1}{ \sqrt{1.113650056 x 10^{-17}}} = 2.99792458 x 10^7$

Units

$1H = 1 \dfrac {kg m^2}{s^2 A^2}$

$1 F = 1 \dfrac {s^4 A^2}{m^2 kg}$

$\dfrac {kg m^2}{s^2 A^2} \dfrac {s^4 A^2}{m^2 kg} = \dfrac {s^2} {1}$

$\dfrac {H F}{m^2} = \dfrac {s^2}{m^2}$

Appendix A

There’s a reason they make a big deal about it being speed in a vacuum. Light can travel through a material and the result is, we calculate a speed that is less than the number shown above. The explanation for this is kind of like a person driving a hundred kilometers per hour on the turnpike but at every rest stop they stop for 10 minutes and so when they’re done driving across the turnpike, the calculation of distance traveled divided by time will give you a number less than a hundred.

Parallel Addition

The addition you learned in grade school is sometimes called Serial Addition. It has a “cousin” called Parallel Addition. You can build circuits to show either form of addition.

SerialAddition(2,3) = 5

$2 + 3 = 5$

ParallelAddition(2,3) = 1.2

$\dfrac {1}{2} + \dfrac{1}{3} = \dfrac {1}{1.2}$

Time Relativity Ratio

The relativity calculation leading to the ratio of times can be reached with less discussion if we postulate a few things rather than explain them. Doing this math correctly could get you 20 to 30 points on a test in physics. Once you are satisfied with the calculation you will probably be curious about the content that was cut out. Appendices will soon be added at the end of the blog to address this.

Postulate 1

Even though we have clocks when you got two different speeds, there is a starting point and stopping point for the experiment.

Postulate 2

A right triangle provides the foundation for the calculation. The sides of the triangle are distances from the story used for the calculation.

Postulate 3

The speed of light is independent of frame of reference and this forces two clocks to record different changes in time.

Postulate 4

We draw the triangle so that a hypotenuse connects a longer vertical side and a short horizontal. The vertical line has time relative to the train. The other two lines move in the horizontal direction because they show movement relative to the ground underneath the train.

Definitions

c- speed of light
v- velocity of the train
d- a distance traveled between the start-and-stop point of the experiment
we use a,o,h as labels for the sides of the right triangle, adjacent, opposite and hypotenuse respectively.

We use the Pythagorean Theorem with the sides of the right triangle:

$d_a^2 + d_o^2 = d_h^2$

$c^2t_a^2 + v^2t_h^2 = c^2t_h^2$

$t_a^2 + \dfrac {v^2}{c^2} t_h^2 = t_h^2$

$t_a^2 = t_h^2 - \dfrac {v^2}{c^2} t_h^2$

$t_a^2 = t_h^2(1 - \dfrac {v^2}{c^2})$

$\dfrac {t_a^2}{t_h^2} =1 - \dfrac {v^2}{c^2}$

$\dfrac {t_a}{t_h} = \sqrt{1 - \dfrac {v^2}{c^2}}$

Appendix A – Perspective

Trains move over the ground, and we introduced the train is moving at velocity ‘v’ to the west. However for the purposes of the calculation it may actually help to imagine the train standing still and the ground moving under the train at a velocity of v the east.

That’s what it actually looks like if you are on the train because to you it seems like the train isn’t moving. That helps explain why the line of the triangle corresponding to the photon of light light moves in a vertical line when moving relative to the train.

To the extent that the ground moves, light on the train moving relative to the ground has to move in the horizontal direction.

Appendix B – Starting and Stopping Points

This might be the most challenging part of the discussion. Previously, anytime we talked about starting and stopping points, we used time. Now we have time moving differently and something else has to govern the starting and stopping points. The person devising the train story might go to elaborate lengths with details to make their starting point and stopping points exist.

Appendix C

Two ideas are shown below. It seems possible that one of them explains the other but we have to decide which came first. Said another way, one has to be the postulate, then the other can be something that we get from the postulate.

Nothing to have a speed greater than the speed of light.
The speed of light (and other things in the same category) is invariant (does not change) to frame of reference.

The argument was made, that if the speed of light cannot change, then two things traveling different distances between a start-and-stop can only do so by having different times.

Component Forces and Inclined Plane

The one real force is the force pulling toward the center of the earth — we calculate this by taking the product of mass, m, and g, the acceleration of gravity. F=mg.

This force is split into two component forces, one in the direction of travel down the inclined plane and the other force being the force going into the plane, the “normal force”.

We by using a right triangle. It should make sense at the component forces will be equal to or less than the actual Force. It should also make sense that if one of the component force is equal to the actual Force than the second component Force has to be zero.

The hypotenuse of the right triangle is the actual Force.

You can select any angle you want for the work we do to get the overall math, but we suggest picking an angle that is very shallow, like 20 degrees, so that most of the force will be going down into the plane and the force in the direction of travel will be much smaller.

