KE of system / different reference frames question

[Humber]

I'd like to think that if the car's acceleration using the Earth for traction makes the world spin faster, etc, when the car eventually stops, it will surely slow the Earth back down by that same (infinitesimal) amount.

;)

Infinitesimal it is. With the given numbers, 5e-18J and a velocity change of 1e-21m/s. That is well wide of the 100kJ claimed, and of the 100W of the o.p.
In reality, the Earth is not an infinitly rigid sphere, and will dissipate energy, just as it does for Earthquakes and nuclear weapons tests.

 

[Dale]

vi,c =initial velocity of car
vf,e = final velocity of earth
The difference is therefore the relative velocity of car and Earth.

No, the initial relative verlocity of the Earth wrt the car would be v_{i,e}-v_{i,c} the final relative velocity would be v_{f,e}-v_{f,c}. Your quantity v_{f,e}+v_{i,c} is not a relative velocity of anything. See the link I posted.

Then you don't have a case for your calculations.

Sure I do, Newton's laws are the basis for all of my calculations. That and the standard definitions for the various quantities.

The reduction of the Earth's momentum, is the basis of your KE claim.
It applies to the second case too. Which is why you will always get 100kJ, regardless of mass. It is an error.

What makes you think it is an error? A small rock and a big rock fall at the same acceleration despite the difference in the mass. Not everything is a function of every mass.

Δp_c*v_f,c
= -10000kg.m/s*10m/s = 100000 kg.m²/s²
= -100kJ

If you are trying to calculate the change in the KE of the car then you are missing a factor of 1/2. Otherwise I don't know what you are trying to calculate.

 

[Dale]

You have calculated 100kJ energy from the ground of which 50kJ is stored in the battery.
That leaves 50kJ unaccounted for.

I don't know if you are being deliberately irritating or you really can't grasp something this basic. I have said this multiple times already:

Earth loses 100 kJ KE
Car gains 50 kJ KE
Battery gains 50 kJ electrochemical energy
All energy is accounted for.

There is no missing energy 50 kJ + 50 kJ = 100 kJ, if you don't believe me then get a calculator and confirm.

Which does not allow 50kJ for the car and 50kJ for the battery, because the system is a Kinetic Energy Recovery System. And you would still have a difference of 50kJ, when compared with the first case.

The KERS recovers 50 kJ of electrochemical energy from the KE of the Earth and the car gains 50 kJ of KE from the KE of the earth. The Earth loses 100 kJ of KE.

Why don't you ponder that a bit until you get it.

 

[Humber]

No, the initial relative verlocity of the Earth wrt the car would be v_{i,e}-v_{i,c} the final relative velocity would be v_{f,e}-v_{f,c}. Your quantity v_{f,e}+v_{i,c} is not a relative velocity of anything. See the link I posted.

They are different on each case.

Sure I do, Newton's laws are the basis for all of my calculations. That and the standard definitions for the various quantities.
What makes you think it is an error? A small rock and a big rock fall at the same acceleration despite the difference in the mass. Not everything is a function of every mass.

And the change in momentum that you said you did not make, though it is quite clear that you did.

If you are trying to calculate the change in the KE of the car then you are missing a factor of 1/2. Otherwise I don't know what you are trying to calculate.

The primary error remains. The result of the elaborated calculation that eliminates the Earth's mass in the process is, and always is;

iΔpc*vf,c
= -10000kg.m/s*10m/s = 100000 kg.m2/s2
= -100kJ

 

[Humber]

I don't know if you are being deliberately irritating or you really can't grasp something this basic. I have said this multiple times already:

Earth loses 100 kJ KE
Car gains 50 kJ KE
Battery gains 50 kJ electrochemical energy
All energy is accounted for.

In the first case, the car will be stationary, with the rest state KE, and with a known amount of charge in the battery.
The above says that in the second case, there will be both charge in the battery and kinetic energy. A clear violation of both the conservation of momentum and energy.
All the result of one arithmetic error, that has been flagged to your attention.
Δpc*vf,c
= -10000kg.m/s*10m/s = 100000 kg.m2/s2
= -100kJ

There is no missing energy 50 kJ + 50 kJ = 100 kJ, if you don't believe me then get a calculator and confirm.

Your calculation says you have made that error.

The KERS recovers 50 kJ of electrochemical energy from the KE of the Earth and the car gains 50 kJ of KE from the KE of the earth. The Earth loses 100 kJ of KE.
Why don't you ponder that a bit until you get it.

Yes, I agree. Your calculation confirms that you have made that error.

