# Help in explaining formula of Kinetic Energy

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### #21 swansont

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Posted 30 July 2013 - 01:56 PM

"If you know calculus" ... that's the nub isn't it! It moves the mysterious appearance of one half to somewhere else i.e. the integration of the variable v being v^2/2

I had the experience a few years ago of trying to explain this to a young person who was confused about this very issue. I too came at it from the mathematical end, which is entirely natural to me. Nevertheless we weren't making any headway and I realised that the maths was obscuring rather than revealing. It's often not helpful to explain something using explanations that are at least if not more complicated, especially for people who are not mathematically inclined. It turned out that we had to go back to basics and explain things which my young tutee had forgotten or not taken in. Things like work done is defined as the force applied in moving (or accelerating) an object over a certain distance, the equivalance of potential and kinetic energy, Newton's second law, the equations of motion etc. No calculus was involved. We got there!

You have to explain to the level of your audience. But when you don't know that level, you have to guess. We never got any feedback that we were talking above or were at Vay's level.

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Posted 31 July 2013 - 05:06 AM

Conservation of what?

While this is true in some cases I don;t think it applies here; the first answer given is completely correct and answers the question without any element of "look how clever I am". It's straightforward math, if you know calculus.

ajb's answer is a bit of an end run around that, because one might as easily ask where the factor of 2 comes from in the equation he uses, and the answer is still going to be that it comes from calculus. Ultimately that's where all the answers lead, because that's where the factor of 2 comes from: the math.

There is really no wiggle room here: the question was answered. If the information is found wanting, it may be that the wrong question was asked.

Conservation of energy of coarse.

Energy of a photon:

E = h*c = 1.9878e-23 / L

c/G = 30

30 / 1.9878e-23 / G  = 1.5e+25

That " may " be a way of examining what 1/2 is.
My relation to 30 is time as in 60 seconds = 1 minute but this is theoretical.

With other examples, it would be nice to see some numbers in the formulas..

If energy is truly conserved, then by all means the energy of a photon should be G itself..
Hence, c is everywhere in all places all at once, mass-less?? I have a hard time believing this by the way.

What does this have to do with KE??? " The---->Constant Energy related to c"
They all seem to be 1/2  connected.

It appears that there is always something " reserved " with all that applies to energy in all its forms.

I really think issues stem from:

Gravity, The Speed of Light, and the h constant, they are all =1 and related to torque and angular momentum which obscures many things.

The reason for this is because their is something spinning around something that is missing or simply not there. I know that sounds dumb.

But inverse square laws are a perfect example!

To further make matters not fully understood the very calculus we use "also" relies on

Trigonometry and Geometry.

With that being said, there appears to be many virtual ways to calculate the same things over and over again to make that 1/2 = 1, but what is this 1 relative to??? Because of inversion there would yet still be another half missing  onto infinity. This is where pi ratio plays the final role.

Since this would be the case then our known pi ratio is 1/2 missing!

Edited by Iwonderaboutthings, 31 July 2013 - 09:09 AM.

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### #23 Griffon

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Posted 31 July 2013 - 06:12 AM

You have to explain to the level of your audience. But when you don't know that level, you have to guess. We never got any feedback that we were talking above or were at Vay's level.

Yes, that is true. Although the level of the question is an indication of a sort.
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### #24 swansont

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Posted 31 July 2013 - 09:23 AM

Conservation of energy of coarse.

The presence of the 1/2 conserves energy, because it makes the work equal to the kinetic energy. If it wasn't there, your kinetic energy would be twice the amount of energy you put into the system. Energy would not be conserved.

Energy of a photon:

E = h*c = 1.9878e-23 / L

c/G = 30

30 / 1.9878e-23 / G  = 1.5e+25

That " may " be a way of examining what 1/2 is.
My relation to 30 is time as in 60 seconds = 1 minute but this is theoretical.

You are over-thinking the problem.

With other examples, it would be nice to see some numbers in the formulas..

If energy is truly conserved, then by all means the energy of a photon should be G itself..
Hence, c is everywhere in all places all at once, mass-less?? I have a hard time believing this by the way.

What does this have to do with KE??? " The---->Constant Energy related to c"
They all seem to be 1/2  connected.

You are seeing connections that aren't really there.

