Why is the kinetic energy equation multiplied by ½?

[rajen0201]

How did you find PF?: We can derive the equation 1/2mV2. But what is the fundamental reason behind mV2(total energy) is divided by half?

I find only derivative solutions to this question. But still unsatisfied.

 

[Vanadium 50]

What kind of answer would satisfy you?

 

[Dale]

We can derive the equation 1/2mV2. But what is the fundamental reason behind mV2(total energy) is divided by half?

I find only derivative solutions to this question. But still unsatisfied.

You will need to be much more specific. Which specific derivations have you seen and why are you still unsatisfied with them? Since you are already aware of a derivation, this is less a question on the validity of the formula and more a question about what would satisfy you. We need more information to answer that.

 

[etotheipi]

In a sense, constant terms out the front of an equation aren't fundamental at all, and are dependent on your choice of system of units. I might define a new unit for energy, the schmoule ($\text{Sc}$), such that $1\, \text{J} \equiv 2\, \text{Sc}$. If I choose to still measure mass in kilograms and velocity in metres per second, I realize that now my formula has to look like $T = mv^2$.

Edit

I have subsequently realized that this is not at totally accurate picture. We can at best recast the equation into the form $$T = \text{Sc}\text{J}^{-1} mv^2$$ where the factor $\text{Sc}\text{J}^{-1}$ equals one half. It was thus incorrect of me to omit the leading units.

The formula is always $T = \frac{1}{2}mv^2$, following by definition from consideration of the work-energy theorem, and this is independent of the chosen system of units.

 

[ZapperZ]

How did you find PF?: We can derive the equation 1/2mV2. But what is the fundamental reason behind mV2(total energy) is divided by half?

I find only derivative solutions to this question. But still unsatisfied.

This is a bit vague. You said that you know the derivation of the kinetic energy equation, but you now want to know the "fundamental reason" for it? What exactly do you mean by "fundamental reason"? A logical, mathematical derivation often trumps intuition and human reasoning. That is as "fundamental" as it gets.

So what exactly are you looking for here?

If you have a problem knowing the fundamental reason for the KE equation, do you also have a a problem understanding the fundamental reason for the kinematical equations derived from Newton's laws? After all, they all came from the same place. Is the "1/2" causing major headache in understanding here?

Zz.

 

[Lnewqban]

How did you find PF?: We can derive the equation 1/2mV2. But what is the fundamental reason behind mV2(total energy) is divided by half?

I find only derivative solutions to this question. But still unsatisfied.

This may be incorrect, but it is the way in which I understand the reason of your question:

It is half the area of the rectangle formed in the velocity versus time graph.
For the body represented in the graph to gain momentum (its velocity to steadily increase), a steady force is applied onto it; therefore, we have a line of constant slope that divides a rectangle exactly in half.

More about the same here:
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Please, see:

 

[gmax137]

But what is the fundamental reason behind mV2(total energy) is divided by half?

Are you under the impression that the "total energy" is $= mv^2$ ? Where did you get that idea? It isn't correct.

 

[A.T.]

It is half the area of the rectangle formed in the velocity versus time graph.

More generally, the form 1/2 c x² often pops up, if a constant was integrated twice.

 

[Herman Trivilino]

The factor of one-half is a convention.

 

[samalkhaiat]

In a sense, constant terms out the front of an equation aren't fundamental at all, and are dependent on your choice of system of units. I might define a new unit for energy, the schmoule ($\text{Sc}$), such that $1\, \text{J} \equiv 2\, \text{Sc}$. If I choose to still measure mass in kilograms and velocity in metres per second, I realize that now my formula has to look like $T = mv^2$.

The factor \frac{1}{2} in the kinetic energy expression, which is a dimensionless number, is independent of any unit system. It comes from T = m \int_{0}^{v} v dv, or more accurately from the classical limit of the relativistic dispersion relation \mathcal{E} - m = m \left( \sqrt{1 + (\frac{p}{m})^{2}} - 1 \right) \approx \frac{1}{2} \frac{p^{2}}{m} . Notice that the (1/2) poped up even though I used a units system in which the speed of light is one c = 1.