 

[jduffy77]

I should clarify something which most people interested in physics will not be familiar with. What Humber means when he says "a calculation confirms an error" is not that an error was made in the traditional sense, like incorrect math or a logical inconsistency. He simply means that the result conflicts with his preconceived expectation of what the result should be. So it is wrong.

 

[Humber]

The correct solution to this problem, when the juvenile runs out of excuses, and recognizes his error, relies on nothing more than the conservation of momentum, which is a law quite independent of frames.

There are only two bodies involved: the car and the Earth so, a very simple case. When the car slows, it changes momentum;

Car gains momentum -p
Earth gains momentum p

p^2/2m = p^2v^2/2m = 1/2mv^2, which is the familiar KE equation.

The car loses KE = p²/2m_car
The Earth gains KE = p²/2m_earth

The total energy is therefore;
p²/2m_car+ p²/2m_earth

Because m_earth is very large ,~6e24kg, the second term is negligible and can be approximated to the first term.

Initially, all of that energy will have come from the fuel of the F1 car, and in part, like the consumer counterpart, KERS is intended to reduce fuel consumption by returning energy that would otherwise be lost.

In a conventionally braked vehicle, the KE of the car's mass is dissipated as heat in the brakes. KERS does not affect momentum transfer to the ground, but stores the KE that is otherwise lost to heat.

That energy may be transferred to a flywheel;
The angular momentum L, of a particle, and the moment of inertia, I
L = r.p = r.mv
L =I.ω
KE = 1/2.I.ω²

In the given example of a 1000kg vehicle braking to a stop from 10m/s, and where the mass of the Earth is 1e25kg, the energy transferred to the ground is;

(100000kg.m/s)^2/2e25kg
= 5e-18J

There is no significant transfer of energy to or from the F1 car and the ground as a result of KERS. The conservation of momentum says that result is independent of frame.

 

[Humber]

The following calculation shows that there is negligible transfer to the ground, and is in agreement, also producing 5e-18J

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

p_{i,e}=M v_{i,e}=0 \ kg \ m/s
f_e=-f_c=1000 \ N
\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s
p_{f,e}=p_{i,e}+\Delta p_e = 10000 \ kg \ m/s
v_{f,e}=p_{f,e}/M= 10^{-21} \ m/s
KE_{i,e}=1/2 M v_{i,e}^2 = 0 \ kJ
KE_{f,e}=1/2 M v_{f,e}^2 = 5\;10^{-18} \ J
All that remains, is to show that result is independent of frame.
Frame independence for the result can be shown by the introduction of an observer, moving relative to the Earth and Car at velocity u.

m = mass of car

ΔKE(car)

= (−m u − p)^2/2m − (−m u)^2/2m

= (m^2 u^2 −2m u·p + p^2)/2m − (m^2 u^2)/2m

= −u·p + p^2/2m


M = mass of Earth.

ΔKE(earth)

=(−M u + p)^2/2M − (−M u)^2/2M

= (M^2 u^2 + 2M u·p + p^2)/2M - (M^2 u^2)/2M

= u.p + p^2/2M

The total energy is once again

p^2/2m + p^2/2M

 

[Dale]

When a car brakes, momentum is transferred from the car to the ground.
From the conservation of momentum;

Car gains momentum -p
Earth gains momentum p

This is true, due to Newton's 3rd law.

The car loses KE = p²/2m_car
The Earth gains KE = p²/2m_earth

This is false, in general.

If some object gains momentum Δp then p_f=p_i+\Delta p

Dividing both sides by the mass m we get v_f=v_i+\Delta v

Substituting that into the expression for the object's final KE we obtain
KE_f=\frac{m}{2} v_f^2 = \frac{m}{2} (v_i+\Delta v)^2 = \frac{m}{2} (v_i^2 + 2 v_i \Delta v +\Delta v^2) = KE_i + v_i \Delta p + \Delta p^2/2m
\Delta KE = KE_f - KE_i = v_i \Delta p + \Delta p^2/2m

This is different from what you wrote because of the v_i \Delta p term. Your neglecting that term is actually your key conceptual error, so I do hope you spend some time examining the brief derivation. If the object is particularly massive, like the earth, then the \Delta p^2/2m term drops out, but the v_i \Delta p term does not.

 

[Dale]

I'd like to think that if the car's acceleration using the Earth for traction makes the world spin faster, etc, when the car eventually stops, it will surely slow the Earth back down by that same (infinitesimal) amount.

That is correct.