Draw a right trangle, and designate the angle that is 20° to be Theta. We’ve already said that the small side represents the force in the direction of travel and noticed that the small side is the opposite side. We know that opposite / hypotenuse equals sine Theta.

$\dfrac {opposite} {hypotenuse} = sin (\theta)$

$opposite = hypotenuse \cdot sin(\theta)$

$ma_{travel} = mg \cdot sin(\theta)$

Appendix A

We could have kept the same right triangle and made each side an acceleration. We would have had g for the actual acceleration, an acceleration in the direction of travel and an acceleration into the plane.

Appendix B

Our work focused on force in the direction of travel led to sine. Similar work focused on force into the plane would have led to cosine.

$ma_{Normal} = mg \cdot cos(\theta)$

Derivative of n^x

We haven’t seen a derivative for $n^x$ and it seemed like this math object might be a good test for our Dark Blue Magic program. Dark Blue Magic is the collection of tools and ideas that students think we would want to have if, for some unexpected reason, we ended up on a deserted island and we had to rediscover all of Algebra and Calculus.

One of the ideas in Dark Blue Magic is that the derivative of a function should be quite similar to the original function. For example, we would expect the derivative of a monomial to be built using monomials. We would expect the derivative of a trigonometric function to be something comprised of trigonometric functions. This idea isn’t a rule; it is a guideline for making guesses.

Our first hypothesis would be that

$\dfrac {d} {dx} n^x = an^y$

Nothing is known about either ‘a’ or ‘y’ and it is possible that one of them could be unnecessary; if this is the case we will have a=1 or y=1.

We used handheld calculators for getting derivatives of specific examples.

For n=3 and x=2
the function would be $f(x)=3^x$
We write the following
Derivative with respect to x for f(x) = 3^x where x=2[/latex] (this will be used several times below)

We would calculate:

$\dfrac {3^{2.000001} - 3^2} {2.000001 - 2} = \dfrac {3^{2.000001} - 3^2} {0.000001}$

Some results are shown below

Derivative with respect to x for $f(x) = 2^x$ where x=0 –> 0.693147
Derivative with respect to x for $f(x) = 2^x$ where x=1 –> 1.386295
Derivative with respect to x for $f(x) = 2^x$ where x=2 –> 2.772590
Derivative with respect to x for $f(x) = 2^x$ where x=3 –> 5.545179
Derivative with respect to x for $f(x) = 2^x$ where x=4 –> 11.090359
Derivative with respect to x for $f(x) = 2^x$ where x=5 –> 22.186723

We aren’t getting integer values. Integer values would be nice. Well, we have the next best thing–when we scrutinize these first few values it looks like one divided by another could be an integer. We know about round-off error so if we get 2.9999, we’ll recognize it as being 3.

We notice a place where one of these nonintegers could be twice another of the nonintegers so we try a division. We see another opportunity and we try that division also.

11.090359/5.545179 = 2
22.186723/5.545179 = 4

After staring at what we have and asking “how can things be tied together?”, we notice that 2 is 2^1 and 4 is 2^2 and that the values for the arguments of the derivative have a difference of 1 or 2, respectively. We work this into the formula below (see comments in Appendix A):

$f(x)=2^x, \dfrac {f'(4)} {f'(3)} = 2 = 2^1 = 2^{4-3}$

$f(x)=2^x, \dfrac {f'(5)} {f'(3)} = 4 = 2^2 = 2^{5-3}$

Does this trend continue?

11.090359/1.386295 = 8

$f(x)=2^x, \dfrac {f'(4)} {f'(1)} = 8 = 2^3 = 2^{4-1}$

11.090359/0.693147 = 16

$f(x)=2^x, \dfrac {f'(4)} {f'(0)} = 16 = 2^4 = 2^{4-0}$

Using the variables j and k for the integers, we can write what we know so far from the above examples:

$\dfrac {f'(j)} {f'(k)} = 2^{j-k}$

Our numbers 0.693147, 1.386295, 2.772590, 5.545179, … are mysterious to us. If we had a noninteger like 3.14159… or 1.414214… then we would probably recognize the decimal as leading us to something like $\pi$ or $\sqrt{2}$ .

Mockingbird students will soon be adding a list of nonintegers, like the ones above, that we believe are a part of our Dark Blue Magic.

ln 1 = 0
ln 2 = 0.69314718056
ln 3 = 1.09861228867

This lets another cat out of the bag, at least for the 0.69314718056 noninteger. Could it be that “1.0986122…” will be the result for math just like we’ve done, having 3 instead of 2?

