Kinetic Energy Paradox: What Went Wrong?

[pankazmaurya]

That's not quite true. The reason Paul would be wrong even if he used the correct velocities is because he hasn't accounted for all parts of the system. The train is definitely not an isolated body - it is experiencing a force from the rails..

if the rails are frictionless then there would be no requirement of force to maintain speed of 100

 

[pankazmaurya]

Very true.

I guess I'm simplifying it as much as possible for the question (the whole ignore all but kinetic energy). Perhaps too much.

are trying to say that if paul knows that in the second section the car takes 3x amount of fuel against 5/3 as calculated by him then he can logically conclude that he is a part of bigger system and is not looking at the whole thing

 

[JaredJames]

from pauls point of veiw the accelerates from -100 to -200 because he is moving in opposite direction of the car

No, -100 to -200 would imply the car is going backwards.

From Paul's point of view the car accelerates from -100 to 0 for the first section, then from 0 to 100 for the second section.

If you took the train as stationary, the car would we coming at the train at +100.

So from the trains perspective (Paul) the car would go 100 to 200 to 300. But, to get the correct speed you would have to minus the trains speed:
Car Observed Speed = 100, Train = 100, Cars Actual Speed = 0
Car Observed Speed = 200, Train = 100, Cars Actual Speed = 100
Car Observed Speed = 300, Train = 100, Cars Actual Speed = 200

So to calculate the energy of the car you would have to compensate for the trains motion.

if the rails are frictionless then there would be no requirement of force to maintain speed of 100

I believe Gokul was referring to a real life situation and not the simplified version in the question.

are trying to say that if paul knows that in the second section the car takes 3x amount of fuel against 5/3 as calculated by him then he can logically conclude that he is a part of bigger system and is not looking at the whole thing

Correct.

 

[pankazmaurya]

No, -100 to -200 would imply the car is going backwards. To get the cars speed in reference to the train, you add -100 to the cars speed.

So when the car is at 0, (0 - 100) = -100
Car = 100, (100 - 100) = 0Correct.

the train is going in opposite direction of the car or accorgdng to paul the car is moving backwards...while calculating the relative speed you havnt taken the sign of velocties...if you rightfully consider the signs then it will be -100-100=-2009according to paul the car is moving backwards HENCE THE NEGATIVE SIGN TO INDICATE THE DIRECTION...

 

[JaredJames]

the train is going in opposite direction of the car or accorgdng to paul the car is moving backwards...while calculating the relative speed you havnt taken the sign of velocties...if you rightfully consider the signs then it will be -100-100=-2009according to paul the car is moving backwards HENCE THE NEGATIVE SIGN TO INDICATE THE DIRECTION...

I've edited the post above, I don't think rcgdlr is correct.

Regardless, your sign convention above is wrong. You have to take the train as a zero value, so instead of the train moving at 100, the train is at 0 which means means the car is coming at the train at +100. -100 would imply the car moving away from the train.

 

[pankazmaurya]

I've edited the post above, I don't think rcgdlr is correct.

Regardless, your sign convention above is wrong. You have to take the train as a zero value, so instead of the train moving at 100, the train is at 0 which means means the car is coming at the train at +100. -100 would imply the car moving away from the train.

well it depends whether the car has passed paul or not ...whatever the case might be we are not interested in that as we have to only calculate the kinetic energy in which the sign makes no diffrence

 

[JaredJames]

Whether it's passed Paul or not is irrelevant.

You have to assign sign convention.

For me (for ease), + implied the car moving in the opposite direction to Paul and - implies it in the same direction as him.

You can swap them over (like you did) if you want, and no it doesn't make a difference.

Now, as per my previous post (#38?) the car is observed doing 100 at the train then 200 then 300, but you have to remove the trains speed leaving you with 0, 100, 200.

I believe rcgdlr is incorrect with -100, 0 100 because it would imply the car going from moving in the same direction as the train to in the same direction.

And also, the kinetic energy from -100 to 100 is not the same as 0 to 200.

From -100 to 100 would imply the car loses KE until it stops then gains it again - which is not what the car is doing.

 

[pankazmaurya]

Whether it's passed Paul or not is irrelevant.

You have to assign sign convention.

For me (for ease), + implied the car moving in the opposite direction to Paul and - implies it in the same direction as him.

You can swap them over (like you did) if you want, and no it doesn't make a difference.

Now, as per my previous post (#38?) the car is observed doing 100 at the train then 200 then 300, but you have to remove the trains speed leaving you with 0, 100, 200.

but i am again stuck up how will paul conclude that he is moving with 100?

 

[JaredJames]

but i am again stuck up how will paul conclude that he is moving with 100?

You're thinking about this too much.

He can't. He has to know his speed and the cars speed to calculate the energy for the car based on his observations. If he doesn't know the speed of the train he can't work out the speed of the car and as such the energy.