 

[Dale]

And the change in momentum that you said you did not make, though it is quite clear that you did.

What exactly are you claiming that I said? Please use the quote feature to show exactly where I said I did not make a change in momentum. If you cannot find where I said it then I didn't say it, you only inferred it. Given this conversation so far, you probably inferred wrongly.

 

[Dale]

The above says that in the second case, there will be both charge in the battery and kinetic energy.

Yes, that is correct, 50 kJ charge in the battery and 50 kJ KE in the car.

A clear violation of both the conservation of momentum and energy.

It is not a violation of either conservation of momentm or conservation of energy.

 

[Dale]

The correct solution to this problem, when the juvenile runs out of excuses, and recognizes his error, relies on nothing more than the conservation of momentum, which is a law quite independent of frames.

I love the arrogance of the ignorant. Keep it coming Humber, it is very amusing.

There are only two bodies involved: the car and the Earth so, a very simple case. When the car slows, it changes momentum;

Car gains momentum -p
Earth gains momentum p

p^2/2m = p^2v^2/2m = 1/2mv^2, which is the familiar KE equation.

The car loses KE = p²/2m_car
The Earth gains KE = p²/2m_earth

Again, this is false, in general. See my post 59 above. This is your key error, so please spend some time studying the derivation.

 

[Humber]

Car gains momentum -p
Earth gains momentum p
p^2/2m = p^2v^2/2m = 1/2mv^2, which is the familiar KE equation.

The car loses KE = p^2/2mcar
The Earth gains KE = p^2/2mearth

This is false, in general.

If some object gains momentum Δp then p_f=p_i+\Delta p

Dividing both sides by the mass m we get v_f=v_i+\Delta v

Substituting that into the expression for the object's final KE we obtain
KE_f=\frac{m}{2} v_f^2 = \frac{m}{2} (v_i+\Delta v)^2 = \frac{m}{2} (v_i^2 + 2 v_i \Delta v +\Delta v^2) = KE_i + v_i \Delta p + \Delta p^2/2m
\Delta KE = KE_f - KE_i = v_i \Delta p + \Delta p^2/2m

This is different from what you wrote because of the v_i \Delta p term. Your neglecting that term is actually your key conceptual error, so I do hope you spend some time examining the brief derivation. If the object is particularly massive, like the earth, then the \Delta p^2/2m term drops out, but the v_i \Delta p term does not.

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

p_{i,e}=M v_{i,e}=0 \ kg \ m/s
f_e=-f_c=1000 \ N
\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s
p_{f,e}=p_{i,e}+\Delta p_e = 10000 \ kg \ m/s
v_{f,e}=p_{f,e}/M= 10^{-21} \ m/s
KE_{i,e}=1/2 M v_{i,e}^2 = 0 \ kJ Change in Earth's KE;
KE_{f,e}=1/2 M v_{f,e}^2 = 5\;10^{-18} \ J

KE_earth =p^2/2M_earth
= (10000kg.m/s)^2/2*1e25
= 5e-18J

 

[Dale]

KE_earth =p^2/2M_earth
= (10000kg.m/s)^2/2*1e25
= 5e-18J

Which works since v_i=0 for the Earth in frame (a). It does not work in frame (b) where v_i \ne 0 for the earth.

 

[Humber]

I love the arrogance of the ignorant. Keep it coming Humber, it is very amusing.

Again, this is false, in general. See my post 59 above. This is your key error, so please spend some time studying the derivation.

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

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

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.

KE = p²/2m = m²v²/2m = 1/2mv²

When you finish falling over yourself and the arithmetic churn, you can stop.

 

[Dale]

Do you agree with the derivation in post 59?

If yes, then do you understand how it was your key mistake?

If no, then which specific step of the derivation do you believe is in error, and what do you believe is the error?

 

[Humber]

Do you agree with the derivation in post 59?

If yes, then do you understand how it was your key mistake?

If no, then which specific step of the derivation do you believe is in error, and what do you believe is the error?

You correct it. I am not going to help you chase your tail.

 

[Dale]

It is correct. Do you agree or not?

It sounds like you can't actually see anything wrong with it but you don't like the conclusion, so you illogically reject it.

 

[Humber]

It is correct. Do you agree or not?
It sounds like you can't actually see anything wrong with it but you don't like the conclusion, so you illogically reject it.

You have now made 4 unforced errors, and contradicted your own result.

When that happens, you will need to find out where your error is. I am quite willing to let you live with it, if you choose to do so.

 

[mender]

It is correct. Do you agree or not?