It appears that there is always something " reserved " with all that applies to energy in all its forms.
I really think issues stem from:

Gravity, The Speed of Light, and the h constant, they are all =1 and related to torque and angular momentum which obscures many things.

No, they are not all =1 (if you want to do a meaningful calculation relating to an observation).
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Posted 1 August 2013 - 01:37 AM

The presence of the 1/2 conserves energy, because it makes the work equal to the kinetic energy. If it wasn't there, your kinetic energy would be twice the amount of energy you put into the system. Energy would not be conserved.

You are over-thinking the problem.

You are seeing connections that aren't really there.

No, they are not all =1 (if you want to do a meaningful calculation relating to an observation).

You are 100% correct on the The------>"meaningful calculation"

But in today's "mentality" I think that perhaps we should consider that what is meaningful to some is pointless to others " hence" if its not broken don't fix it mentality.When connecting Classical Physical Formulas with QM operators we see things that don't pair up and or brake down. Even if it is a tiny tiny micro amount of discrepancy it still is an issue for a larger part of the whole.

In a Digitized Binary Sense this deal of 1/2 "seems" to be incompatible with how Calculus defines derivatives.

Hence again: But in today's "mentality" 2013 not the 1900s..

To add computer systems don't recognize ratios they stack floating point variables much like dx within Calculus.

For those whom may not know this the h constant is the number that rules technology and was used in the making of the microprocessor. Not sure about the life thing though!

Planck's Constant: The Number That Rules Technology, Reality, and Life

http://www.pbs.org/w...ancks-constant/

Another great link  on : Number Bases: Introduction / Binary Numbers

http://www.purplemat...es/numbbase.htm

So how did I go from KE, Pi Ratio, Time And Computer Science????????

Again their all connected to energy and time.

They all seem to be un-avoidable when used separately

I know that sounded dumb but think about it.

Edited by Iwonderaboutthings, 1 August 2013 - 01:42 AM.

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### #26 swansont

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Posted 1 August 2013 - 09:34 AM

You are 100% correct on the The------>"meaningful calculation"

But in today's "mentality" I think that perhaps we should consider that what is meaningful to some is pointless to others " hence" if its not broken don't fix it mentality.When connecting Classical Physical Formulas with QM operators we see things that don't pair up and or brake down. Even if it is a tiny tiny micro amount of discrepancy it still is an issue for a larger part of the whole.

In a Digitized Binary Sense this deal of 1/2 "seems" to be incompatible with how Calculus defines derivatives.

How so? What is the antiderivative of x?

So how did I go from KE, Pi Ratio, Time And Computer Science????????
Again their all connected to energy and time.

They all seem to be un-avoidable when used separately;)
I know that sounded dumb but think about it.

Fundamental constants show up everywhere, because they are fundamental constants. This misconception sounds like it's related to the fallacy of ignoring a common cause. KE, time, and computer science are connected because they are all described by physics, which uses math, which is where pi enters the picture.

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### #27 Amaton

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Posted 2 August 2013 - 03:48 AM

While this is true in some cases I don;t think it applies here; the first answer given is completely correct and answers the question without any element of "look how clever I am". It's straightforward math, if you know calculus.

ajb's answer is a bit of an end run around that, because one might as easily ask where the factor of 2 comes from in the equation he uses, and the answer is still going to be that it comes from calculus. Ultimately that's where all the answers lead, because that's where the factor of 2 comes from: the math.

That's true, but is it relevant to the nature of the question? I mean... Vay could've simply been asking for a mathematical derivation. But what if, and this is a common case with such questions, Vay was looking for something less blunt, on the more conceptual side i.e. an intuitive reasoning behind the math.

For example, a student asks "Why is d/dx sin x = cos x?" Now, one professor could set up the difference quotient and perform the dry symbolic work. The other professor instead elaborates on his colleague's work using the geometric interpreation. And he shows the student how limiting the angle measure relates the angle's functions by infinitesimal rate of change. That's intuitive background.

Yes, the 1/2 factor comes down to the calculus, but why? This is what I believe Wilmot McCutchon tried to explain in post #6.

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Posted 2 August 2013 - 07:08 AM

How so? What is the antiderivative of x?

Fundamental constants show up everywhere, because they are fundamental constants. This misconception sounds like it's related to the fallacy of ignoring a common cause. KE, time, and computer science are connected because they are all described by physics, which uses math, which is where pi enters the picture.