 

[Dale]

In the train's frame of reference the car is traveling in the negative direction, but speed is the magnitude of direction, so the speed is only increasing (from 100-200 in the first leg and 200-300 in the second leg).

The laws of physics work the same in all reference frames. So it is not necessary to consider the ground's reference frame, the problem may be worked entirely in the train's reference frame. As mentioned above, the car is not an isolated system, and the behavior of the ground is very different in the two scenarios.

 

[JaredJames]

In the train's frame of reference the car is traveling in the negative direction, but speed is the magnitude of direction, so the speed is only increasing (from 100-200 in the first leg and 200-300 in the second leg).

Yes, but the energy for going from 200 to 300 (trains reference) =/= 100 to 200 (actual change). So you have to compensate for the trains speed.

 

[pankazmaurya]

You're thinking about this too much.

He can't. He has to know his speed and the cars speed to calculate the energy for the car based on his observations. If he doesn't know the speed of the train he can't work out the speed of the car and as such the energy.

and if he calculates that then he can boldly say that there is fault in the kinetic energy therom

 

[pankazmaurya]

In the train's frame of reference the car is traveling in the negative direction, but speed is the magnitude of direction, so the speed is only increasing (from 100-200 in the first leg and 200-300 in the second leg).

The laws of physics work the same in all reference frames. So it is not necessary to consider the ground's reference frame, the problem may be worked entirely in the train's reference frame. As mentioned above, the car is not an isolated system, and the behavior of the ground is very different in the two scenarios.

how is he behavior of ground different in both case

 

[JaredJames]

and if he calculates that then he can boldly say that there is fault in the kinetic energy therom

If he doesn't allow for the kinetic energy of the train, his results for the car will be incorrect.

His reference frame didn't allow for the trains speed and so his energy requirements for the car were wrong.

 

[Dale]

how is he behavior of ground different in both case

In one frame it is moving in the other it is not. In the frame where the ground is moving it has a very large KE which can be transferred to the car. In the frame where the ground is not moving its KE is 0.

 

[JaredJames]

This is just getting messy.

Paul didn't allow for the trains motion and so his values for speed aren't correct for working out energy requirements. The car is never moving at 300 relative to the ground and so the energy expendature of the car is never that required to get to 300.

If you want to work it out for the car moving at 300, you can but you have to allow for the trains energy as well.

The fact that pauls initial reading is 100 when he knows the car is stopped should tell him that the train is moving at 100.

Aside from that there's not much more to say.

 

[Dale]

How does the train's KE have any bearing on the problem? The train is not interacting with the car. How does the problem change if the train is made of very heavy or very light material?

 

[JaredJames]

How does the train's KE have any bearing on the problem? The train is not interacting with the car. How does the problem change if the train is made of very heavy or very light material?

It doesn't, which is the point.

If the train is doing 100 and you ignore this - as paul did - then it appears the car has gone 200 to 300, which from the trains perspective it may have. But we're talking about the fuel used by the car, and the car hasn't used the fuel to get to 300. It used the fuel to get to 200. Any additional apparent speed is down to the trains motion.

If you took the car to be traveling at 300 from the trains view, then yes it would have the relevant KE. But, again we're not talking about that.

The question is regarding the cars fuel use in stage 2. So we're looking at things from the cars point of view and any effects the train grants to the car are irrelevant so far as the cars fuel use goes - so you must allow for any effects the train has.

The car during stage 2 is traveling at 200 from it's point of view. It's fuel got it to that point. Now, the fuel did not get it to 300 - which means you can't work out the energy requirement to 300 and claim its the fuel required for stage 2. The fuel requirement to go from 100 to 200 =/= 200 to 300 so Paul working out the latter is not correct for how much fuel is required by the car for stage 2. He is ignoring the fact the train is allowing the car to travel at 300 - not the fuel of the car.

 

[Gokul43201]

if the rails are frictionless then there would be no requirement of force to maintain speed of 100

Nope. The [STRIKE]rails are[/STRIKE] road is considered to be frictionless only in the sense that there is no energy loss due to dissipative processes. If the [STRIKE]rails were[/STRIKE] road was truly frictionless (no rolling friction), then it would be impossible for the [STRIKE]train[/STRIKE] car to accelerate. You need an external force to accelerate the [STRIKE]train[/STRIKE] car. Unless you correctly factor in the work done by all external forces, you are not properly applying the work energy theorem.

Draw the free body diagram for the [STRIKE]train[/STRIKE] car. Are there no external forces acting on it? If there are, can you say that the energy of the [STRIKE]train[/STRIKE]car+fuel system should be conserved? If there aren't, how does the [STRIKE]train[/STRIKE] car accelerate?