It sounds like you can't actually see anything wrong with it but you don't like the conclusion, so you illogically reject it.

You're pretty good at this; some people take quite a few more posts to see that.

 

[mender]

When that happens, you will need to find out where your error is. I am quite willing to let you live with it, if you choose to do so.

This is known as an assumed close.

 

[Humber]

You're pretty good at this; some people take quite a few more posts to see that.

He has contradicted his own result, and made several other errors in the process.
He will need to correct those, not I.

 

[Humber]

This is known as an assumed close.

And you are known as a "troll", I believe?

 

[my_wan]

vi,c =initial velocity of car
vf,e = final velocity of earth
The difference is therefore the relative velocity of car and Earth.
They are not the same in each case.

How can a relative velocity not be the same in each case? If we are on separate meteors then whatever velocity I say you have is exactly the same velocity you say I have. There can be no difference. "Difference" by definition means to subtract, not add. "And" is not explicitly an addition or subtraction operator, it merely associates two terms for which an operator must be provided.

 

[mender]

And you are known as a "troll", I believe?

The black pot speaketh!

 

[mender]

He has contradicted his own result, and made several other errors in the process.
He will need to correct those, not I.

Sounds like it may be a good time to establish a baseline.

Humber, what's 2 + 2?

 

[Dale]

You have now made 4 unforced errors, and contradicted your own result.

When that happens, you will need to find out where your error is. I am quite willing to let you live with it, if you choose to do so.

None of this answers the post you quoted; you are merely attempting to dodge the question.

Do you agree or disagree with the proof of post 59?

 

[my_wan]

Sounds like it may be a good time to establish a baseline.

Humber, what's 2 + 2?

Dale is correct that you either must provide the basis of your disagreement or chalk up such statements above to diversionary tactics. Dale attempted to establish a baseline with his post. You can either work with it or choose another baseline that helps to actually make your point clearer. Diversions, such as above, which the only apparent purpose is to obscure the baseline of the discussion do not fit anywhere.

 

[Humber]

How can a relative velocity not be the same in each case? If we are on separate meteors then whatever velocity I say you have is exactly the same velocity you say I have. There can be no difference. "Difference" by definition means to subtract, not add. "And" is not explicitly an addition or subtraction operator, it merely associates two terms for which an operator must be provided.

That is the point. They should be the same, but aren't.

Case 1
v_{i,c}=p_{f,c}/m= 10 \ m/s Initial velocity of car
v_{f,c}=p_{f,c}/m= 0 \ m/s Final velocity of car

v_{i,e}=p_{f,e}/M= 0 \ m/s Initial velocity of Earth
v_{f,e}=p_{f,e}/M= 10^{-21} \ m/s Final velocity of Earth

Obviously, when the car is at rest, the final velocities of car and Earth should be the same,
and -10e-21m/s, if the Car has accelerated the Earth in the opposite direction.
Total change for the car = 10m/s + 10e-21m/sCase 2;
v_{i,c}=p_{f,c}/m= 0\ m/s Initial velocity of car
v_{f,c}=p_{f,c}/m= -10 \ m/s Final velocity of car

v_{i,e}=p_{f,e}/M= -10m/s \ m/s Initial velocity of Earth
v_{f,e}=p_{f,e}/M= -9.999999999999999999999 \ m/s Final velocity of Earth

This makes no sense at all.

 

[Humber]

Dale is correct that you either must provide the basis of your disagreement or chalk up such statements above to diversionary tactics. Dale attempted to establish a baseline with his post. You can either work with it or choose another baseline that helps to actually make your point clearer. Diversions, such as above, which the only apparent purpose is to obscure the baseline of the discussion do not fit anywhere.

I don't need to do that. The error is quite clear.
The change is 1e-18J in the first case and agrees with KE =p^2/2m, but not the revised calculation. The error there, is rather obvious.

 

[Humber]

Sounds like it may be a good time to establish a baseline.

Humber, what's 2 + 2?

You are trolling, and have nothing to contribute, but simply want to avoid the consequences for your ddw cart, when it is proven that there is no ground energy.
I have done that by two different means, and now you want to spread your bluff denial, and emulation of Spork, to here.

 

[mender]

Speaking of contributions, do you have pictures of your carts? RP said you made a couple of new ones to go with your cow-cart.

 

[rcgldr]

there is no ground energy.

DDWFTTW like devices rely on the difference in speed between two media. Which media that KE is extracted from depends on the frame of reference.

