KE of system / different reference frames question

In summary, the conversation discusses the concept of using the ground as a source of energy for a moving object, such as in the case of KERS. It explains how the energy in a given frame of reference can be repartitioned and how this can be misunderstood if one is thinking too literally. The conversation also addresses the issue of where the energy for KERS comes from, with one person arguing that it comes from the car's fuel and another suggesting it comes from the kinetic energy of the car itself. Overall, the conversation highlights the importance of considering reference frames when discussing energy and its sources.
  • #1
jduffy77
40
0
I am re-posting this question here in a new thread as Humber mistakenly posted it in a two year old thread.

Humber said:
This post appeared on a ddwfttw forum:

uncool said:
For those who actually care, relative to any frame other than that of the ground, the ground does have energy. It is possible to get energy from the ground - in fact, this is exactly what happens, for example, in KERS. The non-ground frame analysis would go as follows:

Say we have a 10 kg object initially moving at 10 m/s relative to the ground. We choose to start analyzing in the frame where this object is initially at rest - so the ground is initially moving at -10 m/s.

We use KERS to brake the object at a rate of -1 m/s^2, or equivalently, a force of -10 N. By Newton's third law, that means that there is a force of 10 N on the Earth.

As the object is at rest relative to this frame, there is no relevant kinetic power on the object itself. However, there is kinetic power on the Earth; using P = F*v, we get that the Earth is losing kinetic energy at a rate of -100 W, so by conservation of energy, the KERS can be storing energy at a rate of 100 W.

Note that this is exactly the same as what we get in the frame of the ground, as there, F = -10 N, v = 10 m/s, so the object is losing kinetic energy at a rate of -100 W, so by conservation of energy, the KERS can be storing energy at a rate of 100 W.


The above idea, seems to be quite commonly accepted amongst those claiming to have physics degrees, but it seems to me to be a the result of literal thinking, and a rather confused idea of what frames of reference means.

The energy for the KERS, initially comes from the car's fuel, some of which ultimately ends up as kinetic energy of the car. It is that energy which is recovered by KERS, and does not come from the ground.

I am interested in hearing options for or against either claim.
 
Physics news on Phys.org
  • #2
I'm not sure what your issue is with "literal thinking", to me the quote is a clear example of not thinking literally. Literal thinkers imagine that energy is innate, so a car that is moving along the road has it, in any frame of reference. But energy is frame dependent-- conservation of energy means the energy stays the same in a given frame, it doesn't mean it is the same in all frames. So choosing different frames always repartitions the energy, and therefore it also changes the language about what the energy is doing or where it is coming from. This is all perfectly normal, it is what you find when you relax the need to think literally.

So there is no problem at all with saying that a car can extract energy from the road when it uses KERS, but the language sounds awkward unless one is being extremely clear about the choice of reference frame. For example, one can have a car that accelerates from rest to 10 m/s, and if we start in a frame already moving at 10 m/s, this will look like a car that uses its fuel energy yet loses all its kinetic energy-- the missing energy shows up in the Earth's kinetic energy in that frame. That's just how it is, in that frame. We can argue that isn't the best frame to think about, or we can argue it is, but it certainly isn't wrong.
 
  • #3
I had to look up KERS and that is Kinetic Energy Recovery System.

It is so far so good up to the 5th paragraph.
The 10kg object can be considered to be at rest in a frame, call it frame A, moving at 10 m/s relative to the earth. But once the braking occurs, the object will have a force applied to it and begin to move relative to that frame. ( Otherwise you are dealing with a moving frame of reference. ) At the end of it all, the object will have a velocity of -10 m/s in frame A.

Suppose there are 2 cars moving side by side. You have a tiny window to look out from car 1 out but all you can see is the car 2 beside you, which would appear to be not moving relative to your reference frame A. Person X on the Earth can see both Car1 and 2 and conclude that both cars are mving at 10 m/2 relative to the earth.
Car2 signals you that he is going to apply KERS. As he does so, to you in Car 1, it would look as if Car2 is accelerating in the backward direction up to a maximun of -10 m/s. You in Car1 would begin to wonder how in the world did Car2 accelerate if Car2 was using KERS.
 
  • #4
The energy for the KERS, initially comes from the car's fuel, some of which ultimately ends up as kinetic energy of the car. It is that energy which is recovered by KERS, and does not come from the ground.

A really bad statement all in all. It is not put together in any coherent fashion. You usually find something like this on a site promoting the newest crackpot idea which is meant to confuse, and not for an explanation of KERS. I cannot tell if the writer is saying that the energy of the fuel is being recovered or the kinetic energy of the vehicle. If he is saying it is the energy of the fuel, then why not go farther back and say it is the energy of the sun trapped in the plant material, from 600 million years ago, that became hydrocarbons through transformation under the Earth's surface.