 

[JaredJames]

Nope. The rails are considered to be frictionless only in the sense that there is no energy loss due to dissipative processes. If the rails were truly frictionless (no rolling friction), then it would be impossible for the train to accelerate. You need an external force to accelerate the train. Unless you correctly factor in the work done by all external forces, you are not properly applying the work energy theorem.

Draw the free body diagram for the train. Are there no external forces acting on it? If there are, can you say that the energy of the train+fuel system should be conserved? If there aren't, how does the train accelerate?

Just to confirm, you know the question relates to the fuel use of a car observed from a moving train?

 

[Gokul43201]

I believe Gokul was referring to a real life situation and not the simplified version in the question.

No, I'm referring to exactly the version of the problem stated in the OP.

Yes, but the energy for going from 200 to 300 (trains reference) =/= 100 to 200 (actual change). So you have to compensate for the trains speed.

There is no such thing as the actual change. You can solve this problem in any inertial frame you choose - you just have to correctly account for all forces (or all parts of the system).

The key here is in recognizing that the Earth is a part of the system, and every time the car accelerates forward, the Earth responds with a minuscule backward acceleration. This tiny change in the Earth's speed multiplied by it's huge mass is needed to conserve momentum. Accordingly, the square of that tiny velocity, multiplied by the Earth's mass accounts for the seemingly paradoxical disappearance of energy.

 

[Gokul43201]

Just to confirm, you know the question relates to the fuel use of a car observed from a moving train?

Oops - I may have mixed up the train with the car, but that doesn't affect the essence of my answer. Replace rails with road as needed, if that is the case.

 

[JaredJames]

Oops - I may have mixed up the train with the car, but that doesn't affect the essence of my answer. Replace rails with road as needed, if that is the case.

Question states ignore all friction, assume fuel goes directly to motion.

There is no such thing as the actual change. You can solve this problem in any inertial frame you choose - you just have to correctly account for all forces (or all parts of the system).

Exactly, Paul doesn't account for the trains involvement and so his figures are off.

The question is talking about the cars fuel use in stage 2. If he doesn't allow for the trains motion, the fuel use he calculates is off.

As I keep saying, the fuel use for the car going from 100 to 200 =/= 200 to 300.

 

[Gokul43201]

Question states ignore all friction, assume fuel goes directly to motion.

Does this make any sense though? Draw the free body diagram of the car. If there are no external forces acting on it, how can it accelerate? Obviously, the effect of burning fuel which turns an axle is manifest in the force the road exerts on the car (the reaction force to what the wheels exert on the road). If you don't want to call it friction, feel free to use some other term (traction?) or make the car ride on a rack and pinion. Whatever the case may be, one can not ignore the force exerted by the road on the car.

EDIT: The only actual way to ignore the role of friction in driving the car and to assume the fuel goes directly into motion would be to fit the car with a rocket engine and decouple the engine from the wheels. But in that case, you have to deal with the KE of the ejected exhaust to solve the problem correctly.

Exactly, Paul doesn't account for the trains involvement and so his figures are off.

Relativity demands that Paul not have to do anything special to account for the train's involvement - he can solve the problem from any inertial reference frame he chooses, but he has to carefully figure out the speeds of all parts of the system, (in this case, the car and the earth) relative to his frame.

Alternatively, he must calculate forces and displacements separately in each frame and use the work energy theorem correctly. Specifically, the work done on the car by the road in the Earth frame is not equal to the work done on the car by the road in the train frame.

 

[JaredJames]

Does this make any sense though? Draw the free body diagram of the car. If there are no external forces acting on it, how can it accelerate? Obviously, the effect of burning fuel which turns an axle is manifest in the force the road exerts on the car (the reaction force to what the wheels exert on the road). If you don't want to call it friction, feel free to use some other term (traction?) or make the car ride on a rack and pinion. Whatever the case may be, one can not ignore the force exerted by the road on the car.

Were you never at school when they said "simplify be ignoring friction losses"? Come on, you're playing dumb now. You know full well they tell you to ignore it to make it easier.

Relativity demands that Paul not have to do anything special to account for the train's involvement - he can solve the problem from any inertial reference frame he chooses, but he has to carefully figure out the speeds of all parts of the system, (in this case, the car and the earth) relative to his frame.

Alternatively, he must calculate forces separately in each frame and use the work energy theorem correctly. Specifically, the force on the car from the road in the Earth frame is not equal to the force on the car from the road in the train frame.

Yes, which is what he fails to do.

I'm not arguing he can't do it, I'm just saying he doesn't - which is where the problem lies. I know you're right though.

 

[Gokul43201]

Were you never at school when they said "simplify be ignoring friction losses"? Come on, you're playing dumb now. You know full well they tell you to ignore it to make it easier.

