Same old question, I still don't get it: E=1/2mv^2 - more E with V?

[ryanabc67]

TL;DR Summary

Q: "does it take more energy to accelerate at higher speeds"
A: The 'usual answer': "Your kinetic energy grows exponentially"....yeah, but how much FUEL do you burn?

I still don't get this and most answers all over kind of miss the point and just kind of recite without really explaining.
Here's the dilemma. Yes, kinetic energy grows exponentially with speed. So a 1kg object changing from 0 to 1m/s increases by .5J. And from 4 to 5m/s increases from 8J to 12.5J = 4.5J.

But the question is NOT how much kinetic energy does a body have as it accelerates. The question is, how much energy does it REQUIRE to accelerate? This is not the same question.

Clearly, speed is totally relative and therefore, so is kinetic energy. If you are running towards me at 4m/s and your friend is stationary when I begin moving towards you both, my kinetic energy increase relative to you is 4.5J but only .5J relative to your friend, so, no, sorry, CHANGE IN KINETIC ENERGY DOES NOT EQUAL CHANGE IN REQUIRED ENERGY.

The question is about how much energy is required to accelerate and given the laws of "relativity" (Galilean), when I'm travelling 4m/s .... relative to me - I'm stationary. No matter how fast I'm travelling, there is no speed that can be assigned to me. Whatever speed 'you' think I'm travelling is just your own opinion and has nothing to do with me. Everyone will have a different opinion on that, depending on what direction they are moving and how fast.

So my speed is always zero and every increase I make is from 0 to 1 (let's assume I pause after each increase for a bit).

What doesn't make sense however, is the following scenario.
Caveat: I'm NOT arguing with the law of conservation of energy. Clearly THE FOLLOWING IS WRONG... I just don't understand why.

So, why is this wrong; what am I missing here...

I'm sitting in space with my jetpack on. I weigh 2kg (not 1kg). I'm stationary and not moving.
I look out and see the earth travelling towards me at 100m/s. And I see the moon right beside it travelling at me at 10m/s.
I'm going to crash into one of them because both the world and moon need more energy and they will absorb my kinetic energy. And believe it or not, they'll pay me for it. They'll pay me $1 for every Joule of energy when I crash.

But I'm kind of broke. It costs me $1 for every Joule of energy I expend, and I only have $2. But I figure, this is still a pretty good deal:
I will turn towards them and accelerate from 0m/s to 1m/s.
So for the earth my kinetic energy just went from 10,000 to 10,201J, and for the moon 100 to 121J.
Nice!!! it only cost me $1 and one joule of energy and in return I'll expend and extra 201J on Earth. I think I'll head towards Earth. That's $201 bucks in my pocket for just $1 cost.

So I figure it's just like with beer, if one is good then 2 must be better, so, I accelerate again.
But this time, what I see is the earth travelling at me at 101m/s. Sure 1m/s was from me, but space is relative, so, I'll just pretend I'm stationary again since no one can actually argue with that, and, so again I accelerate from 0 to 1m/s towards the Earth.
But this time my kinetic energy goes up to 10, 404J, or a whopping increase of 203J, or $203. Big party when I get out of the hospital.

So, yeah, I crash, go unconscious and wake up with pretty nurses all around. And I decide, hey, if this worked once, why not do it again?
So I spend $200 on a new jetpack and fly away from Earth back where I came from at 1m/s. That only costs me a buck.
(I'm travelling "with" the Earth here so I only need a little bit of speed. Also, we can ignore the cost of gravity since it turned out I crashed much harder on the way in because of gravity but that nullifies the cost of leaving against it).

So I wait a day or so until I'm good and far away, just coasting at a constant 1m/s. Then I turn around and try to do it all over again - but, hey wait, the Earth is only coming at me at -1m/s. This time I need to accelerate all the way up to about 100m/s. This was a bad idea and I learn my lesson about the law of conservation of energy.

But, this story strayed far from the normal explanations of this. And this leads to a case where the kinetic energy is just stored up in the universe but could all be used up if everything just crashed into everything else. What I want clarification on is the 1J turning into 203J and the idea that every 1m/s increase costs 1J here....which makes me wonder if instead I chose a span of .00001m/s to increase by instead. It seems maybe the kinetic energy formula has nothing at all to do with the cost of acceleration and how much energy it requires.