The fuel provides energy to the car to move. The chemical energy is transformed into kinetic energy. KERS would recover some of the kinetic energy.
 
  • #5
I think the point missed in the old DDWFTTW thread is the source of the force at the tires. For a KERS vehicle, the force at the tires equals the mass of the vehicle times it's rate of acceleration (or deceleration) (ignoring rolling resistance and aerodynamic drag). For a DDWFTTW vehicle, the force at the tires equals the torque used to drive the propeller divided by the radius of the tires, which is independent of the mass of the vehicle.
 
Last edited:
  • #6
256bits said:
Otherwise you are dealing with a moving frame of reference.
Not a problem since for any given situation, there is only one non-moving frame of reference and an infinite number of moving ones.

I think what you're meaning is an accelerating frame of reference.
 
  • #7
Ken G said:
I'm not sure what your issue is with "literal thinking", to me the quote is a clear example of not thinking literally. Literal thinkers imagine that energy is innate, so a car that is moving along the road has it, in any frame of reference. But energy is frame dependent-- conservation of energy means the energy stays the same in a given frame, it doesn't mean it is the same in all frames. So choosing different frames always repartitions the energy, and therefore it also changes the language about what the energy is doing or where it is coming from. This is all perfectly normal, it is what you find when you relax the need to think literally.

So there is no problem at all with saying that a car can extract energy from the road when it uses KERS, but the language sounds awkward unless one is being extremely clear about the choice of reference frame. For example, one can have a car that accelerates from rest to 10 m/s, and if we start in a frame already moving at 10 m/s, this will look like a car that uses its fuel energy yet loses all its kinetic energy-- the missing energy shows up in the Earth's kinetic energy in that frame. That's just how it is, in that frame. We can argue that isn't the best frame to think about, or we can argue it is, but it certainly isn't wrong.
That is indeed the thinking I was taking about. That "all things are relative".

Acceleration is not one of those things, and the fuel in that tank, is another.
Can you justify that the fuel consumed, is capable of accelerating the Earth by same amount it accelerates the car?
 
  • #8
256bits said:
A really bad statement all in all. It is not put together in any coherent fashion. You usually find something like this on a site promoting the newest crackpot idea which is meant to confuse, and not for an explanation of KERS. I cannot tell if the writer is saying that the energy of the fuel is being recovered or the kinetic energy of the vehicle. If he is saying it is the energy of the fuel, then why not go farther back and say it is the energy of the sun trapped in the plant material, from 600 million years ago, that became hydrocarbons through transformation under the Earth's surface.

The fuel provides energy to the car to move. The chemical energy is transformed into kinetic energy. KERS would recover some of the kinetic energy.

The first few statements are not mine. The claim there, is that the Earth can transfer 100W to the KERS.
But, regardless of its origins, the fuel is the store of some energy form somewhere. It could equally be a battery. When the car accelerates, some of that energy becomes KE in the car's mass, and that is the energy that is recovered, and not something that comes from the ground.
 
  • #9
rcgldr said:
I think the point missed in the old DDWFTTW thread is the source of the force at the tires. For a KERS vehicle, the force at the tires equals the mass of the vehicle times it's rate of acceleration (or deceleration) (ignoring rolling resistance and aerodynamic drag). For a DDWFTTW vehicle, the force at the tires equals the torque used to drive the propeller divided by the radius of the tires, which is independent of the mass of the vehicle.

Yes. F=ma.
When the car brakes, momentum is transferred from the car, to the ground, dp/dt = F.
For a KERS to work, that same force is present, but the KE of the car is not dissipated as heat in the brake pads and disks, but becomes stored in a mass as angular KE, or by direct conversion to electrical energy and so stored in a battery.
Apart form some work at the tyres and surface deformation, which is a result of F, and energy that can't be recovered, there is no energy from the ground.
 
  • #10
Humber said:
The first few statements are not mine.

No, those are correct. The incorrect one he mentioned was not put forth in a coherent fashion was yours though.

Humber said:
The claim there, is that the Earth can transfer 100W to the KERS.
But, regardless of its origins, the fuel is the store of some energy form somewhere. It could equally be a battery. When the car accelerates, some of that energy becomes KE in the car's mass, and that is the energy that is recovered, and not something that comes from the ground.

As was already pointed out; whether or not the energy comes from the ground depends on the reference frame chosen to analyze.
 
  • #11
Humber said:
That is indeed the thinking I was taking about. That "all things are relative".

Acceleration is not one of those things, and the fuel in that tank, is another.
Can you justify that the fuel consumed, is capable of accelerating the Earth by same amount it accelerates the car?

No one has made that claim Humber. You are trying to use a straw man argument.

Why don't you lay out your claim here that all energy must be calculated relative to the Earth and see how that flies here. Or as you claimed elsewhere, that no energy can come from the ground.
 