Can there be a way " if possible" that pi ratio in it's absolute value can be found????????

What is the process????

Everything eventually needs an upgrade when dealing with science, math and all that applies to the natural world described with the numerics of human psychology. Reasoning behind this is due to calculations being set in a backwards sense, not aimed for the future but thats one way of looking at this. Seems like a for loop if you ask me.

It is because of this flaw and premise that we still in our day and age use " numbers" linked to the laws of nature that determine " some " relation and or interaction." Though highly respected to me scientist are, I feel they should be more open minded to approaches that others may encounter and be able to solve derivatives without the need for " standard calculations."

Anti-Derivative of x uses 1 again as a numerator, here is a good link.

http://cims.nyu.edu/...derivatives.pdf

The whole issue is based on what is this one? if you ask me it is still the constant of 2 as in .5*2=1

BUT! In binaries this constant of 2 is not always the case. In computer science numbers as:

0 1 2 3 4 5 6 7 8 9 all eventually move over as 0=1 2=3 3=4 and so fourth.

They seem to be dynamic and static again there are 2 things going on with these numbers..

I want to emphasize that binary systems do not understand ratios, this is why the " floating point" was created I assume.

How can that be related to physical science???????????

A quanternion method should be most suitable for this Anti-Derivative of x , in one where field axioms and other algebraic rules coupled with calculus not need apply. But that would only be an extrapolation of this sort.

Edited by Iwonderaboutthings, 2 August 2013 - 07:11 AM.

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### #29 swansont

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Posted 2 August 2013 - 08:49 AM

That's true, but is it relevant to the nature of the question? I mean... Vay could've simply been asking for a mathematical derivation. But what if, and this is a common case with such questions, Vay was looking for something less blunt, on the more conceptual side i.e. an intuitive reasoning behind the math.

For example, a student asks "Why is d/dx sin x = cos x?" Now, one professor could set up the difference quotient and perform the dry symbolic work. The other professor instead elaborates on his colleague's work using the geometric interpreation. And he shows the student how limiting the angle measure relates the angle's functions by infinitesimal rate of change. That's intuitive background.

Yes, the 1/2 factor comes down to the calculus, but why? This is what I believe Wilmot McCutchon tried to explain in post #6.

I don't disagree, but if this is the case, I think it's a matter of having asked the wrong question, or wording it incorrectly. Which changes this from "none of you answered the question" to "none of you are telepathic"

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### #30 Bryce

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Posted 15 October 2016 - 01:07 AM

is this true, the 1/2 is because the change in velocity measured in m/s will not be the actual distance covered , such as 20m/s was the resultant change in velocity over say 4 seconds. so to times 20 by 4 would yield a result of 80 meters but it would actually only be 40 meters traveled during the time the force was applied to stop it. it is only 40 meters because the average speed while changing from 20m/s to 0m/s is 10m/s and 10 x 4 seconds is 40.

This video may help explain what i am looking for.

To the OP is this simple enough ? its how my brain works. no math, just simple explaining of why.

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### #31 Sriman Dutta

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Posted 15 October 2016 - 04:27 AM

Bryce, your reply shows the equation S=ut+1/2at^2 .

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### #32 Bryce

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Posted 15 October 2016 - 05:29 AM

Hi Sriman, I am just a bus driver. I like Physics a bit, get headaches trying to translate formulas so give up most times. I probably should not be here, but the question intrigued me while considering the Oberth effect again recently, KSP lol. I also like RC aircraft and got into the physics of lift, Good old L= Cl 1/2 roh V^2 S , so the question of why we use half had been bugging me for some time (About 10years or more) but I never really found a suitable answer and gave up looking and just accepted it, I recall from my schooling back in the late 80's that it was from what I described. so given this information about me and how relatively dumb I am, can I truthfully ask.

Is your comment suggesting my post is useless and not related to Kinetic energy and that it has nothing to do with the Formula KE=1/2*a*v^2 and ultimately not the reason it contains the "1/2" in the equation.

Edited by Bryce, 15 October 2016 - 05:32 AM.

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### #33 Sriman Dutta

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Posted 15 October 2016 - 10:04 AM

Hi Bryce, I personally highly appreciate your participation in the discussion. But I think your formula is different from the OP. The others may clarify it. And a warm WELCOME To SFN.

