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Help in explaining formula of Kinetic Energy


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#1 Vay

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Posted 17 September 2011 - 01:35 AM

So the formula for Kinetic energy is KE=1/2*a*v^2

Where does the 1/2 come from?

And the square is because the work required exponentially increases?

KE = work, so that means work = 1/2*a*v^2? Therefore, 1/2*a*v^2= force x distance? Can anyone explain the help relationship?

Edited by Vay, 17 September 2011 - 01:36 AM.

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#2 Schrödinger's hat

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Posted 17 September 2011 - 01:51 AM

So the formula for Kinetic energy is KE=1/2*a*v^2

Where does the 1/2 come from?

And the square is because the work required exponentially increases?

KE = work, so that means work = 1/2*a*v^2? Therefore, 1/2*a*v^2= force x distance? Can anyone explain the help relationship?


Well let's say you have constant force on an object accelerating from rest.
f=ma, so ma is constant.
 E=\int f dx=\int m\frac{d^2x}{dt^2}dx
But dx = \frac{dx}{dt}dt
So  E = \int m\frac{dv}{dt}v dt
Which by chain rule is:
 E = \int mv dv = \frac{1}{2}mv^2 + C

You can also come to the same conclusion starting with gravitational potential energy (mgh) and using the algebraic kinematics equations (v^2-u^2=2*g*h)
So starting at u=0
mv^2=2mgh=2E ->E=1/2mv^2

Edited by Schrödinger's hat, 17 September 2011 - 01:53 AM.

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#3 ajb

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Posted 17 September 2011 - 09:58 AM

You can get at it without calculus, Quite rightly you say that

"work done = change in kinetic energy".

So let us have a constant force F acting over a distance s acting on a particle of mass m. Then

Fs = \Delta KE.

Now use Newton's law

\Delta KE = m a s

where a is the acceleration of the particle.

Now use

v^{2}= u^{2} + 2as,

where v is our "final velocity" and u our "initial velocity" as defined by the distance. Now let us further assume that the particle was initially at rest and then is accelerated by the force over the distance. Then u=0 and we define the KE at velocity zero to be zero. Then

KE = \frac{1}{2}mv^{2},

for an arbitrary velocity v.
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#4 mathematic

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Posted 17 September 2011 - 11:38 PM

Another approach is using special relativity. E=Mc^2.
M=m/√(1-(v/c)^2)
Therefore E=mc^2 + (mv^2)/2 + ....
The first term is the rest energy and the second term is the kinetic energy.

Here M = total mass, m = rest mass.
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#5 ajb

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Posted 18 September 2011 - 05:51 PM

Another approach is using special relativity. E=Mc^2.
M=m/√(1-(v/c)^2)
Therefore E=mc^2 + (mv^2)/2 + ....
The first term is the rest energy and the second term is the kinetic energy.

Here M = total mass, m = rest mass.


This is an over complication and the notion of total mass has fallen out of popularity in modern physics, but OK. However I don't think it help Vay very much.

Anyway, In my opinion we should define the kinetic energy in special relativity as

KE=  \sqrt{m^{2}c^{4} + p^{2}c^{2}} - mc^{2},

which is the total energy minus the rest energy, I assume we have no external potential to worry about.

Then we should think about the "Newtonian" limit in which is momenta is small . So, let us expand the KE about p=0 and to first order you get

KE = \frac{p^{2}}{2m},

which is what we want. You just put p = mv in to the above and you end up at the "classical" kinetic energy term.

As an aside one also has the ultra relativistic limit in which p\rightarrow \infty or equivalently m \rightarrow 0. In this case we get to first order in expanding the KE about very large p

KE = pc,

which is of course what we would expect.

Edited by ajb, 18 September 2011 - 06:40 PM.

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#6 Wilmot McCutchen

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Posted 18 September 2011 - 08:46 PM

I don't see an answer to Vay's question. He is an elementary physics student who needs the big picture. Vay asks a reasonable question: where does the 1/2 come from? He probably won't understand an explanation expressed with unfamiliar symbols and variables. Relativity theory is probably not familiar either.

His question got me wondering, since there seems to be no ready answer. When we encounter classical mechanics as part of required coursework, time pressure leaves no leisure for such ruminations. Maybe others on this forum, who have more deeply considered this question, can provide a better answer than my ignorant speculation, but in good faith, here goes my conjecture:

Momentum is always conserved, so you can add it moment to moment as the swarm of matter proceeds along a path. Summing an infinite quantity of infinitesimal momenta over a period of time and over a distance (in a definite direction, since we are considering kinetic energy rather than a diffuse form like heat) we integrate, to arrive at the expression for kinetic energy: 1/2* kg*m^2/s^2. Integration of momentum (kg*m/s) is what gives the 1/2 by the rules of calculus.