 

[russ_watters]

Long post, and I didn't read until the end, I stopped after the running scenario: frame dependent isn't the same as not conserved and changing the scenario in the middle of a problem will of course give wonky results. The answer is what the equation says at face value: accelerating to twice the speed takes 4x the energy.

 

[ryanabc67]

I suspect your response is that it says at face value; you didn't read it.

 

[jbriggs444]

I suspect your response is that it says at face value; you didn't read it.

Nonetheless, @russ_watters got it right. Energy is conserved but not invariant. You say that you understand this but you go on to demonstrate that you do not.

It is not possible to simply expend energy to change your velocity. You must transfer momentum as well. Whatever you are transferring momentum to will gain or lose energy as a result.

You can pick a reference frame where you are moving fast so that the energy gain you get from pushing off from a co-moving object is large. Yay for efficiency!!

But if you turn around and look at the result for that co-moving object, it will have lost energy. Boo for efficiency :-(

TANSTAAFL. You cannot get something for nothing.

 

[Ibix]

TL;DR Summary: Q: "does it take more energy to accelerate at higher speeds"
A: The 'usual answer': "Your kinetic energy grows exponentially"....yeah, but how much FUEL do you burn?

CHANGE IN KINETIC ENERGY DOES NOT EQUAL CHANGE IN REQUIRED ENERGY.

Yes it does. Remember that if you start running, the Earth changes speed too. Factor that energy change in too and you'll find your lost energy.

Consider two bodies, masses $m$ and $M$ that are initially at rest. They push off each other, acquiring velocities $v$ and $V$ respectively. Conservation of momentum tells you that $v=-MV/m$. What was the energy change? Since the initial KE was zero, it's just $\frac 12mv^2+\frac 12MV^2$.

Now analyse again, in a frame where the initial velocities were $u$. The final velocities are $V+u$ and $v+u$. What's the energy change? $$\begin{eqnarray*}
\Delta E&=&\frac 12m(v+u)^2+\frac 12M(V+u)^2-\frac 12mu^2-\frac 12Mu^2\\
&=&\frac 12mv^2+\frac 12MV^2+mvu+MVu\\
&=&\frac 12mv^2+\frac 12MV^2+m(-MV/m)u+MVu\\
&=&\frac 12mv^2+\frac 12MV^2
\end{eqnarray*}$$This is the same as before. But you will not get the same answer if you disregard the energy given to one object - that is where you are going wrong, as noted by both posters above.

 

[jack action]

So, why is this wrong; what am I missing here...

Every time you accelerate, you must lose mass.

I'm sitting in space with my jetpack on. I weigh 2kg (not 1kg). I'm stationary and not moving.

Once you begin accelerating, you won't be 2 kg of mass anymore. You will have to push some part of you (the fuel in the jet pack) in the opposite direction, forever lost to you, now going at a certain speed away from you.

When you are on Earth, your mass is your combined masses, not just 2 kg of you. You can push yourself off the Earth. The Earth - or part of it - will be the mass acquiring momentum away from you.

 

[Dale]

I just don't understand why.

It is wrong because you neglected the conservation of momentum.

As others have said, energy is conserved and it is also frame variant. But conservation of energy is not the only important conservation law. You must also consider conservation of momentum.

When you accelerate in your jet pack you will release a tremendous amount of energy. In a frame where you have a low velocity the vast majority of that energy will go into the exhaust. In earths frame more goes into you and less into the exhaust. You cannot neglect conservation of momentum in any frame and get the right answer

 

[russ_watters]

Every time you accelerate, you must lose mass.


Once you begin accelerating, you won't be 2 kg of mass anymore. You will have to push some part of you (the fuel in the jet pack) in the opposite direction, forever lost to you, now going at a certain speed away from you.

Pros and cons to different scenarios, but one reason I prefer cars to rockets here is there is little(gas) or no(electric) change in mass.

Also, others pointed out the Newton's 3rd Law implications but my gut tells me(without calculating) the "missing energy" is negligible for a car that acts against the Earth.