In the case of a Brennan_torpedo, from a water based frame of reference, the retracted wires are the source of KE. Change the frame of reference so that the wires are attached to "non-moving" posts, with water flowing downstream. Using the posts as a frame of reference, then the water is the source of the KE.

Getting back to a DDWFTTW cart, from an air (true wind) based frame of reference, the ground (earth) is the source of KE.

 

[Humber]

Speaking of contributions, do you have pictures of your carts? RP said you made a couple of new ones to go with your cow-cart.

Is that the topic of the OP? Again, you are trolling, and exposing yourself as a zealot.

 

[mender]

No; trolling would be to evade answering questions, even simple ones, to keep a discussion from reaching its inevitable (to most people at least) conclusion.

Need I pay homage to you, the master, before you provide a couple of pictures of your cart? And it was you who brought up the topic of carts.

 

[Humber]

DDWFTTW like devices rely on the diffrence in speed between two media. Which media that KE is extracted from depends on the frame of reference.

In the case of a Brennan_torpedo, from a water based frame of reference, the retracted wires are the source of KE. Change the frame of reference so that the wires are attached to "non-moving" posts, with water flowing downstream. Using the posts as a frame of reference, then the water is the source of the KE.

Getting back to a DDWFTTW cart, from an air (true wind) based frame of reference, the ground (earth) is the source of KE.

Is that the topic of this thread? And the torpedo, as the reference states, was directly powered by on-shore steam engines. That has nothing at all to do with relative motion.

 

[Humber]

Need I pay homage to you, the master, before you provide a couple of pictures of your cart? And it was you who brought up the topic of carts.

mender;3692808

 

[mender]

Back to the topic then: answer Dalespam's question.

None of this answers the post you quoted; you are merely attempting to dodge the question.

Do you agree or disagree with the proof of post 59?

 

[mender]

You can continue to show that you have no control at all of the zeal that forces you to make a fool of yourself.

Come on now, stay on topic and answer the question!

 

[Humber]

Come on now, stay on topic and answer the question!

Sounds like it may be a good time to establish a baseline.

Humber, what's 2 + 2?

Speaking of contributions, do you have pictures of your carts? RP said you made a couple of new ones to go with your cow-cart.

No; trolling would be to evade answering questions, even simple ones, to keep a discussion from reaching its inevitable (to most people at least) conclusion.

Need I pay homage to you, the master, before you provide a couple of pictures of your cart? And it was you who brought up the topic of carts.

Come on now, stay on topic and answer the question!

Back to the topic then: answer Dalespam's question.

The black pot speaketh!

How many times do you need to hear that the source of energy is frame dependent before you'll accept reality?

You are embarrassing yourself.

 

[Subductionzon]

Humber, you know better than that. A simple Galilean transformation shows you that the Brennan torpedo being powered by a steam driven cable is the same as a Brennan torpedo going downstream faster than the stream. Again it is all about relative velocities.

 

[Humber]

Humber, you know better than that. A simple Galilean transformation shows you that the Brennan torpedo being powered by a steam driven cable is the same as a Brennan torpedo going downstream faster than the stream. Again it is all about relative velocities.

And about cables. They are an engineering problem. The advantage of Brennan's system, was that it only needed two cables. Rudder control and drive on the same pair. That demanded a differential drive for the prop, and so two counter-rotating props. A complex arrangement that paid off in the distance the torpedo could travel, within the limits of cable capacity.

And the topic is KERS and ground energy.

 

[mender]

And the topic is KERS and ground energy.

How many times do you need to hear that the source of energy is frame dependent before you'll accept reality?

 

[my_wan]

That is the point. They should be the same, but aren't.

Ok, looking over your response I see a more reasonable attempt to respond.

Consider this analogy. We are outside and separate from the space shuttle and get in a fight. We knock each other back and start flying away from the space shuttle at 5 m/s. But wait, we are flying away from each other at 10 m/s. In the car/earth analogy the one of us is the 5 m/s velocity, and associated k_e, momentum, etc. The other one of us is the energy that went into the battery. So the total energy is accounted for with a k_e of the car and the k_p of the battery. Both exactly half just like our fight resulted in half the velocity for each of us relative to the space shuttle.

You are attempting to conserve k_e in your debate with DaleSpam without accounting for where the k_p that went into the battery came from.

 

[Humber]

Ok, looking over your response I see a more reasonable attempt to respond.