I'm absolutely not playing dumb. There is a very clear distinction between ignoring losses due to friction (which is what we should do in this problem) and ignoring friction itself (which we can not do). In a pure rolling friction there is no dissipation (much like static friction), so by assuming a rolling friction without any kinetic component, we are indeed ignoring losses due to friction without actually ignoring the rather important role that friction does play.

Why don't you answer my earlier question: if there is no friction, where in your free body diagram is the external force that causes the car to accelerate?

 

[JaredJames]

I'm absolutely not playing dumb. There is a very clear distinction between ignoring losses due to friction (which is what we should do in this problem) and ignoring friction itself (which we can not do). In a pure rolling friction there is no dissipation (much like static friction), so by assuming a rolling friction without any kinetic component, we are indeed ignoring losses due to friction without actually ignoring the rather important role that friction does play.

Why don't you answer my earlier question: if there is no friction, where in your free body diagram is the external force that causes the car to accelerate?

Ah I see, I thought you meant friction loses.

 

[rcgldr]

Trying to get this thread back on track. If the road itself is used as a frame of reference, then there is no perceived change in the kinetic energy of the earth, and all of it goes into the car. Also the application of force to accelerate the car is occurring between the tires and the road, so the road is the appropriate frame of reference.

If a frame of reference with a speed relative to the road is used, the the energy change of the road/earth needs to be taken into account for the total change in kinetic energy of Earth and car to equal the chemical or electrial potential energy of the fuel converted into kinetic energy by the engine.

This same subject has been brought up twice recently. Take a look at post #3 of this thread:

https://www.physicsforums.com/showthread.php?t=464859

 

[JaredJames]

Perfect rcgldr.

I think we're all on the same track now.

So yeah, you have to take into account the train - as demonstrated in post 3 (of that thread) there I believe.

Perhaps you could bring your first post there over because that would answer this nicely.

 

[rcgldr]

So yeah, you have to take into account the train - as demonstrated in post 3 (of that thread) there I believe.

Not so much the train itself, but the trains frame of reference, since it's speed relative to the road and surface of the Earth is non-zero. The mass of the train can be be ignored when considering the total mass of the earth, since it's such a relatively tiny component of the total mass.

 

[JaredJames]

Not so much the train itself, but the trains frame of reference, since it's speed relative to the road and surface of the Earth is non-zero. The mass of the train can be be ignored when considering the total mass of the earth, since it's such a relatively tiny component of the total mass.

Trust me, my wording may not be clear but I do understand. You only have to read back through my posts and you'll see I'm discussing the trains frame of reference - it's speed relative to the car has to be taken into account etc etc.

I think this has been answered well enough now.

 

[Dale]

It doesn't, which is the point.

If the train is doing 100 and you ignore this - as paul did - then it appears the car has gone 200 to 300, which from the trains perspective it may have. But we're talking about the fuel used by the car, and the car hasn't used the fuel to get to 300. It used the fuel to get to 200. Any additional apparent speed is down to the trains motion.

If you took the car to be traveling at 300 from the trains view, then yes it would have the relevant KE. But, again we're not talking about that.

The question is regarding the cars fuel use in stage 2. So we're looking at things from the cars point of view and any effects the train grants to the car are irrelevant so far as the cars fuel use goes - so you must allow for any effects the train has.

The car during stage 2 is traveling at 200 from it's point of view. It's fuel got it to that point. Now, the fuel did not get it to 300 - which means you can't work out the energy requirement to 300 and claim its the fuel required for stage 2. The fuel requirement to go from 100 to 200 =/= 200 to 300 so Paul working out the latter is not correct for how much fuel is required by the car for stage 2. He is ignoring the fact the train is allowing the car to travel at 300 - not the fuel of the car.

You are missing the point of the question. The point is that Newton's laws are invariant under Galilean boosts, this is known as Galilean relativity. You can do the analysis just fine in the train's reference frame, you do not need to do the analysis in the ground's reference frame. The physics is the same. The only difference is the KE of the ground and the speed of the car.

The KE of the train is not relevant because the train does not interact with the car at all, consider a super-massive train and a super-light train it does not change the problem at all. Now, consider a super-massive ground (normal) and a super-light ground, the problem changes significantly because the car interacts with the ground.

The car gains more energy in the train's frame than in the ground's frame. That additional energy comes from the KE of the earth.

EDIT: never mind, I see we all agree from the intervening posts

 

[JaredJames]

DaleSpam,

The car's fuel use is the same in either case correct? Of course.

In which case, no matter which method you use (correctly) you will end up with the same required fuel for the car. Do we have that? No.

I'd point out that in the quote you used from me I note there would be a difference in KE of the car for the relevant speeds.

 

[Dale]

In which case, no matter which method you use (correctly) you will end up with the same required fuel for the car. Do we have that? No.

Correct. My point was only that the reason is because Paul negelcted the KE of the earth, not because Paul neglected the KE of the train. The KE of the train is irrelevant.