  • #12
256bits said:
Otherwise you are dealing with a moving frame of reference.

mender said:
Not a problem since for any given situation, there is only one non-moving frame of reference and an infinite number of moving ones.

I think what you're meaning is an accelerating frame of reference.

Mender old bean! You've been missed at TR. Did you ever get around to do that test pulling your cart to wind speed and then releasing it along with some packing popcorn? I would still love to see a video of that.

Also looking forward to 256bits response. And what about a decelerating FoR?
 
  • #13
jduffy77 said:
No, those are correct. The incorrect one he mentioned was not put forth in a coherent fashion was yours though.
As was already pointed out; whether or not the energy comes from the ground depends on the reference frame chosen to analyze.

Which is a contradiction of both my statement, and that of 256bits, which is paraphrasing of mine.
When the car accelerates, momentum is transferred from the ground to the car, and when it brakes, the other way. That is conservation of momentum at work, which leaves you to explain if the ground powers the KERS, why fuel is needed to transfer momentum one way, but not the other. And, if you can possibly manage it, please keep your personal remarks and references to the cart to yourself.
 
  • #14
Humber said:
Which is a contradiction of both my statement, and that of 256bits, which is paraphrasing of mine.
When the car accelerates, momentum is transferred from the ground to the car, and when it brakes, the other way. That is conservation of momentum at work, which leaves you to explain if the ground powers the KERS, why fuel is needed to transfer momentum one way, but not the other. And, if you can possibly manage it, please keep your personal remarks and references to the cart to yourself.

Please check your quotes. The two sentences you have quoted were mine but the body of your post is responding to someone else.
 
  • #15
Happy New Year, RP!
RCP said:
Mender old bean! You've been missed at TR. Did you ever get around to do that test pulling your cart to wind speed and then releasing it along with some packing popcorn? I would still love to see a video of that.
Nothing yet and not likely in the foreseeable future, too much real work to do! I haven't stopped in because I didn't think there was anything left but some stirring of the ashes!

ETA: wow - part 30??!
 
Last edited:
  • #16
Humber said:
When the car accelerates, some of that energy becomes KE in the car's mass, and that is the energy that is recovered, and not something that comes from the ground.
In some valid frames, when the car accelerates it loses KE, and can even come to rest. When it is at rest there is 0 KE. Furthermore, KE can never be negative. So, how do you propose to recover energy from something with 0 energy and whose energy cannot go negative?

Have you even attempted to calculate the energy in a moving reference frame and see how it works out?
 
  • #17
mender said:
Happy New Year, RP!

Nothing yet and not likely in the foreseeable future, too much real work to do! I haven't stopped in because I didn't think there was anything left but some stirring of the ashes!

ETA: wow - part 30??!

Best wishes for the new year back atcha mender. Yep, Cartville is still accelerating, and FYI humber has now posted pictures of 3 different carts he made. (Film at 11! :smile:) Remember how positive everyone was he would never make one?

And if you or anyone has time, there's still my question above about the FoRs applicable for deceleration?
 
  • #18
Although off topic from the orignal DDWFTTW thread ... ignoring any losses due to aerodynamic drag, rolling resistance, conversion of energy, ... , then considering a car and Earth as a closed system, then energy (and momentum) of this closed system is conserved.

KERS - extracts energy from Earth + car closed system and adds that energy to a device that stores energy (kinetic or potential).

Usage of fuel - extracts chemical potential energy from fuel and adds that energy to the kinetic energy of Earth + car closed system.

For both of these cases, the amount of energy converted is independent of the frame of reference, as long as the frame of reference is inertial.

Getting back on topic from the original DDWFTTW thread, someone was wondering if a DDWFTTW vehicle was somehow extracting it's own kinetic energy (wrt ground). The responses that followed tried to explained that a DDWFTTW vehicle extracts energy by slowing down air (wrt ground), and that a DDWFTTW vehicle requires a true wind (wrt ground) in order to operate.
 
  • #19
DaleSpam said:
In some valid frames, when the car accelerates it loses KE, and can even come to rest. When it is at rest there is 0 KE.
KE is the work of acceleration.
http://en.wikipedia.org/wiki/Kinetic_energy
"The kinetic energy of an object is the energy which it possesses due to its motion.[1] It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest."
The accleration of a 640kg car, cannot be equated with the acceleration of the Earth.

DaleSpam said:
Furthermore, KE can never be negative.So, how do you propose to recover energy from something with 0 energy and whose energy cannot go negative?
Your first statement is correct, because KE is related to the square of velocity, but then again, momentum isn't. There is certainly -p and +p. That allows conservation to work.

If it were necessary for there to be "-ve KE" to "recover +ve KE," then no transfer could ever take place under any circumstances. The answer is the the KE does not need to be negative, because all cases lie between zero and some non-zero number.