Consider this analogy. We are outside and separate from the space shuttle and get in a fight. We knock each other back and start flying away from the space shuttle at 5 m/s. But wait, we are flying away from each other at 10 m/s. In the car/earth analogy the one of us is the 5 m/s velocity, and associated k_e, momentum, etc. The other one of us is the energy that went into the battery. So the total energy is accounted for with a k_e of the car and the k_p of the battery. Both exactly half just like our fight resulted in half the velocity for each of us relative to the space shuttle.

You are attempting to conserve k_e in your debate with DaleSpam without accounting for where the k_p that went into the battery came from.

That has nothing at all to do with the matter at hand, and there is clearly a problem with the relative velocities in each case, as I quite clearly showed you. They are not my numbers. One wonders why you raised the matter, but now want to drop it, in favour of another.

 

[rcgldr]

DDWFTTW like devices rely on the difference in speed between two media. Which media that KE is extracted from depends on the frame of reference.

Is that the topic of this thread?

Yes, from the first post:

This post appeared on a ddwfttw forum: "For those who actually care, relative to any frame other than that of the ground, the ground does have energy. ... KERS ... initial energy from fuel

It doesn't matter what the initial energy source was.

Asume an inertial frame of reference at some initial fixed speed and direction wrt ground. In that frame, the ground (earth) has energy. If a vehicle accelerates or decelerates by applying a force to the ground, then the energy of the ground (earth) is changed wrt the frame of reference. In the cases where the speed of the ground (earth) is increased wrt frame of reference, then energy is added to the ground (earth). In the cases where the speed of the ground (earth) is decreased wrt frame of reference, then energy is extracted from the ground (earth). I don't understand why there is an issue with this concept.

 

[Humber]

This is true, due to Newton's 3rd law.

This is false, in general.

If some object gains momentum Δp then p_f=p_i+\Delta p

Dividing both sides by the mass m we get v_f=v_i+\Delta v

Substituting that into the expression for the object's final KE we obtain
KE_f=\frac{m}{2} v_f^2 = \frac{m}{2} (v_i+\Delta v)^2 = \frac{m}{2} (v_i^2 + 2 v_i \Delta v +\Delta v^2) = KE_i + v_i \Delta p + \Delta p^2/2m
\Delta KE = KE_f - KE_i = v_i \Delta p + \Delta p^2/2m

This is different from what you wrote because of the v_i \Delta p term. Your neglecting that term is actually your key conceptual error, so I do hope you spend some time examining the brief derivation. If the object is particularly massive, like the earth, then the \Delta p^2/2m term drops out, but the v_i \Delta p term does not.

Momentum is conserved between the car and the Earth.
The car gains -p, and the Earth gains p, so in the above case, that is Δp. It is p_i, which drops out.
In your first post you obtained a result based upon that fact, which is why you also derived
1e-18J, which is the same result as obtained from KE = p²/2m_earth.

Momentum is always conserved, and is frame independent, so that is a unique and frame independent result.

 

[Humber]

Yes, from the first post:

It doesn't matter what the initial energy source was.

Asume a frame of reference at some initial fixed speed and direction wrt ground. In that frame, the ground (earth) has energy. If a vehicle accelerates or decelerates by applying a force to the ground, then the energy of the ground (earth) is changed wrt the frame of reference. In the cases where the speed of the ground (earth) is increased wrt frame of reference, then energy is added to the ground (earth). In the cases where the speed of the ground (earth) is decreased wrt frame of reference, then energy is extracted from the ground (earth). I don't understand why there is an issue with this concept.

Then there is a futher inconsistency. In the first post, there is an equal change of power
of 100W, or 100J/s, but Dalespam's solution, has twice that change for the ground energy.
You need ground energy, to support the cart, and that will cause you to contradict yourselves, time and time again.
Momentum is conserved, and that says that there is no ground energy, but for a negligible amount. The energy transferred to the ground, due to surface deformation, will swamp that.

 

[my_wan]

That has nothing at all to do with the matter at hand, and there is clearly a problem with the relative velocities in each case. One wonders why you raised the matter, but now want to drop that, in favour of another.

This is merely a reframing of the exact same problem. Perhaps not useful, and you are free to choose another but certainly not dropping the original matter. So let's go back to the original.

You have calculated 100kJ energy from the ground of which 50kJ is stored in the battery.
That leaves 50kJ unaccounted for.

Nope, it doesn't. Think of it this way then. Wouldn't slowing down fro 100 m/s be the same as slowing down by 50 m/s twice? Yet m50^2 + m50^2 \neq m100^2. Slowing down to half your speed take a lot more energy than half your energy. Look up the "Merton mean speed theorem" if you want to know why it is half the total acceleration to stop the car.