Why is the kinetic energy equation multiplied by ½?

[rajen0201]

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.

The ultimate source of energy is electromagnetic energy.
ALso Ibix said: "This will not lead to equal shares of the kinetic energy" Is it not violating Newton's third law? The multiplication of Earth's MeVe^2 must equal to mv^2 of the ball to satisfy Newton's third law.
example:1 of an astronomer away from his space station with no string/rope attached. to get back into the station, he throws items such as a hammer in opposite direction so, as per Newton's third law, he finally servives. What amount of energy he spent?
example:2 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.

 

[Ibix]

Then how Newton's third law satisfied here?

Requiring the conservation of momentum satisfies it. Newton's third law does not require that equal energy be supplied to both bodies, just that equal forces be applied.

 

[Ibix]

The multiplication of Earth's MeVe^2 must equal to mv^2 of the ball to satisfy Newton's third law.

No. Newton's third law says that the applied forces, and hence the momentum changes, must be equal and opposite for the two bodies. This does not mean that the energy changes must be equal.

 

[jack action]

Like @Ibix said, Newton's third law required only the forces to be equal. Energy Power is force times velocity and velocities can be different. So even if the energy power is conserved, it may be split in any proportion between the two.

The best example would a ball bouncing back on a wall. No matter what, the wall is fixed, thus its velocity is zero before and after the hit. Any force felt by the wall, multiplied by zero velocity gives zero energy power. Therefore whatever energy the ball comes with, it goes back with it (only in a different direction).

Edit: I incorrectly used the definition of power for energy. But the principle of conservation of energy extends to power, so I think my point is still valid. Or one could change the term 'velocity' for 'displacement' and the point would still be valid.

 

[rajen0201]

Requiring the conservation of momentum satisfies it. Newton's third law does not require that equal energy be supplied to both bodies, just that equal forces be applied.

Force is the rate at which momentum changes with respect to time (F = dp/dt). Note that if p = mv and m is constant, then F = dp/dt = m*dv/dt = ma so, that is force. So, conservation of momentum also satisfy Newton's third law.

 

[rajen0201]

Requiring the conservation of momentum satisfies it. Newton's third law does not require that equal energy be supplied to both bodies, just that equal forces be applied.

Force is the rate at which momentum changes with respect to time (F = dp/dt). Note that if p = mv and m is constant, then F = dp/dt = m*dv/dt = ma so, that is force. So, conservation of momentum also satisfies Newton's third law.

 

[rajen0201]

No. Newton's third law says that the applied forces, and hence the momentum changes, must be equal and opposite for the two bodies. This does not mean that the energy changes must be equal.

Here, energy doesn't change over time. it must same(1/2mV2+mgh) at every time segment to conserve the law of energy. the energy given 1/2mV2 to ball must be the same at any location.

 

[Ibix]

Here, energy doesn't change over time. it must same(1/2mV2+mgh) at every time segment to conserve the law of energy. the energy given 1/2mV2 to ball must be the same at any location.

You seem to be confusing separate issues here. Let's start with your boat example, a simpler case.

Let the two boats have masses $m$ and $M$ and they are both initially at rest. The man stands in one boat and pushes the other, so the boat of mass $m$ is doing velocity $v$ and the boat of mass $M$ is doing velocity $V$. We'll neglect water resistance here.

The initial momentum was zero because nothing was moving. The final momentum must also, therefore, be zero. That is,$$\begin{eqnarray*}
0&=&mv+MV\\
v&=&-\frac MmV
\end{eqnarray*}$$Newton's third law is satisfied by this - the same force was applied to the boats for the same time giving them the same momentum with opposite signs. But this does not mean they have the same energy. Let the kinetic energy of the boats be $E_m=mv^2/2$ and $E_M=MV^2/2$. But we know that $v=-MV/m$, so we can write$$\begin{eqnarray*}
E_m&=&\frac 12 m\left(\frac MmV\right)^2\\
&=&\frac 12\frac MmMV^2\\
&=&\frac MmE_M
\end{eqnarray*}$$So we find that the boats do have equal momenta (whatever their mass) but do not have equal energies (unless their masses are equal). Note that I have not applied the conservation of energy here - the energy that the boats gain is supplied by chemical energy stored in the man's body, but there is no need to discuss this to get the result above.

Are you following so far?

 

[gmax137]

But we know that $v=-MV/v$, so we can write

Typo here; you mean
"But we know that $v=-MV/m$, so we can write..."

 

[Ibix]

Typo here; you mean
"But we know that $v=-MV/m$, so we can write..."

Thanks - corrected.