 

[etotheipi]

The factor \frac{1}{2} in the kinetic energy expression, which is a dimensionless number, is independent of any unit system.

I'll assume you're right about this, I'm just not too sure why. If I make up a weird system of units identical to the SI units except that the unit of mass is now $u_M = n\text{kg}$, won't the formula for kinetic energy change? It seems to me that it would become $T = \frac{n}{2}mv^2$, if $T$ is still measured in joules.

 

[samalkhaiat]

I'll assume you're right about this, I'm just not too sure why. If I make up a weird system of units identical to the SI units except that the unit of mass is now $u_M = n\text{kg}$, won't the formula for kinetic energy change? It seems to me that it would become $T = \frac{n}{2}mv^2$, if $T$ is still measured in joules.

It does not matter which unit you measure the energy with. Joule, eV or even Kg are irrelevant units. You can have an object with 1000 “unit of mass” (any unit of mass), but the half will still be in the expression for T. In orher words: the half is not there because we measure the energy by Joules.

 

[etotheipi]

It does not matter which unit you measure the energy with. Joule, eV or even Kg are irrelevant units. You can have an object with 1000 “unit of mass” (any unit of mass), but the half will still be in the expression for T. In orher words: the half is not there because we measure the energy by Joules.

That makes sense. I must be overlooking something whilst messing around with these units. I got the idea because it's possible for certain formulae to change form in different systems of units (e.g. EM in Gaussian vs SI units), however that's perhaps distinct to this. I'll have another look at this in the morning!

 

[Vanadium 50]

. If I make up a weird system of units identical to the SI units except that the unit of mass is now uM=nkgu_M = n\text{kg}, won't the formula for kinetic energy change?

Sure. So will the expression for momentum, and force, and probably a zillion others. This turns a set of simple equations into a horrible mess. One that still has relative factors of 1/2 floating around.

 

[jack action]

First, there are no defintions for energy, only difference in energy is defined:
$$
\Delta E =F\Delta x
$$
This can easily be transformed to:
$$\begin{split}
\Delta E &=F\Delta x \\
&=m\frac{\Delta v}{\Delta t}\Delta x \\
&=m\frac{\Delta x}{\Delta t}\Delta v \\
&= m\bar{v}\Delta v
\end{split}
$$
It is important to note that $\frac{\Delta x}{\Delta t}$ is the average velocity between the two states. Therefore, the difference in energy between state $1$ and state $2$:
$$\begin{split}
E_2 - E_1 &= m\bar{v}\Delta v \\
&= m\frac{v_1+ v_2}{2}(v_2 - v_1) \\
&= m\frac{v_2^2- v_1^2}{2} \\
&= \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2
\end{split}
$$
We should stop here, but one could extrapolate this to:
$$
\begin{split}
E_1 &= \frac{1}{2}mv_1^2 \\
E_2 &= \frac{1}{2}mv_2^2
\end{split}
$$
Or in the general case:
$$E = \frac{1}{2}mv^2$$
Thus the $\frac{1}{2}$ comes from the halving done when calculating the average velocity.

 

[etotheipi]

Sure. So will the expression for momentum, and force, and probably a zillion others. This turns a set of simple equations into a horrible mess. One that still has relative factors of 1/2 floating around.

This was what I thought. The equation for work might just change to e.g. $W = 2Fd$ and then when you integrate up to get $T$ you lose that factor of $\frac{1}{2}$. GPE becomes $2mgh$, everything scales accordingly. The relative factors of $\frac{1}{2}$ would still be there, but not in the defining equation for $T$. It was my understanding that a change in the system of units can result in a change in the dimensionless coefficients in physical laws.

However this appears to disagree with what @samalkhaiat suggested. And if in doubt it's probably me that's in the wrong , so I'd need to have another look when less tired.

 

[samalkhaiat]

$T$. It was my understanding that a change in the system of units can result in a change in the dimensionless coefficients in physical laws.

That is the biggest mistake you can do. Nothing more wrong than saying 1 = 2. You can always play with dimensionfull quantities (following well defined rules), but you cannot do the same with dimensionless quantities which are just real numbers.