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### #34 Bryce

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Posted 15 October 2016 - 10:39 AM

Thank's Sriman, looks like I am stuck with my Mathematical blindness. the answers are there but I just don't have the time or intellectual fortitude to understand it

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### #35 Sriman Dutta

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Posted 15 October 2016 - 10:54 AM

Bryce never think that you are intellectually poor. Keep learning and strive forward.....

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### #36 sethoflagos

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Posted 15 October 2016 - 05:52 PM

Bryce, your reply shows the equation S=ut+1/2at^2 .

It is the same 'half'. And it comes from being the arithmetic mean of zero and one(!).

Imagine a pneumatic 'gun' that applies a precise 1 Newton force to a 1 kg mass projectile through its barrel  for precisely 1 second.

From Newton's second law, and the definition of the Newton, we know that the acceleration will be 1 m/s2 and therefore the projectile will leave the barrel at 1 m/s.

But the total distance travelled during linear acceleration is the time multiplied by the arithmetic mean of initial and final velocities. As the projectile starts at rest with respect to the equipment(!), and leaves at 1 m/s, the average velocity (again with respect to the equipment) is 0.5 m/s and therefore the scientist, being an excellent Newtonian, has designed his barrel to be 0.5 metres in length. The same 'half'.

As for the kinetic energy, the scientist tells me with confidence that from the definition of the Joule, a constant force of 1 Newton acting over a distance of half a metre imparts an energy of 0.5 Joules (the same 'half').

But I disagree with him.

I've been carefully watching his experiment with the aid of my trusty laser ruler and stopwatch, while walking towards the gun at the speed of 1 m/s. We agree on the mass of the projectile; we agree on the force applied; we agree on the time period the force was applied; we even agree on the rate of acceleration; but we disagree on initial and final projectile velocities (which I have measured to be 1 and 2 metres per second respectively); and also the distance over which the force acts, which I carefully measured as 1.5 metres.

So I believe that the projectile has gained not 0.5 but 1.5 Joules of kinetic energy (NB the arithmetic mean of 1 & 2!)

He of course quickly points out to me that if we can both agree that a projectile travelling at one metre per second has a kinetic energy of 0.5 Joules (his final value, my initial one), Then my final kinetic energy would be 0.5 + 1.5 = 2 Joules.

So a doubling of velocity quadruples the kinetic energy per kilogramme (providing the mv2), and the constant of proportionality is the same 'half' it has always been.

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### #37 Bryce

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Posted 15 October 2016 - 10:39 PM

Awww man, I nearly understood that until you mentioned Proportionality, it was entertaining for the lay like me.

I decided to look up that word (Proportionality), It turns out I understand it as coefficient, I see that the use of the word proportional can be tricky, Bugger.... it took me a while to learn to use the word Velocity speed correctly and acceleration set me back completely, Satellites accelerating around the earth all the time O.O

I guess now, I do have the ability to learn, But it seems I only learn at a rate proportional to the effort i put in, i only wish i had some sort of exponent to help that along.

OK> i need to develop a mathematical vocabulary, my words for this month (yes a month lol) are

proportional, Coefficient

exponent

PS

Lift = Cl 1/2r V^2 S

I think I understand why the lift formulas Proportionate "Cl" works, basically it has to do with the shape and the way it causes an acceleration of air particles (mostly in the change of direction) though this is vague and nowhere near complete, I can say that the Cl value is proportional to the amount of curvature on mostly the upper surface, this Cl Proportionate will also change when you change the AoA, because, you have effectively changed the curvature relative to the air flow. So there is an acceleration hidden here in this value, and the lift is the result of some sort of Kinetic interaction. so the whole formula has the 1/2 value incorporated in it, and from what Seth was saying, the half value is a constant of proportion. so would that make Cl the second constant of proportion or ??

So I have typed all this out in an effort to understand, I want to know, now it's time to post it, but, I am a bit scared, should I just delete it or post it. if I don't post it I may not learn, but it's not your duty to help or to teach me at a dulled down level at which I stand, but there may be some willing to help and point out errors, so why not post it.

OK, Balls out, here I go.

Do you all use proportionate and proportional interchangeably or ??

Did I make many mistakes there with the use of the word "Proportional". any other errors/misconceptions

I'm off now looking for the Fairy land Physics thought experiment room. Always liked them

Edited by Bryce, 15 October 2016 - 10:42 PM.