DaleSpam said:
Have you even attempted to calculate the energy in a moving reference frame and see how it works out?
Momentum transfer to the ground is necessary to force, but not to the motion itself, but wheels do need that force, so momentum is indeed transferred as the car gains KE via acceleration.
The car gains momentum p, and the Earth -p.
total energy = p^2/2mcar+ p^2/2mearth
mearth = 6e24Kg, so the second term is negligible.

ETA:
It is not a matter of arbitrarily choosing frames because changes in momentum are not frame dependent. One would be at a loss to explain how the battery in an electric car loses charge, if it where just a matter of choosing which frame it lost it in.
 
Last edited:
  • #20
Humber said:
That is indeed the thinking I was taking about. That "all things are relative". Acceleration is not one of those things, and the fuel in that tank, is another.
Sure, but the statements you objected to did not say anything about either the acceleration or the fuel, they talked about recovering energy from the Earth, which is perfectly correct in the frame of the car's velocity.
Can you justify that the fuel consumed, is capable of accelerating the Earth by same amount it accelerates the car?
No, nor is there any need to justify that, because no one thinks the Earth is being accelerated as much as the car. That comes from a=F/m, the F is an action/reaction pair so is the same on the car and the Earth. Not the "a." There is only a very small "a" for the Earth, but it suffices to remove exactly the correct amount of kinetic energy from the Earth to make the problem work out in any reference frame.

What is happening here is that you are confused about the differences between energy, momentum, and acceleration, and your confusion is causing you to contradict the statements made by people who do understand these differences. I'm not sure you can move forward until you understand those basic concepts better, and then you can get to the point where you understand that stories about what energy is doing always depend on the reference frames from which they are told. This is all elementary to us.
 
  • #21
Humber said:
KE is the work of acceleration.
http://en.wikipedia.org/wiki/Kinetic_energy
"The kinetic energy of an object is the energy which it possesses due to its motion.[1] It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest."
Exactly. So if the car is already at rest in some frame of reference then the "work needed to accelerate [it] from rest to its stated velocity" is 0. In other words, it has no KE.

Humber said:
Your first statement is correct, because KE is related to the square of velocity, but then again, momentum isn't. There is certainly -p and +p. That allows conservation to work.
Momentum and energy are different quantities, they are not interchangeable. Both are individually conserved. You cannot use conservation of momentum instead of conservation of KE, both are needed.

Humber said:
Momentum transfer to the ground is necessary to force, but not to the motion itself, but wheels do need that force, so momentum is indeed transferred as the car gains KE via acceleration.
The car gains momentum p, and the Earth -p.
total energy = p^2/2mcar+ p^2/2mearth
mearth = 6e24Kg, so the second term is negligible.
The change in the second term is only negligible in the frame where the Earth is at rest, see below.

Humber said:
It is not a matter of arbitrarily choosing frames because changes in momentum are not frame dependent. One would be at a loss to explain how the battery in an electric car loses charge, if it where just a matter of choosing which frame it lost it in.
But it is just a matter of arbitrarily choosing frames. Newton's laws are invariant under changes in reference frame, so you can arbitrarily choose your frame and apply Newton's laws. So it had better be possible to explain the same situation from multiple reference frames or you are violating Newton's laws.

Here is how it works:

Let m = 1000 kg be the mass of the car and let M = 1E25 kg be the mass of the earth. There is a horizontal force of -1000 N (to the right is positive) acting on the car from the road for a period of 10 s. Calculate the initial and final values for kinetic energy, KE, and momentum, p, for both the Earth and the car in (a) the frame where the Earth is initially at rest and the car is initially moving at 10 m/s and (b) the frame where the Earth is initially moving at -10 m/s and the car is at rest.

(a) [itex]p_{i,c}=m v_{i,c}=10000 \ kg \ m/s[/itex]
[itex]\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s[/itex]
[itex]p_{f,c}=p_{i,c}+\Delta p_c = 0 \ kg \ m/s[/itex]
[itex]v_{f,c}=p_{f,c}/m= 0 \ m/s[/itex]
[itex]KE_{i,c}=1/2 m v_{i,c}^2 = 50 \ kJ[/itex]
[itex]KE_{f,c}=1/2 m v_{f,c}^2 = 0 \ kJ[/itex]

[itex]p_{i,e}=M v_{i,e}=0 \ kg \ m/s[/itex]
[itex]f_e=-f_c=1000 \ N[/itex]
[itex]\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s[/itex]
[itex]p_{f,e}=p_{i,e}+\Delta p_e = 10000 \ kg \ m/s[/itex]
[itex]v_{f,e}=p_{f,e}/M= 10^{-21} \ m/s[/itex]
[itex]KE_{i,e}=1/2 M v_{i,e}^2 = 0 \ kJ[/itex]
[itex]KE_{f,e}=1/2 M v_{f,e}^2 = 5\;10^{-18} \ J[/itex]

So in the first reference frame the energy for charging the battery comes from the KE of the car which goes down by 50 kJ. The Earth gains a negligible amount of energy.