 

[etotheipi]

Perhaps this is it; the formula in my example only changes because we are still measuring energy in joules, and joules is derived from the kilogram. So I'm cheating. If I were meticulous in redefining all quantities in terms of my new base quantities, and the new unit of energy were 1 new mass unit times 1 square metre times second to the negative two, then I think you would retain the factor of $\frac{1}{2}$. This is in agreement with your argument.

 

[tech99]

We see the same thing with LI^2 / 2 and CV^2 / 2.
I thought it was because half the energy of the prime mover is dissipated as heat during the energy transfer.

 

[samalkhaiat]

Perhaps this is it; the formula in my example only changes because we are still measuring energy in joules, and joules is derived from the kilogram. So I'm cheating. If I were meticulous in redefining all quantities in terms of my new base quantities, and the new unit of energy were 1 new mass unit times 1 square metre times second to the negative two, then I think you would retain the factor of $\frac{1}{2}$. This is in agreement with your argument.

Yes, for example, you can never get rid of the 2 in the 1/r^{2} law.

 

[rajen0201]

Are you under the impression that the "total energy" is $= mv^2$ ? Where did you get that idea? It isn't correct.

For example, take a ball and throw it radially upward. you are pushing the Earth same time as per Newton's third law. So, total energy(E=MC^2 chemical energy) used by man throwing ball must be divide in two equal and opposite part. Other example is when we push a empty boat from other boat, both boat will away with respect to their mass. Is this logic is correct?

 

[rajen0201]

I'll assume you're right about this, I'm just not too sure why. If I make up a weird system of units identical to the SI units except that the unit of mass is now $u_M = n\text{kg}$, won't the formula for kinetic energy change? It seems to me that it would become $T = \frac{n}{2}mv^2$, if $T$ is still measured in joules.

For example, take a ball and throw it radially upward. you are pushing the Earth same time as per Newton's third law. So, total energy(E=MC^2 chemical energy) used by man throwing ball must be divide in two equal and opposite part. Other example is when we push a empty boat from other boat, both boat will away with respect to their mass. Is this logic is correct?

 

[rajen0201]

First, there are no defintions for energy, only difference in energy is defined:
$$
\Delta E =F\Delta x
$$
This can easily be transformed to:
$$\begin{split}
\Delta E &=F\Delta x \\
&=m\frac{\Delta v}{\Delta t}\Delta x \\
&=m\frac{\Delta x}{\Delta t}\Delta v \\
&= m\bar{v}\Delta v
\end{split}
$$
It is important to note that $\frac{\Delta x}{\Delta t}$ is the average velocity between the two states. Therefore, the difference in energy between state $1$ and state $2$:
$$\begin{split}
E_2 - E_1 &= m\bar{v}\Delta v \\
&= m\frac{v_1+ v_2}{2}(v_2 - v_1) \\
&= m\frac{v_2^2- v_1^2}{2} \\
&= \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2
\end{split}
$$
We should stop here, but one could extrapolate this to:
$$
\begin{split}
E_1 &= \frac{1}{2}mv_1^2 \\
E_2 &= \frac{1}{2}mv_2^2
\end{split}
$$
Or in the general case:
$$E = \frac{1}{2}mv^2$$
Thus the $\frac{1}{2}$ comes from the halving done when calculating the average velocity.

Is it not possible that E is itself 1/2E ? For example, take a ball and throw it radially upward. you are pushing the Earth same time as per Newton's third law. So, total energy(E=MC^2 chemical energy) used by man throwing ball must be divide in two equal and opposite part. Other example is when we push a empty boat from other boat, both boat will away with respect to their mass. Is this logic is correct?

 

[PeroK]

For example, take a ball and throw it radially upward. you are pushing the Earth same time as per Newton's third law. So, total energy(E=MC^2 chemical energy) used by man throwing ball must be divide in two equal and opposite part. Other example is when we push a empty boat from other boat, both boat will away with respect to their mass. Is this logic is correct?