 

[snorkack]

Vis viva was originally defined as mv² by Leibniz.
The expression of mv²/2 was offered by Bernoulli and occasionally since but was made prevalent by works of Coriolis and Poncelet in 1819-1839.

What reasons did Coriolis and Poncelet give to prove that mv²/2 was better than the traditional mv²?

 

[rajen0201]

You seem to be confusing separate issues here. Let's start with your boat example, a simpler case.

Let the two boats have masses $m$ and $M$ and they are both initially at rest. The man stands in one boat and pushes the other, so the boat of mass $m$ is doing velocity $v$ and the boat of mass $M$ is doing velocity $V$. We'll neglect water resistance here.

The initial momentum was zero because nothing was moving. The final momentum must also, therefore, be zero. That is,$$\begin{eqnarray*}
0&=&mv+MV\\
v&=&-\frac MmV
\end{eqnarray*}$$Newton's third law is satisfied by this - the same force was applied to the boats for the same time giving them the same momentum with opposite signs. But this does not mean they have the same energy. Let the kinetic energy of the boats be $E_m=mv^2/2$ and $E_M=MV^2/2$. But we know that $v=-MV/m$, so we can write$$\begin{eqnarray*}
E_m&=&\frac 12 m\left(\frac MmV\right)^2\\
&=&\frac 12\frac MmMV^2\\
&=&\frac MmE_M
\end{eqnarray*}$$So we find that the boats do have equal momenta (whatever their mass) but do not have equal energies (unless their masses are equal). Note that I have not applied the conservation of energy here - the energy that the boats gain is supplied by chemical energy stored in the man's body, but there is no need to discuss this to get the result above.

Are you following so far?

You seem to be confusing separate issues here. Let's start with your boat example, a simpler case.

Let the two boats have masses $m$ and $M$ and they are both initially at rest. The man stands in one boat and pushes the other, so the boat of mass $m$ is doing velocity $v$ and the boat of mass $M$ is doing velocity $V$. We'll neglect water resistance here.

The initial momentum was zero because nothing was moving. The final momentum must also, therefore, be zero. That is,$$\begin{eqnarray*}
0&=&mv+MV\\
v&=&-\frac MmV
\end{eqnarray*}$$Newton's third law is satisfied by this - the same force was applied to the boats for the same time giving them the same momentum with opposite signs. But this does not mean they have the same energy. Let the kinetic energy of the boats be $E_m=mv^2/2$ and $E_M=MV^2/2$. But we know that $v=-MV/m$, so we can write$$\begin{eqnarray*}
E_m&=&\frac 12 m\left(\frac MmV\right)^2\\
&=&\frac 12\frac MmMV^2\\
&=&\frac MmE_M
\end{eqnarray*}$$So we find that the boats do have equal momenta (whatever their mass) but do not have equal energies (unless their masses are equal). Note that I have not applied the conservation of energy here - the energy that the boats gain is supplied by chemical energy stored in the man's body, but there is no need to discuss this to get the result above.

Are you following so far?

You seem to be confusing separate issues here. Let's start with your boat example, a simpler case.

Let the two boats have masses $m$ and $M$ and they are both initially at rest. The man stands in one boat and pushes the other, so the boat of mass $m$ is doing velocity $v$ and the boat of mass $M$ is doing velocity $V$. We'll neglect water resistance here.

The initial momentum was zero because nothing was moving. The final momentum must also, therefore, be zero. That is,$$\begin{eqnarray*}
0&=&mv+MV\\
v&=&-\frac MmV
\end{eqnarray*}$$Newton's third law is satisfied by this - the same force was applied to the boats for the same time giving them the same momentum with opposite signs. But this does not mean they have the same energy. Let the kinetic energy of the boats be $E_m=mv^2/2$ and $E_M=MV^2/2$. But we know that $v=-MV/m$, so we can write$$\begin{eqnarray*}
E_m&=&\frac 12 m\left(\frac MmV\right)^2\\
&=&\frac 12\frac MmMV^2\\
&=&\frac MmE_M
\end{eqnarray*}$$So we find that the boats do have equal momenta (whatever their mass) but do not have equal energies (unless their masses are equal). Note that I have not applied the conservation of energy here - the energy that the boats gain is supplied by chemical energy stored in the man's body, but there is no need to discuss this to get the result above.

Are you following so far?

"The initial momentum was zero because nothing was moving. The final momentum must also, therefore, be zero."
How final momentum zero? both boats will continue to move with (infinite time like they are in full vacuum space) constant speed as there are no other forces considered. Momentum must be related to time as per F = dp/dt = m*dv/dt = ma
Force = mass x gravity value on earth. now multiply with height h = mgh that is work for object. same is applicable to Earth case. Here, gravity of object on Earth to be considered. So, mgh Earth =mgh object. And 1/2mv2 object = 1/2MV2 Earth.