(b) [itex]p_{i,c}=m v_{i,c}= 0 \ kg \ m/s[/itex]
[itex]\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s[/itex]
[itex]p_{f,c}=p_{i,c}+\Delta p_c = -10000 \ kg \ m/s[/itex]
[itex]v_{f,c}=p_{f,c}/m= -10 \ m/s[/itex]
[itex]KE_{i,c}=1/2 m v_{i,c}^2 = 0 \ kJ[/itex]
[itex]KE_{f,c}=1/2 m v_{f,c}^2 = 50 \ kJ[/itex]

[itex]p_{i,e}=M v_{i,e}= -10^{26} \ kg \ m/s[/itex]
[itex]f_e=-f_c=1000 \ N[/itex]
[itex]\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s[/itex]
[itex]p_{f,e}=p_{i,e}+\Delta p_e = -9.999999999999999999999 \; 10^{25} \ kg \ m/s[/itex]
[itex]v_{f,e}=p_{f,e}/M= -9.999999999999999999999 \ m/s[/itex]
[itex]KE_{i,e}=1/2 M v_{i,e}^2 = 5\;10^{26} \ kJ[/itex]
[itex]KE_{f,e}=1/2 M v_{f,e}^2 = 4.999999999999999999999\;10^{26} \ J[/itex]

So in the second reference frame the energy for charging the battery comes from the KE of the Earth which goes down by 100 kJ. The car gains 50 kJ of kinetic energy.
 
  • #22
RCP said:
FYI humber has now posted pictures of 3 different carts he made. (Film at 11! :smile:) Remember how positive everyone was he would never make one?
Can you save me some time and provide links to those pictures?

RCP said:
And if you or anyone has time, there's still my question above about the FoRs applicable for deceleration?

Not sure exactly what you're asking for, RP, but I'll give it a shot if you elaborate.
 
  • #23
DaleSpam said:
Exactly. So if the car is already at rest in some frame of reference then the "work needed to accelerate [it] from rest to its stated velocity" is 0. In other words, it has no KE.
Momentum and energy are different quantities, they are not interchangeable. Both are individually conserved. You cannot use conservation of momentum instead of conservation of KE, both are needed.
Momentum is always conserved, kinetic energy may or may not be.

DaleSpam said:
The change in the second term is only negligible in the frame where the Earth is at rest, see below.
But it is just a matter of arbitrarily choosing frames. Newton's laws are invariant under changes in reference frame, so you can arbitrarily choose your frame and apply Newton's laws. So it had better be possible to explain the same situation from multiple reference frames or you are violating Newton's laws.
And so by the same reasoning, one may not equate the acceleration of a car, with that of the Earth.

DaleSpam said:
Here is how it works:
Let m = 1000 kg be the mass of the car and let M = 1E25 kg be the mass of the earth. There is a horizontal force of -1000 N (to the right is positive) acting on the car from the road for a period of 10 s. Calculate the initial and final values for kinetic energy, KE, and momentum, p, for both the Earth and the car in (a) the frame where the Earth is initially at rest and the car is initially moving at 10 m/s and (b) the frame where the Earth is initially moving at -10 m/s and the car is at rest.

(a) [itex]p_{i,c}=m v_{i,c}=10000 \ kg \ m/s[/itex]
[itex]\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s[/itex]
[itex]p_{f,c}=p_{i,c}+\Delta p_c = 0 \ kg \ m/s[/itex]
[itex]v_{f,c}=p_{f,c}/m= 0 \ m/s[/itex]
[itex]KE_{i,c}=1/2 m v_{i,c}^2 = 50 \ kJ[/itex]
[itex]KE_{f,c}=1/2 m v_{f,c}^2 = 0 \ kJ[/itex]

[itex]p_{i,e}=M v_{i,e}=0 \ kg \ m/s[/itex]
[itex]f_e=-f_c=1000 \ N[/itex]
[itex]\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s[/itex]
[itex]p_{f,e}=p_{i,e}+\Delta p_e = 10000 \ kg \ m/s[/itex]
[itex]v_{f,e}=p_{f,e}/M= 10^{-21} \ m/s[/itex]
[itex]KE_{i,e}=1/2 M v_{i,e}^2 = 0 \ kJ[/itex]
[itex]KE_{f,e}=1/2 M v_{f,e}^2 = 5\;10^{-18} \ J[/itex]

So in the first reference frame the energy for charging the battery comes from the KE of the car which goes down by 50 kJ. The Earth gains a negligible amount of energy.
That is to say, the the 50kJ of KE the car gained from some unmentioned source of energy,
has been captured and stored in a battery. The amount transferred to the ground is
p^2/2Mearth = 5e-18J. The transfer of that energy, has been achieved by deceleration of
the car. The force is dp/dt = 1000N.
The total energy is 50kJ + 5e-18J. The initial source supplying the 50kJ, may be another battery. It supplied 50kJ to the car, and all but 5e-18J was recovered.