So now I'm wondering, where did the 1/2 go in E = MC^2?
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#7 ajb

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Posted 18 September 2011 - 08:58 PM

I don't see an answer to Vay's question. He is an elementary physics student who needs the big picture. Vay asks a reasonable question: where does the 1/2 come from? He probably won't understand an explanation expressed with unfamiliar symbols and variables.


The most direct answer without knowing any calculus, but just accepting the basic equations of motion in one dimension is the one I gave using work done.

... 1/2 by the rules of calculus.


Yes, this is really the origin of the half, whatever approach you use.
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#8 DrRocket

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Posted 18 September 2011 - 09:43 PM

I don't see an answer to Vay's question. He is an elementary physics student who needs the big picture. Vay asks a reasonable question: where does the 1/2 come from? He probably won't understand an explanation expressed with unfamiliar symbols and variables. Relativity theory is probably not familiar either.

His question got me wondering, since there seems to be no ready answer. When we encounter classical mechanics as part of required coursework, time pressure leaves no leisure for such ruminations. Maybe others on this forum, who have more deeply considered this question, can provide a better answer than my ignorant speculation, but in good faith, here goes my conjecture:

Momentum is always conserved, so you can add it moment to moment as the swarm of matter proceeds along a path. Summing an infinite quantity of infinitesimal momenta over a period of time and over a distance (in a definite direction, since we are considering kinetic energy rather than a diffuse form like heat) we integrate, to arrive at the expression for kinetic energy: 1/2* kg*m^2/s^2. Integration of momentum (kg*m/s) is what gives the 1/2 by the rules of calculus.

So now I'm wondering, where did the 1/2 go in E = MC^2?


ajb and Schrodinger's hat each provided explanations accessible to beginning physics students, and in the case of ajb an explanation devoid of calculus that could be understood by most high school students.

The path to education involves some work on one's part to elevate one's understanding to the level required by clear concise explanations, such as those provided. Expecting answers to be dumbed down to one's current level of ignorance only prolongs that undesirable state.
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#9 The Crow

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Posted 25 October 2011 - 05:28 AM

Mr. Wilmot has perhaps forgotten that the mc^2 in einstien's relation includes mass energy which is redundant in classical mechanics. The kinetic energy is energy solely due to motion of a body, so it is natural to get it from the quantity that represents amount of motion in a body, that is , momentum. however, momentum is not conserved in presence of net external forces, which is basically present in all mechanics, so the better derivation is by the work-energy theorem.
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#10 Michael Lee

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Posted 21 June 2013 - 04:22 AM

Perhaps it will help if we derive the formula without using calculus.

This works only with uniform acceleration.

 

Fd = KE

F = ma

 

so,  mad = KE

 

Draw a graph with time on the x axis and velocity on the y axis, then plot

a line at the origin with a slope equal to the acceleration of the body.  This

tells you how fast the body is moving at a given moment of time.  Now d is equal

to the area under the line for a given interval of time  and it's equal to vt/2.  

 

So, KE = ma(vt/2)

 

KE = m ta v/2        ta = v

 

KE = 1/2 m v2


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#11 Michael Lee

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Posted 3 July 2013 - 01:42 AM

It's nonsense, sorry.


Edited by Michael Lee, 3 July 2013 - 02:10 AM.

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#12 hypervalent_iodine

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Posted 3 July 2013 - 01:45 AM

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#13 Iwonderaboutthings

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Posted 27 July 2013 - 08:37 PM

So the formula for Kinetic energy is KE=1/2*a*v^2

Where does the 1/2 come from?

And the square is because the work required exponentially increases?

KE = work, so that means work = 1/2*a*v^2? Therefore, 1/2*a*v^2= force x distance? Can anyone explain the help relationship?

The 1/2 and exponential increases has to do with precession.  The exponent drives the force and the 1/2 precesses to a completion, within " this" formula.. The 1/2 thus become =1 squared due to the 1/2 and the exponent looping and completion within the boundaries rather limits of the formula.


I don't see an answer to Vay's question. He is an elementary physics student who needs the big picture. Vay asks a reasonable question: where does the 1/2 come from? He probably won't understand an explanation expressed with unfamiliar symbols and variables. Relativity theory is probably not familiar either.

His question got me wondering, since there seems to be no ready answer. When we encounter classical mechanics as part of required coursework, time pressure leaves no leisure for such ruminations. Maybe others on this forum, who have more deeply considered this question, can provide a better answer than my ignorant speculation, but in good faith, here goes my conjecture:

Momentum is always conserved, so you can add it moment to moment as the swarm of matter proceeds along a path. Summing an infinite quantity of infinitesimal momenta over a period of time and over a distance (in a definite direction, since we are considering kinetic energy rather than a diffuse form like heat) we integrate, to arrive at the expression for kinetic energy: 1/2* kg*m^2/s^2. Integration of momentum (kg*m/s) is what gives the 1/2 by the rules of calculus.