The bigger "missing energy" would be in comparing cars colliding head-on vs rear-ender at the same closing speed. That "missing energy" is resolved with conservation of momentum for an inelastic collission: the final speed of the cars in a rear-ender isn't zero.(As others have said)

 

[jbriggs444]

Also, others pointed out the Newton's 3rd Law implications but my gut tells me(without calculating) the "missing energy" is negligible for a car that acts against the Earth.

This is incorrect. It depends on what reference frame one adopts.

If one adopts a reference frame in which the road is moving then an accelerating car is doing non-negligible work on the moving Earth.

If you add it up, any unexpectedly high gain in kinetic energy because the road is moving is exactly compensated for by a loss in kinetic energy of the moving Earth.

 

[russ_watters]

This is incorrect. It depends on what reference frame one adopts.

If one adopts a reference frame in which the road is moving then an accelerating car is doing non-negligible work on the moving Earth.

I'm not sure what you mean by that; by "moving", do you mean accelerating while holding the car stationary?

 

[jbriggs444]

I'm not sure what you mean by that; by "moving", do you mean accelerating while holding the car stationary?

I mean adopting a reference frame where the Earth is moving.

For instance, one could adopt a reference frame in which the Sun is stationary and the Earth is moving at its orbital speed of some 107,000 kilometers per hour.

If a car is accelerating eastward at midnight (in the direction of the orbital motion), the force of the tires on the road will be draining energy from the Earth at a high rate. Power = force times velocity. The car gains kinetic energy rapidly in this scenario while the Earth loses kinetic energy rapidly.

 

[Dale]

Also, others pointed out the Newton's 3rd Law implications but my gut tells me(without calculating) the "missing energy" is negligible for a car that acts against the Earth.

It is only negligible in the inertial frame where the earth is initially at rest. In any other inertial frame it must be included.

 

[russ_watters]

I mean adopting a reference frame where the Earth is moving.

For instance, one could adopt a reference frame in which the Sun is stationary and the Earth is moving at its orbital speed of some 107,000 kilometers per hour.

If a car is accelerating eastward at midnight (in the direction of the orbital motion), the force of the tires on the road will be draining energy from the Earth at a high rate. Power = force times velocity. The car gains kinetic energy rapidly in this scenario while the Earth loses kinetic energy rapidly.

Fair enough, I didn't think of that.

Would you agree, though, that for a frame where they are at rest to start, the difference would be negligible or am I missing something there too?
[Edit: I see Dale confirms.]

 

[jbriggs444]

Fair enough, I didn't think of that.

Would you agree, though, that for a frame where they are at rest to start, the difference would be negligible or am I missing something there too?

The instantaneous power at the start of the scenario in their their shared initial rest frame is zero for both. Zero power going into the Earth. Zero power going into the car. Yes.

As the scenario progresses, the Earth has negligible rearward velocity while the car has non-negligible forward velocity, both measured against the inertial frame that we have chosen to adopt. So yes, the power going into the car is non-negligible and the power going into the Earth is negligible in this frame.

So yes, I agree with what I think you were trying to say.

 

[Ibix]

Yes, kinetic energy grows exponentially with speed.

Also, side note, kinetic energy grows quadratically with speed, not exponentially. Exponential would be something like $E\propto 2^v$, but we have $E\propto v^2$.

 

[Mister T]

Yes, kinetic energy grows exponentially with speed.

No. Kinetic energy is a quadratic function of speed, not exponential.

If, for example, you make the speed $3$ times larger then the kinetic energy is $3^2$ times larger.

Why is understanding this simple relationship an issue?

 

[Mister T]

So my speed is always zero and every increase I make is from 0 to 1

You can't be always zero if you are also sometimes 1. Your proposal is self-contradictory.

 

[sophiecentaur]

Every time you accelerate, you must lose mass.

Once you begin accelerating, you won't be 2 kg of mass anymore.

Whilst this is technically true (of course) there are methods of propulsion which involve 'approximately' zero mass change, for instance being towed by a tractor craft. The rocket equation is not essential here but ion drive or photon drive systems can approximate to no change in mass.