DaleSpam said:
(b) [itex]p_{i,c}=m v_{i,c}= 0 \ kg \ m/s[/itex]
[itex]\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s[/itex]
[itex]p_{f,c}=p_{i,c}+\Delta p_c = -10000 \ kg \ m/s[/itex]
[itex]v_{f,c}=p_{f,c}/m= -10 \ m/s[/itex]
[itex]KE_{i,c}=1/2 m v_{i,c}^2 = 0 \ kJ[/itex]
[itex]KE_{f,c}=1/2 m v_{f,c}^2 = 50 \ kJ[/itex]

[itex]p_{i,e}=M v_{i,e}= -10^{26} \ kg \ m/s[/itex]
[itex]f_e=-f_c=1000 \ N[/itex]
[itex]\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s[/itex]
[itex]p_{f,e}=p_{i,e}+\Delta p_e = -9.999999999999999999999 \; 10^{25} \ kg \ m/s[/itex]
[itex]v_{f,e}=p_{f,e}/M= -9.999999999999999999999 \ m/s[/itex]
[itex]KE_{i,e}=1/2 M v_{i,e}^2 = 5\;10^{26} \ kJ[/itex]
[itex]KE_{f,e}=1/2 M v_{f,e}^2 = 4.999999999999999999999\;10^{26} \ J[/itex]

So in the second reference frame the energy for charging the battery comes from the KE of the Earth which goes down by 100 kJ. The car gains 50 kJ of kinetic energy.

And now, the total energy is 100kJ, of which 50kJ was recovered by the car.
Edit.
Oh, I see. In one case the Earth gains momentum and in the other, it loses it, but the total energy is now 100kJ from the Earth + 50kJ, that initially supplied the car ?
There are not two events, but one, and the conservation of momentum still applies, no matter which frame is employed.
 
Last edited:
  • #24
mender said:
Can you save me some time and provide links to those pictures?



Not sure exactly what you're asking for, RP, but I'll give it a shot if you elaborate.

The above is a violation of the conservation of momentum.
 
  • #25
Humber said:
Momentum is always conserved, kinetic energy may or may not be.
Correct, and in the example KE is not conserved in either frame. The KE goes down by 50 kJ in both frames. This 50 kJ of missing KE is the source of the energy put into the battery in both frames.

Humber said:
And so by the same reasoning, one may not equate the acceleration of a car, with that of the Earth.
I agree. And if you will look back at the math you will see that I never did equate them. That was your statement, never mine.

Humber said:
That is to say, the the 50kJ of KE the car gained from some unmentioned source of energy,
has been captured and stored in a battery. The amount transferred to the ground is
p^2/2Mearth = 5e-18J. The transfer of that energy, has been achieved by deceleration of
the car.
That is all correct.

Humber said:
And now, the total energy is 100kJ, of which 50kJ was recovered by the car.
No, in this one the total energy 5E26 J. The change in the KE of the Earth is -100 kJ, the change in the KE of the car is 50 kJ, and the remaining 50 kJ is again captured by the battery. The only difference being that the source of the energy was the KE of the earth, not the KE of the car.

Humber said:
Oh, I see. In one case the Earth gains momentum and in the other, it loses it, but the total energy is now 100kJ from the Earth + 50kJ, that initially supplied the car ?
No, the car had 0 kJ initially, it gained 50 kJ of KE from the earth.

Humber said:
There are not two events, but one, and the conservation of momentum still applies, no matter which frame is employed.
Correct, as does the conservation of energy.
 
  • #26
DaleSpam said:
Correct, and in the example KE is not conserved in either frame. The KE goes down by 50 kJ in both frames. This 50 kJ of missing KE is the source of the energy put into the battery in both frames.
In the first case, all but 5e-18J.is conserved.
In the second 100kJ comes from the Earth, and the 50kJ is said to be transferred to the car, when at some point, the car gained that energy from another source. The total is 150kJ,
so only 50kJ is conserved.

DaleSpam said:
I agree. And if you will look back at the math you will see that I never did equate them. That was your statement, never mine.
Then it was an omission. Without acceleration, there will be no KE change for the Earth.
The same force is applied to each object and so equal and opposite.

DaleSpam said:
No, in this one the total energy 5E26 J.
Which came from where? The only energy source, is that which internally powered the car.
That is a massive amount of energy, with no pedigree.