Momentum is equal and opposite, but the ball has almost all the energy. You can see this from:
$$mv = MV \ \Rightarrow \ V = \frac m M v$$
Hence:
$$KE_{Earth} = \frac 1 2 MV^2 = \frac 1 2 M (\frac m M v)^2 = (\frac m M) \frac 1 2 mv^2 = (\frac m M) KE_{ball}$$
And the Earth has almost no KE compared to the ball, because of its huge mass $M$.

 

[Ibix]

Is this logic is correct?

No. You can't have "equal and opposite" parts of energy, and if you could the total would be zero. The ball and the Earth will have equal and opposite momenta, but this will not lead to equal shares of the kinetic energy. If the ball has mass $m$ and the Earth has mass $m_E$ then the ball gets $m_E/m$ times more kinetic energy than the Earth in their center of mass frame.

 

[etotheipi]

For example, take a ball and throw it radially upward. you are pushing the Earth same time as per Newton's third law. So, total energy(E=MC^2 chemical energy) used by man throwing ball must be divide in two equal and opposite part. Other example is when we push a empty boat from other boat, both boat will away with respect to their mass. Is this logic is correct?

This is quite confused. $mc^2$ is not chemical energy, and it doesn't come into play in classical mechanics. And furthermore $mc^2$ is not the same as double $\frac{1}{2}mv^2$.

Newton's third law pertains to forces. It generally tells you nothing about how energy is divided up (except in a few specific cases). In a system of two interacting bodies, one could be held at rest and the other moved (in which case only work is done on the latter), or they could both move, etc.

 

[Ibix]

This is quite confused. $mc^2$ is not chemical energy, and it doesn't come into play in classical mechanics. And furthermore $mc^2$ is not the same as double $\frac{1}{2}mv^2$.

Actually, there will be a tiny reduction in the mass of the reagents if a chemical reaction releases energy. It's immeasurably tiny (a few kJ divided by $c^2$ is almost nothing), and I agree that it isn't relevant at all to this discussion.

 

[etotheipi]

Actually, there will be a tiny reduction in the mass of the reagents if a chemical reaction releases energy. It's immeasurably tiny (a few kJ divided by $c^2$ is almost nothing), and I agree that it isn't relevant at all to this discussion.

Sure, I meant it in the sense rest energy is not the same thing as chemical energy, although the latter might well contribute to the former

 

[sophiecentaur]

For example, take a ball and throw it radially upward. you are pushing the Earth same time as per Newton's third law. So, total energy(E=MC^2 chemical energy) used by man throwing ball must be divide in two equal and opposite part.

This is not correct. Momentum is conserved so that the Earth'smomentum change is the same as minus the ball's Momentum change. But the Energy is not 'shared' in any way. For a large Earth, the Energy transferred is Zero. Kinetic Energy is proportional to v², remember so the 'share' is the square of the ratio of speeds.

 

[rajen0201]

Perhaps this is it; the formula in my example only changes because we are still measuring energy in joules, and joules is derived from the kilogram. So I'm cheating. If I were meticulous in redefining all quantities in terms of my new base quantities, and the new unit of energy were 1 new mass unit times 1 square metre times second to the negative two, then I think you would retain the factor of $\frac{1}{2}$. This is in agreement with your argument.

Is it not possible that E is itself 1/2E ? For example, take a ball and throw it radially upward. you are pushing the Earth same time as per Newton's third law. So, total energy(E=MC^2 chemical energy) used by man throwing ball must be divide in two equal and opposite part. Other example is when we push a empty boat from other boat, both boat will away with respect to their mass. Is this logic is correct?

No. You can't have "equal and opposite" parts of energy, and if you could the total would be zero. The ball and the Earth will have equal and opposite momenta, but this will not lead to equal shares of the kinetic energy. If the ball has mass $m$ and the Earth has mass $m_E$ then the ball gets $m_E/m$ times more kinetic energy than the Earth in their center of mass frame.

Then how Newton's third law satisfied here? take another example, A man seating in a boat without paddles. but there is another boat near him, So, he pushes it. So, it allows him to move in the opposite direction. how much energy did he spend? Here man energy is ultimately chemical energy that converts in mechaniical.