I have a mental picture for thinking about kinetic energy. KE can be looked on as the amount of damage an impact will cause. That means you have to bring in the reference frame and the rocket is not just up there with 1000J of KE. You have to specify what it hits (or could hit) for that energy to be a specific value and the relative velocity in a specific frame introduces itself.

 

[Ibix]

Whilst this is technically true (of course) there are methods of propulsion which involve 'approximately' zero mass change, for instance being towed by a tractor craft.

Agreed. However, all methods will involve the transfer of momentum and energy between bodies, which is the key fact needed toresolve the OP's confusion.

 

[jbriggs444]

Whilst this is technically true (of course) there are methods of propulsion which involve 'approximately' zero mass change, for instance being towed by a tractor craft.

To add to what @Ibix has said about this, even when the reaction mass is approximately zero, that is still where the discrepancy in an energy analysis can be found.

A 1000.001 kg craft at rest expelling 1 gram of exhaust at one million meters per second will attain forward velocity of 1 meter per second and will then have 500 J of kinetic energy. That is a gain of 500 J

A 1000.001 kg craft already at 1 meter per second expelling 1 gram of exhaust at one million meters per second will attain forward velocity of 2 meters per second and will have 2000 J of kinetic energy. That is a gain of almost 1500 J (off by 0.0005 J since we did not count the initial kinetic energy of that gram of fuel we added).

The difference in payload energy gains is approximately 1000 J. Or 999.9995 with the more careful accounting. Our original poster would wonder where the extra 1000 J came from.

In the first case, we ended with that one gram of exhaust moving at 1,000,000 meters per second rearward. That's 500,000,000 (500 megajoules) of energy.

In the second case we ended with one gram of exhaust moving at 999,999 meters per second rearward. That's 499,999,000.0005 J of kinetic energy.

The difference in the exhaust energy gain is exactly 999.9995 J, just as the careful accounting calls for.

 

[sophiecentaur]

that is still where the discrepancy in an energy analysis can be found.

Sorry but numerical examples leave me confused. Symbolic algebra keeps the information healthy all the way. Numbers are anonymous and just one zero can be easily lost or found.

My point was that wherever the energy comes from, the amount needed will be the same. You don't need a reaction propulsion system to show how the KE gained is frame dependent. I could have used a two stage rocket system and the KE gained by the upper stage could be independent of the propellant system. A force applied applied over a given distance produce an increase in energy of ΔE = Fx. The rocket equation result is not relevant.
Stage two of the analysis could involve a rocket but it's not needed first time around.

 

[jbriggs444]

Sorry but numerical examples leave me confused. Symbolic algebra keeps the information healthy all the way. Numbers are anonymous and just one zero can be easily lost or found.

Certainly. Tastes differ. Algebra feels nice to some. And some of us take comfort in seeing numbers match what the algebra says should be happening. One of my weaknesses is that I fall in the latter camp much of the time.

My concern was with a statement that the exhaust mass could be "approximately zero". This could invite someone to handwave the exhaust away and find themselves facing exactly the confusion expressed by the OP: Same energy input. Apparently different energy output.

Of course, all of us regulars are perfectly well aware that the exhaust stream is the place to look for the explanation of this.

My point was that wherever the energy comes from, the amount needed will be the same. You don't need a reaction propulsion system to show how the KE gained is frame dependent.

The KE gained by the payload is frame dependent.
The total KE gained is invariant.

 

[sophiecentaur]

My concern was with a statement that the exhaust mass could be "approximately zero".

But that assumes a rocket which is far more complicated than the OP asks for. Rocket proportion is much more complicated than an Earth-born experiment. If you can supply a Force and it has an effect then that is the simplest situation. Surely this thread (at least it started off that way) is to show that KE is frame dependent. It's an old chestnut and it's turned up in may different guises over the years.

 

[Philip Koeck]

TL;DR Summary: Q: "does it take more energy to accelerate at higher speeds"

I don't know whether this has been discussed:

You have E = ½ m v².
Therefore dE/dv = m v

This means that the amount of energy needed to change the velocity by 1 m/s, for example, increases linearly with v.