DaleSpam said:
The change in the KE of the Earth is -100 kJ, the change in the KE of the car is 50 kJ, and the remaining 50 kJ is again captured by the battery. The only difference being that the source of the energy was the KE of the earth, not the KE of the car.
But the prior to the car gaining the 50kJ from an external source, it had 0J. Therefore, the largest change that can occur when the car returns to 0m/s is 50kJ.

DaleSpam said:
No, the car had 0 kJ initially, it gained 50 kJ of KE from the earth.
It was "given" 50kJ, and then that was transferred to the recovery battery.
And in one case, the car loses momentum and the Earth gains it, while in the second, the car gains it and the Earth loses it. Changing frames, does not change the event.
 
Last edited:
  • #27
DaleSpam, I am sorry that we inflicted Humber on this site. He never will understand frames of reference. He believes that if you have a frame of reference where the Earth is moving that you would have had to accelerate the Earth to that speed and he will ask you where the energy came from as shown by this quote of his:

Originally Posted by DaleSpam View Post

"No, in this one the total energy 5E26 J."

"Which came from where? The only energy source, is that which internally powered the car.
That is a massive amount of energy, with no pedigree.

Many many people have tried to explain the concept of frames of reference but no one has ever got through to him. I wish you the best of luck.
 
  • #28
Humber said:
Then it was an omission. Without acceleration, there will be no KE change for the Earth.
The same force is applied to each object and so equal and opposite.
Are you familiar with Newton's laws? The force is equal and opposite (Newton's 3rd law) but the acceleration is not (Newton's 2nd law). The accelerations are not equal, I never said that they are equal, that was only your statement in posts 19 and 23. I am not sure if your mistake was a simple verbal goof or if you do not understand the difference between force and acceleration.

Humber said:
Which came from where? The only energy source, is that which internally powered the car.
That is a massive amount of energy, with no pedigree.
I have news for you, the kinetic energy of the Earth does not come from some car's power source. The Earth has had kinetic energy for billions of years, long before the car came around. It can just be considered a given in the problem, but technically it came from the gravitational collapse of the dust cloud from which the solar system formed.

Humber said:
And in one case, the car loses momentum and the Earth gains it, while in the second, the car gains it and the Earth loses it.
Correct, but not complete. In one case, (a), the car loses momentum and energy and the Earth gains the momentum and the battery gains the energy. In the other case, (b), the Earth loses momentum and energy and the car gains the momentum and both the car and the battery gain the energy (split evenly). Both are equally valid ways to describe the same situation.
 
Last edited:
  • #29
Subductionzon said:
DaleSpam, I am sorry that we inflicted Humber on this site. ... I wish you the best of luck.
No worries, I will continue as long as it remains enjoyable, and then I will find something else to do.
 
  • #30
DaleSpam said:
Are you familiar with Newton's laws? The force is equal and opposite (Newton's 3rd law) but the acceleration is not (Newton's 2nd law).
The accelerations are not equal, I never said that they are equal, that was only your statement in posts 19 and 23. I am not sure if your mistake was a simple verbal goof or if you do not understand the difference between force and acceleration.
Then you will need to provide an example of kinetic energy exchange, where only one of the objects is accelerated. Indeed the same force is present, but the different masses of car and Earth will result in different accelerations, but not zero for either.

DaleSpam said:
I have news for you, the kinetic energy of the Earth does not come from some car's power source. The Earth has had kinetic energy for billions of years, long before the car came around. It can just be considered a given in the problem, but technically it came from the gravitational collapse of the dust cloud from which the solar system formed.
That's right, so you will need to show good cause, why it should now change velocity in order to power the car.

DaleSpam said:
Correct, but not complete. In one case, (a), the car loses momentum and energy and the Earth gains the momentum and the battery gains the energy.
In the other case, (b), the Earth loses momentum and energy and the car gains the momentum and both the car and the battery gain the energy (split evenly). Both are equally valid ways to describe the same situation.

Case 1: Car loses 50kJ, Earth gains 5e-18J. ( Recovery of the car's 50kJ intial energy.)
Case 2 :Car gains 50kJ, Earth loses 100kJ. ( Generation of energy.)
 
Last edited:
  • #31
Subductionzon said:
DaleSpam, I am sorry that we inflicted Humber on this site. He never will understand frames of reference. He believes that if you have a frame of reference where the Earth is moving that you would have had to accelerate the Earth to that speed and he will ask you where the energy came from as shown by this quote of his:



Many many people have tried to explain the concept of frames of reference but no one has ever got through to him. I wish you the best of luck.

You have a brother here, SZ. What was his forum name? I can't recall it.
 
  • #32
DaleSpam said:
No worries, I will continue as long as it remains enjoyable, and then I will find something else to do.

The first case contains only a small error, so I think we can agree that result is OK.