For 2 we had:

Derivative with respect to x for $f(x) = 2^x$ where x=0 –> 0.693147

Let’s do the work for

Derivative with respect to x for $f(x) = 3^x$ where x=0

$\dfrac {3^{0.000001} - 3^0} {0.000001 - 0} = \dfrac {3^{0.000001} - 3^2} {0.000001}$ = 1.09861289221

The trend continues. Let’s try to build something more general given what we now know. Let’s look at our general formula again:

$\dfrac {f'(j)} {f'(k)} = 2^{j-k}$

Let’s bring up the denominator so we can isolate f'(j) on the left side:

$f'(j) = 2^{j-k} f'(k)$

We could make it simpler by setting k to zero:

$f'(j) = 2^{j-0} f'(0)$

$f'(j) = 2^{j} f'(0)$

Let’s go ahead and replace j with x:

$f'(x) = 2^{x} f'(0)$

We know that for $f(x)=n^x$ , f'(0)=ln(n). Also, that 2 is actually n–we were doing work with n=2 when we started building this general example:

if $f(x)=n^x$ then $f'(x) = ln(n) n^{x}$

For the work in this monograph, only integers were used. We are happy with the success we have to this point, but we will do some work with nonintegers before we fax a copy of this work out to our friends at another academy (a soon-coming Appendix B will have such an example).

Oh, one very important thing! Assume this actually was the first time that this was done–it would be very important to have something written up showing the whole thing with a date and our name on it! In fact, that’s one of the things our friends at other academies are supposed to do–get mail from us with our name and a date, and keep it so later it can be used as proof that we had developed and divulged the idea!

Appendix A

Ever so often, you can read through a textbook and it looks like the person who figured it out was like Spock on an episode of Star Trek, and he figured it out in five minutes, just in time to save the Enterprise.

Some work was done fast, but in other stories, it took person months or years to decide what the next step should be. Mathematicians with years of experience have superhighways in their brains between commonly linked ideas to help them, and as a student starting out, you have a path through the grass. Keep reading stuff and trying stuff–build your own highways! : – )

Appendix B

n=e=2.71828
x= $pi$ =3.14159
$n^x = e^\pi = 23.140582$

$\dfrac {2.71828^{3.14160} - 2.71828^{3.14159}} {3.14160 - 3.14159}$ = 23.14068

OK- That is close enough! We are ready to write this up in a letter and send it to a friend!

About Mockingbird Academy

(warning: it is a fiction story)

Mockingbird Academy promotes student-driven learning as the major force of education, with adult involvement providing a necessary, minor assistance.

The small size of our group makes it possible for parents to participate extensively without disagreements and political fights.

We believe our strict admission standards have resulted not only in a Student Body with better students, but also, a School Board with better parents.

A Circular Game in Calculus

You’ve probably heard of circular reasoning. You probably are familiar with the dilemma of the chicken and the egg, “which came first?”

Before beginning, the purpose here is to create a bit of amusement and to get the reader to think about math proofs.

The issue here is the calculation of the integral of a monomial to get the area.

You have a choice: do you want to use a known area to confirm a calculation, or do you want to use a confirm calculation to prove that a reported value for the area is correct?

C- we can test and confirm the calculation
A- the area value is known

Herein lies the dilemma:

As a math student you have type of experience using both C–>A and C to get A. The textbook gives you a confirmed calculation so we can say that C is true… (warning background music begins to play)

Then one day, skeptics find some errors in the calculus textbook and a judge rules that the calculus textbook is fallible, and we can’t use it to validate calculus calculations (see Appendix B). We must prove a calculation in some way before we can use it.

Suddenly, someone offers you a lot of money–$$$$–if you can prove that the calculation for the derivative of a monomial is correct. Money? Okay I now have your complete attention.

If we have the area, that makes A true and that, with A–>C makes C true.

Let me give a quick example of this, maybe we do something like we have a printer that prints out the curve shape and print out a square we know the area of the square so we can correlate paperweight to pay for area and from that we can get an estimate of what the area is. Now if we show that the calculation gives an answer which is really really close to what we get from doing our paper weigh experiment–and this can be done using an electronic balance that goes to five digits–we might convince the person offering the money that we have done enough for their purposes, and thus they should pay us.

Appendix A

I’m getting an uneasy feeling about parts of this. Propositional logic may not fit with what I’m trying to do.

I went back and rewrote A and C. I did this because…

A –> C

The above “can” work, but it isn’t a guaranteed. It’s entirely possible that we made a mistake and we don’t have the correct calculation, and when we take our known area and test it with our calculation, it doesn’t work.

I think in propositional logic you write an implication (p –> q) only if EVERY time the p is true then the q is true.

Appendix B

The story is unfair in a way because textbooks show a proof for an idea along with introduced in it and tell me what the calculation is. Those studebt with more advancement mathematics can follow the proof and this makes it easier to accept the calculation

Truth by Definition

We don’t believe we invented the concept, surely someone else thought of it first.

The idea of Truth by definition is, if you create something, it is what you created, someone can’t say it’s wrong. Now it might not be useful, you might make mistakes when you use it, but it is what you made it.

We ran into this when we couldn’t find explanations for some things, and it occurred to us, maybe the authors don’t show a derivation because it was created, it simply is what they made it to be.

As an example, we don’t derive addition, we simply do it. Now we can point to how the idea develops when you start doing things like combining counts, and we could argue that addition comes from counting– we might have some fun on a project where we show things that help to give credibility to stuff that hasn’t been proved.