I'm assuming that we're accelerating a small mass and we are fixed on a very large mass, such as the earth, so the large mass only changes its momentum, but essentially doesn't change it's kinetic energy. All the energy I put in goes to the small mass.

Is that what you were looking for?

 

[Ibix]

but essentially doesn't change it's kinetic energy. All the energy I put in goes to the small mass.

The difference between the two parts I've bolded becomes significant when the larger mass is in motion because of the cross term in the squared velocity. Failing to account for that is where the OP goes wrong, I think.

 

[sophiecentaur]

I don't know whether this has been discussed:

You have E = ½ m v².
Therefore dE/dv = m v

This means that the amount of energy needed to change the velocity by 1 m/s, for example, increases linearly with v.

I'm assuming that we're accelerating a small mass and we are fixed on a very large mass, such as the earth, so the large mass only changes its momentum, but essentially doesn't change it's kinetic energy. All the energy I put in goes to the small mass.

Is that what you were looking for?

That's the scenario in my mind. I think that rockets etc. were brought into the thread too soon. A BB gun on a train is a simple example. When the train is stationary, the energy of the BB is all due to the spring (Platform frame) and the BB will make a particular depth of dent in target on the platform. If the train accelerates to a high speed and the gun is fired (forward) at another target on the trackside, the BB will have two contributions to its KE (Platform frame), the energy from the locomotive and the energy from the spring. It will make a bigger dent.

 

[jbriggs444]

If the train accelerates to a high speed and the gun is fired (forward) at another target on the trackside, the BB will have two contributions to its KE (Platform frame), the energy from the locomotive and the energy from the spring. It will make a bigger dent.

This simple explanation does not seem to me to get to the heart of the scenario. I like to emphasize that the total kinetic energy supplied by the spring is an invariant quantity and how that can be the case.

Indeed, we have two contributions to the BB's final frame-relative kinetic energy. We have the kinetic energy of the BB due to the engine's prior efforts. And we have the incremental kinetic energy due to the spring.

But the incremental kinetic energy supplied by the spring will depend on the velocity of the train. The spring itself provides an invariant amount of kinetic energy. This energy is doled out as work done at each of the two spring ends. At the business end, the faster the train goes, the more work the spring does during its expansion. At the train end, the faster the train goes, the more work the spring absorbs.

Assume an ideal spring so that we do not have to quibble about how much energy goes into accelerating the spring itself.

 

[jack action]

That's the scenario in my mind. I think that rockets etc. were brought into the thread too soon. A BB gun on a train is a simple example. When the train is stationary, the energy of the BB is all due to the spring (Platform frame) and the BB will make a particular depth of dent in target on the platform. If the train accelerates to a high speed and the gun is fired (forward) at another target on the trackside, the BB will have two contributions to its KE (Platform frame), the energy from the locomotive and the energy from the spring. It will make a bigger dent.

The BB gun on a train is still a rocket example. The train and BB gun form a single object.

• When the bullet goes away, it "loses" mass, i.e. the train. You may look at it the opposite way: the train "loses" mass (the bullet) while decelerating;
• The train will slow down: you could stop a moving train if the bullet is big enough - or fast enough;
• If the train is stationary, it will back up after firing the gun.

Don't let the interaction between the train and the Earth - a much much bigger mass - fool you. Or the interaction between the target hit by the bullet and the Earth as well. No gain, no loss for the Earth in the end.

 

[sophiecentaur]

The BB gun on a train is still a rocket example

Momentum is conserved so you have to be right in principle. But doesn't that approach close the door on all simple momentum questions that we started our Physics education? The Earth is 'big enough' to get started with and that frame is the only one we need, initially

Considering the 'dents' (the damage done by KE of the BB). It will be the sharing of momenta between BB and train / earth that determines the difference in damage for the two cases.

 

[sophiecentaur]

the difference in damage for the two cases.

There has to be a point at which the BB gets the vast majority of the energy, because of the $v^2$ ratio. In differential calculus we have no problem about the formal way we use "the limit as δx approaches zero" to prove how we differentiate (dy/dx) all the well known functions from first principles.