The second requires some clarification, please.
DaleSpam said:
(b) [itex]p_{i,c}=m v_{i,c}= 0 \ kg \ m/s[/itex]
[itex]\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s[/itex]
[itex]p_{f,c}=p_{i,c}+\Delta p_c = -10000 \ kg \ m/s[/itex]
[itex]v_{f,c}=p_{f,c}/m= -10 \ m/s[/itex]
[itex]KE_{i,c}=1/2 m v_{i,c}^2 = 0 \ kJ[/itex]
[itex]KE_{f,c}=1/2 m v_{f,c}^2 = 50 \ kJ[/itex]

[itex]p_{i,e}=M v_{i,e}= -10^{26} \ kg \ m/s[/itex]
[itex]f_e=-f_c=1000 \ N[/itex]
[itex]\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s[/itex]
[itex]p_{f,e}=p_{i,e}+\Delta p_e = -9.999999999999999999999 \; 10^{25} \ kg \ m/s[/itex]
[itex]v_{f,e}=p_{f,e}/M= -9.999999999999999999999 \ m/s[/itex]
[itex]KE_{i,e}=1/2 M v_{i,e}^2 = 5\;10^{26} \ kJ[/itex]
[itex]KE_{f,e}=1/2 M v_{f,e}^2 = 4.999999999999999999999\;10^{26} \ J[/itex]

So in the second reference frame the energy for charging the battery comes from the KE of the Earth which goes down by 100 kJ. The car gains 50 kJ of kinetic energy.

In this case, the observer, must be in a reference frame that is relative to the centers of mass of the car and planet. The final velocity of the car relative to the Earth, has changed from 10m + 10e-21m/s of the first case, to -9.999999999999999999999m/s in the second case.

Another concern, involves repetition of the event. The reclaimed energy could be used to accelerate the car back to almost 10m/s, because in the first case, the loss of KE to the ground is only 5e-18J or 5e-19W. That is negligible, so can be ignored when the car accelerates, should it be recoverable in that case.
But, in the second case, the Earth loses 100kJ each time the car is braked, and that is (approx) 10kW
Over 200 cycles, a mere 2000s, the Earth has lost ( supplied) almost 1MW. it would seem that a cheap source of energy has gone unexploited.
 
Last edited:
  • #33
Humber said:
Indeed the same force is present, but the different masses of car and Earth will result in different accelerations, but not zero for either.
Good, it sounds like you understand.

Humber said:
That's right, so you will need to show good cause, why it should now change velocity in order to power the car.
I did, see post 21.

Humber said:
Case 1: Car loses 50kJ, Earth gains 5e-18J. ( Recovery of the car's 50kJ intial energy.)
Case 2 :Car gains 50kJ, Earth loses 100kJ. ( Generation of energy.)
Yes. And in each case there is 50 kJ net change in KE which is available for charging the battery.
 
  • #34
Humber said:
The second requires some clarification, please.
Sure, I would be glad to clarify.

Humber said:
The final velocity of the car relative to the Earth, has changed from 10m + 10e-21m/s of the first case, to -9.999999999999999999999m/s in the second case.
Differences in velocity are frame invariant in Newtonian physics. So you are right to be concerned about this.

In frame (a)
[itex]v_{f,e}-v_{f,c}=10^{-21} \ m/s - 0 \ m/s = 10^{-21} \ m/s[/itex]

In frame (b) note the number of 9's behind the decimal point
[itex]v_{f,e}-v_{f,c}= -9.999999999999999999999 \ m/s - (-10 \ m/s) = 10^{-21} \ m/s[/itex]

So the final velocity of the car relative to the Earth is the same in both frames.

Humber said:
Another concern, involves repetition of the event. The reclaimed energy could be used to accelerate the car back to almost 10m/s, ...
But, in the second case, the Earth loses 100kJ each time the car is braked, and that is (approx) 10kW
Over 200 cycles, a mere 2000s, the Earth has lost ( supplied) almost 1MW. it would seem that a cheap source of energy has gone unexploited.
Just as the Earth lost 100 kJ (50 kJ to the car's KE and 50 kJ to the battery), as the car went from 0 to 10 m/s the Earth will gain 100 kJ (50 kJ from the car's KE and 50 kJ from the battery) as the car goes from 10 to 0 m/s. There is no free lunch here.
 
Last edited:
  • #35
DaleSpam said:
Yes. And in each case there is 50 kJ net change in KE which is available for charging the battery.

There is a change of 100kJ for the Earth in the second case, but the car gains only 50kJ.
Does the remaining 50kJ go as heat?
 

Similar threads

Replies
13
Views
1K
Replies
5
Views
3K
Replies
12
Views
1K
  • Mechanics
Replies
7
Views
952
  • Introductory Physics Homework Help
Replies
3
Views
223
Replies
19
Views
1K
Replies
13
Views
704
  • Introductory Physics Homework Help
Replies
6
Views
884
Replies
21
Views
1K
Replies
7
Views
2K
Back
Top