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### #38 sethoflagos

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Posted 16 October 2016 - 01:13 AM

Awww man, I nearly understood that until you mentioned Proportionality, it was entertaining for the lay like me.

I decided to look up that word (Proportionality), It turns out I understand it as coefficient, I see that the use of the word proportional can be tricky, Bugger.... it took me a while to learn to use the word Velocity speed correctly and acceleration set me back completely, Satellites accelerating around the earth all the time O.O

I guess now, I do have the ability to learn, But it seems I only learn at a rate proportional to the effort i put in, i only wish i had some sort of exponent to help that along.

OK> i need to develop a mathematical vocabulary, my words for this month (yes a month lol) are

proportional, Coefficient

exponent

PS

Lift = Cl 1/2r V^2 S

I think I understand why the lift formulas Proportionate "Cl" works, basically it has to do with the shape and the way it causes an acceleration of air particles (mostly in the change of direction) though this is vague and nowhere near complete, I can say that the Cl value is proportional to the amount of curvature on mostly the upper surface, this Cl Proportionate will also change when you change the AoA, because, you have effectively changed the curvature relative to the air flow. So there is an acceleration hidden here in this value, and the lift is the result of some sort of Kinetic interaction. so the whole formula has the 1/2 value incorporated in it, and from what Seth was saying, the half value is a constant of proportion. so would that make Cl the second constant of proportion or ??

So I have typed all this out in an effort to understand, I want to know, now it's time to post it, but, I am a bit scared, should I just delete it or post it. if I don't post it I may not learn, but it's not your duty to help or to teach me at a dulled down level at which I stand, but there may be some willing to help and point out errors, so why not post it.

OK, Balls out, here I go.

Do you all use proportionate and proportional interchangeably or ??

Did I make many mistakes there with the use of the word "Proportional". any other errors/misconceptions

I'm off now looking for the Fairy land Physics thought experiment room. Always liked them

Yes, the aerofoil lift equation is that form because of the relationship  q=0.5 r u^2  where u is air speed, r, air density, and q the dynamic pressure (the additional theoretical pressure you could get if you brought the air to a standstill). The right hand expression is the kinetic energy of the airstream. So the same 'half'.

Don't worry about too much about the coefficient/proportionality thing. A mathematician might possibly be offended by the way I used the term, but I'm certainly not going to lose sleep over it. Except the capital 'C' that comes with your lift coefficient does stand for 'coefficient', so perhaps best to stick with that one,

PS No need to be shy about your level of understanding. It's pretty impressive. I wish a few of my clients were as quick on the uptake!

Edited by sethoflagos, 16 October 2016 - 02:11 AM.

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### #39 studiot

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Posted 16 October 2016 - 11:55 AM

I think that teaching proportion or proportionality (nouns which name the phenomenon) and proportionate (old fashioned) or proportional (modern) - the adjectives used to describe some noun has sadly fallen out of fashion.

I find that for most folks using proportion is actually easier and more natural to understand and use than more explicit mathematics.

Certainly in Newton's day pretty well all work was stated in terms of proportion so if it was good enough for that genius it is more than good enough for me.

Most of the discussion about the 1/2 has centered on propelling a body to a specific speed.

A much easier derivation of the KE is found by considering instead the constant force needed to bring a body to rest from some velocity v.

Consider a body of mass m moving with steady velocity, that is brought to rest by the application of a constant force P.

P produces a constant retardation, f so that P = mf, in a distance x

Applying the simple equation (final velocity)2 = (initial velocity)2 - 2 (acceleration) (distance) and noting that the final velocity is zero at rest.

0 = v2 - 2fx

Thus fx = 1/2 v2

But the work done by the particle against force P is

Px = mfx

Therefore mfx = 1/2 mv2

This quantity is defined as the kinetic energy

Edited by studiot, 16 October 2016 - 09:58 AM.

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### #40 sethoflagos

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Posted 16 October 2016 - 04:16 PM

If I'd thought for a little longer, I might have come up with this derivation:

Calculate the change in kinetic energy of a body of mass m accelerating from u to v in time t

Force = rate of change of momentum = m*(v-u)/t

Distance = time x average velocity = t*(v+u)/2

Delta KE = force*distance = 0,5*m*(v+u)*(v-u)

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