Thinking about this with hindsight and re-reading the thread, I would say it's quite valid to ignore the 'lost energy' for introductory work. Or maybe we should avoid all such situations in early work on such problems. How does one avoid the conclusion that . . . . .

TL;DR Summary:

And this leads to a case where the kinetic energy is just stored up in the universe

which makes it seems to be like some magic at work in the conservation laws. But we can't have that, can we?

 

[A.T.]

Momentum is conserved so you have to be right in principle. But doesn't that approach close the door on all simple momentum questions that we started our Physics education? The Earth is 'big enough' ...

Momentum is different from energy here. See also this thread:
https://www.physicsforums.com/threa...vs-momentum-conservation.1059279/post-6989294

The Earth is never 'big enough' to neglect the momentum it receives, and still have momentum conserved, on the scale of the much smaller object's momentum change. Not even in the Earth's initial rest frame you can get away with that.

The Earth is 'big enough' to neglect the energy it receives in its initial rest frame, and still have energy approximately conserved in that frame, because in that frame the work done on the Earth is negligible, compared to the work done on the much smaller object.

But in frames where the Earth is moving, the work done on the Earth is not negligible, compared to the work done on the much smaller object. See for example my question about the Brennan torpedo:

And here is a question to ponder: In case A2 (rest frame of the cable), since the motor cannot power the torpedo via a static cable, where does the energy from the motor go to?

 

[sophiecentaur]

The Earth is never 'big enough' to neglect the momentum it receives,

That philosophy would limit greatly the questions you could set on momentum. You would basically be stuck with snooker balls colliding or asteroids being deflected by a nuke.You would have many examples of the good being the enemy of the perfect.

Are you saying something more consequential than the v squared idea?

the Brennan torpedo:

Never came across this one. I have to say, the idea and picture made me think of crossed lines when fishing from two boats. But only one torpedo would be used.
Intuitively (as with the faster than the wind vehicle) it won't work but the torpedo and sand yacht demos show how wrong we (I) can be.
"Where does the energy go?" The water is moving. The ground is moving (in the cable frame). Work is done by the motor on the cable because it is 'pulled'. Work is done on the torpedo due to the reaction of the water on the prop. I'm not sure if that's an answer - maybe you can come up with a punch line for it.

 

[Philip Koeck]

That philosophy would limit greatly the questions you could set on momentum. You would basically be stuck with snooker balls colliding or asteroids being deflected by a nuke.

@A.T. wrote: "The Earth is never 'big enough' to neglect the momentum it receives, and still have momentum conserved, on the scale of the much smaller object's momentum change."

What @A.T. says (quoted above) is rather important, if I got it right.
Even if a very light object bounces off an extremely heavy wall this is true.
In an ideally elastic, head on collision the light object would have the same speed after the collision as it had before, just in the opposite direction. So the kinetic energy of the object is unchanged and the wall hasn't received any kinetic energy, but the wall experiences a change of momentum, which is twice the small objects initial momentum. Think of kinetic gas theory for the ideal gas.

The above is only true in the wall's rest frame, though. If the wall is moving with respect to the observer things change (as I am realizing now. Never thought about it before.).

 

[sophiecentaur]

the wall experiences a change of momentum, which is twice the small objects initial momentum.

YEBBUT even the velocity will often be tiny and the KE share even tinier.
I was just thinking about the good old Coefficient of Restitution, which sidesteps the deeper problems but gives you (gave me) a good leg-up for an introduction.

 

[Philip Koeck]

YEBBUT even the velocity will often be tiny and the KE share even tinier.
I was just thinking about the good old Coefficient of Restitution, which sidesteps the deeper problems but gives you (gave me) a good leg-up for an introduction.

I have to correct myself a bit. I was thinking of an infinitely heavy wall. In fact the momentum received by the wall will be a tiny bit smaller than twice the ball's initial momentum and the kinetic energy change of the wall will not be exactly zero, only almost.

What's important I think is that the kinetic energy change of the wall is negligible in such a scenario, but the change in momentum is not. The change in momentum for the wall is twice the initial momentum of the ball, so it's "big" in that sense.

PS: The light object has become a ball.
PPS: I corrected typo "wall" to "ball" above