 

[A.T.]

"The initial momentum was zero because nothing was moving. The final momentum must also, therefore, be zero."
How final momentum zero? both boats will continue to move ...

Momentum is a vector. Equal but opposite vectors cancel to zero.

 

[Ibix]

"The initial momentum was zero because nothing was moving. The final momentum must also, therefore, be zero." How final momentum zero? both boats will continue to move with (infinite time like they are in full vacuum space) constant speed as there are no other forces considered.

The momentum of each boat is non-zero, yes. The sum of the two, however, is always zero because of the conservation of momentum. That is, the momentum of one boat ($mv\neq 0$) and the momentum of the other ($MV\neq 0$) added together are zero.

 

[etotheipi]

Others have already said that momentum is a vector quantity but it's perhaps not immediately obvious from the formula $p = mv$ you see written. The reality is that $p$ and $v$ are actually components of the vector in a given direction (not magnitudes!) which can be negative (e.g. if the vector points in the opposite direction). It might be clearer to write the formula as $p_x = mv_x$ to make clear that they are vector components!

 

[weirdoguy]

but it's perhaps not immediately obvious from the formula p=mvp = mv you see written.

That is why an arrow above vector quantities was invented: $\vec{p}=m\vec{v}$

 

[etotheipi]

That is why an arrow above vector quantities was invented: $\vec{p}=m\vec{v}$

Indeed. But I for one found it very confusing for a long time that all my teachers would say "velocity is a vector quantity, it's represented by an arrow and it has a magnitude and a direction!" but then proceeded to treat it exactly as you would treat a scalar in all the bog-standard calculations.

Of course, what they were actually doing was manipulating the components, which are scalars. But this is far from clear at least in textbooks up to A Level standard!

 

[hutchphd]

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.

This would be true here if we "squealed the tires" and immediately got them up to final speed and "burned rubber" until the car made final speed. Then half the energy would go to heating the tires (just like charging a Capacitor from a fixed voltage source).

 

[russ_watters]

Vis viva was originally defined as mv² by Leibniz.
The expression of mv²/2 was offered by Bernoulli and occasionally since but was made prevalent by works of Coriolis and Poncelet in 1819-1839.

What reasons did Coriolis and Poncelet give to prove that mv²/2 was better than the traditional mv²?

The value of any equation or derived quantity is in what you can do with it. According to the wiki on vis viva, Leibniz discovered it simply by noticing that mv² was a conserved quantity. But that knowledge in itself has limited value if you can't easily use it to make connections with other principles. vis viva is pretty much just stand-alone and not very useful because it lacks those connections.

I don't know about Coriolis and Poncelet, but Bernoulli would have had to derive the correct relationship between different types of energy in order to use it in Bernoulli's equation. But what, exactly, these guys were thinking is a matter of history, not science. The path of discovery isn't always linear, so their reasons may not be relevant or even necessarily correct. So we don't have to know why they preferred KE to vis viva, we only have to know why we do.

 

[etotheipi]

vis viva is pretty much just stand-alone and not very useful because it lacks those connections.

Couldn't we pick up a copy of Goldstein, meticulously pore over every page replacing every $T$ with a $\frac{V}{2}$ (using $V \equiv \text{vis viva}$, perhaps a bad choice... ), and still have a coherent picture of mechanics with all the connections still in-tact?

 

[russ_watters]

Couldn't we pick up a copy of Goldstein, meticulously pore over every page replacing every $T$ with a $\frac{V}{2}$ (using $V \equiv \text{vis viva}$ ), and still have a coherent picture of mechanics with all the connections still in-tact?

I can't tell if you're joking or not, but without seeing the context I'd say "maybe", but why bother adding steps to a problem/equation that add no value? Instead of multiplying by 1/2 you could multiply it by 42 and name it TA, but if it doesn't do anything useful, nobody is going to use it. Again; the value in an equation is in what you can do with it.

 

[etotheipi]

I can't tell if you're joking or not, but without seeing the context I'd say "maybe", but why bother adding steps to a problem/equation that add no value? Instead of multiplying by 1/2 you could multiply it by 42 and name it TA, but if it doesn't do anything useful, nobody is going to use it. Again; the value in an equation is in what you can do with it.

I agree . I'm speculating here (please don't ban me ), but I would have thought one reason is that the equations are nicer. Conservation of mechanical energy is $T + U = k$, and the work energy theorem is $W = \Delta T$. They are easier to remember and feel more natural than if we had a factor of a half floating around. I don't think using vis viva would cause any problems, but kinetic energy seems to be the more natural choice in terms of "niceness".