So now I'm wondering, where did the 1/2 go in E = MC^2?

1/2 by the rules of calculus

Very well said!


Edited by Iwonderaboutthings, 27 July 2013 - 08:38 PM.

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#14 swansont

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Posted 28 July 2013 - 12:45 AM

The 1/2 and exponential increases has to do with precession.  The exponent drives the force and the 1/2 precesses to a completion, within " this" formula.. The 1/2 thus become =1 squared due to the 1/2 and the exponent looping and completion within the boundaries rather limits of the formula.

The 1/2 has nothing to do with precession.
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#15 Iwonderaboutthings

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Posted 29 July 2013 - 02:19 AM

The 1/2 has nothing to do with precession.

Can you proof this mathematically????

 

I'm sorry but I hold firm on this 1/2 "makes" something missing..

That is a simple math fact below.

 

1/2 = 0.5
 
2*0.5 = 1
 
1^2 = 1

 

UNLESS of coarse you mean that .5 = a ratio

"torque-free precession"

Then I can understand this.

 

Could this be conservation? E=mc squared then?

 

Lorents transformations uses 1 as a numerator, but still does not explain what this 1 is = too.

It simply divides this by a denominator denominations.

 

http://hyperphysics....tiv/ltrans.html

 

I think it is very clear to many that 1/2 is not fully understood.

However science does deal with precession, predictions and closer than average accuracy at the mercy of

"WAVES."

 

For this it would then be deductive and logical to "assume" " precession" for this 1/2 usage as waves.

 

There is a larger part missing in the whole of cycles which = some type of precession.

 

Rather this be harmonic ocsillation, music theory, color theory or what ever..

 

 

There is some thing...

 

 

 


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#16 swansont

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Posted 29 July 2013 - 09:15 AM

Can you proof this mathematically????

 
If you integrate Fdx for a constant force (F=ma), you get kinetic energy. There is no rotational motion in the problem, so there is no precession
 
 

I'm sorry but I hold firm on this 1/2 "makes" something missing..
That is a simple math fact below.
 
1/2 = 0.5
 
2*0.5 = 1
 
1^2 = 1
 
UNLESS of coarse you mean that .5 = a ratio
"torque-free precession"
Then I can understand this.
 
Could this be conservation? E=mc squared then?
 
Lorents transformations uses 1 as a numerator, but still does not explain what this 1 is = too.
It simply divides this by a denominator denominations.
 
http://hyperphysics....tiv/ltrans.html
 
I think it is very clear to many that 1/2 is not fully understood.
However science does deal with precession, predictions and closer than average accuracy at the mercy of
"WAVES."
 
For this it would then be deductive and logical to "assume" " precession" for this 1/2 usage as waves.
 
There is a larger part missing in the whole of cycles which = some type of precession.
 
Rather this be harmonic ocsillation, music theory, color theory or what ever..
 
 
There is some thing...


The 1/2 is because integration is the area under a curve. Conceptually, it's related to the area of a triangle being 1/2 base*height. A triangle is half of a rectangle.
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#17 Iwonderaboutthings

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Posted 29 July 2013 - 11:03 PM

 
If you integrate Fdx for a constant force (F=ma), you get kinetic energy. There is no rotational motion in the problem, so there is no precession
 
 

The 1/2 is because integration is the area under a curve. Conceptually, it's related to the area of a triangle being 1/2 base*height. A triangle is half of a rectangle.

What about conservation? wink.png


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#18 Griffon

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Posted 30 July 2013 - 08:59 AM

I don't see an answer to Vay's question. He is an elementary physics student who needs the big picture. Vay asks a reasonable question: where does the 1/2 come from? He probably won't understand an explanation expressed with unfamiliar symbols and variables. Relativity theory is probably not familiar either.

I agree with you. I sometimes wonder whether there's an element of 'look how clever I am' in answers on science boards. Ajb's answer was about the right level I'd have thought. And now there are a range of answers which, if the OP is willing to spend time on them, he should be able to glean an understanding at several levels.

Regarding the fact that KE is proportional to velocity squared, I have long felt this to be unintuitive. You might naively expect that doubling the velocity would provide an object with twice the energy rather than four times. But that is just the way it is from the definitions of Work (force x distance) and Newton's second law (Force = mass x acceleration), as contained in ajb's simple explanation.
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#19 swansont

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Posted 30 July 2013 - 09:29 AM

What about conservation? wink.png


Conservation of what?

I agree with you. I sometimes wonder whether there's an element of 'look how clever I am' in answers on science boards. Ajb's answer was about the right level I'd have thought. And now there are a range of answers which, if the OP is willing to spend time on them, he should be able to glean an understanding at several levels.

 

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.


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#20 Griffon

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Posted 30 July 2013 - 12:27 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! :-)

Edited by Griffon, 30 July 2013 - 12:27 PM.

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