 

[Vanadium 50]

Couldn't we pick up a copy of Goldstein, meticulously pore over every page replacing every TT with a V2\frac{V}{2} (using V≡vis vivaV \equiv \text{vis viva}, perhaps a bad choice... ), and still have a coherent picture of mechanics with all the connections still in-tact?

Sure, but you're going to end up putting a bunch of 2's in. You'll have twomentum and probably twomass in places.

The 1/2 comes from a real integral. It's there for a reason. Sure, you can "get rid" of it by a judicious multiplication by 2, but now you have extra 2's floating around, and instead of arguing why there is a 1/2 over here, we'll be arguing why there is a 2 over there.

 

[gmax137]

The 1/2 comes from a real integral.

Right, the 1/2 comes from the fact that the area of a triangle is 1/2 base * height. That's really all there is to it.

 

[etotheipi]

Sure, but you're going to end up putting a bunch of 2's in. You'll have twomentum and probably twomass in places.

Actually I wasn’t being anywhere near that clever. For explanatory purposes I define “frinetic energy, F” to be be $\frac{3}{2}mv^2$, and don’t change anything else. I discover that $\frac{F}{3}+ U$ is a conserved quantity, and that the change in frinetic energy is thrice the work done. And that frinetic energy is $\frac{3p^2}{2m}$. Anyway yes this would be stupid.

I think you’ve all heard enough from me anyway about this subject manner, so I will proceed to be quiet.

 

[Stephen Tashi]

We can derive the equation 1/2mV2.

"$(1/2)mV^2$" isn't an equation unless you set it equal to something.

So the question is why we define something to be equal to $(1/2)mV^2$.

Suppose we define "Work" by the equation $W = F D$ where $F$ is a constant force exerted on an object as it moves over a distance $D$. If we want to invent a concept called "Energy" and have the "Energy" done by the "Work" equal to the "Energy" present in the object after the work is done, we must have $E = W =$ some function of the objects motion.

Suppose the object has mass m and is initially at rest. Can you calculate it's velocity after the force $F$ has acted upon it for a distance $D$?

If you use the equation $F = MA$ in that calculation, you are doing something that depends on a choice of compatible physical units. For example, if $M$ is in kilograms and $A$ is in meters/sec^2 then $F$ must be measured in Newtons. If you measured $F$ in pounds, you'd have to use the more general equation $F = cMA$ where $c$ is a constant different than 1.

 

[russ_watters]

I agree . I'm speculating here (please don't ban me ), but I would have thought one reason is that the equations are nicer. Conservation of mechanical energy is $T + U = k$, and the work energy theorem is $W = \Delta T$. They are easier to remember and feel more natural than if we had a factor of a half floating around. I don't think using vis viva would cause any problems, but kinetic energy seems to be the more natural choice in terms of "niceness".

Those are rolled-up/simplified equations that just use the names of the terms, not the terms themselves. There is no half factor in them at all, regardless of how you define them. It's when you use the detailed equations that those factors come into play, and you can't get rid of them. If you remove a 1/2 somewhere you have to add a 2 somewhere else. Or, if as I'd prefer: 42/2 MV² = 42 mgh (for example)

 

[snorkack]

But what, exactly, these guys were thinking is a matter of history, not science. The path of discovery isn't always linear, so their reasons may not be relevant or even necessarily correct. So we don't have to know why they preferred KE to vis viva, we only have to know why we do.

The problem is that we seem to have forgotten, except "it was done that way before".

 

[russ_watters]

The problem is that we seem to have forgotten, except "it was done that way before".

Huh? Who has forgotten what? We know exactly why the 1/2 appears in the kinetic energy expression, and it most certainly is not "it was done that way before". Heck, that's basically never a valid approach and we only ever see it taken by the people who complain that they don't like the way things are done now (who hope/speculate that something they don't know contains an error someone should find if they dug through a history book). Science is "live" in that it is constantly being updated to incorporate what works best. The fact that this issue hasn't changed in a long time because it works best the way it is (it's a pretty simple issue) doesn't mean we have forgotten why it's done this way; it's practiced and demonstrated constantly that the way it is done works best. Your view on this issue appears to me to be exactly backwards.

 

[snorkack]

One of the things Coriolis did, in 1829, was to apply the word "travail" to the expression W=Fs.
That Fs is a relevant quantity was, however, long known. Already in Archimedes´ lever law in 3rd century BC, and generalized by Solomon de Caus in 1615.
Well before Newton and his laws.
What was Fs called before Coriolis renamed it "travail"?