Kinetic Energy Paradox: What Went Wrong?

[pankazmaurya]

watz gung wrong here

 

[zhermes]

What do you think?

 

[JaredJames]

watz gung wrong here

Your spelling. Please not the rules on English and text speak.

This looks like homework. Could you provide your opinion on the matter?

 

[pankazmaurya]

Your spelling. Please not the rules on English and text speak.

This looks like homework. Could you provide your opinion on the matter?

if this looks like homework to you then you should better know its answer

 

[pankazmaurya]

What do you think?

there is something very basic in this problem that anyone is not able to catch

 

[JaredJames]

if this looks like homework to you then you should better know its answer

Yes, it looks like homework and as such the PF rules require your attempt first.

Once we've got that I'll gladly offer my answer to the problem.

 

[pankazmaurya]

Yes, it looks like homework and as such the PF rules require your attempt first.

Once we've got that I'll gladly offer my answer to the problem.

though i m not able to get the problem clearly but i have understood that there is a blunder in considering the train frames of reference to apply the work energy thereom

 

[JaredJames]

though i m not able to get the problem clearly but i have understood that there is a blunder in considering the train frames of reference to apply the work energy thereom

And what do you think the blunder is?

 

[pankazmaurya]

And what do you think the blunder is?

that is wat I am not geting.now ur chance...

 

[JaredJames]

that is wat I am not geting.now ur chance...

I'd lose the attitude, I'm not going to tell you the answer. The rules of PF are that I cannot do your work for you and I have no intention of doing so. I can only provide you guidance.

So let's get you started.

What is the formula for kinetic energy? What are the two relevant formulas for work done?

 

[pankazmaurya]

I'd lose the attitude, I'm not going to tell you the answer. The rules of PF are that I cannot do your work for you and I have no intention of doing so. I can only provide you guidance.

So let's get you started.

What is the formula for kinetic energy? What are the two relevant formulas for work done?

its simple u can't tell the answer becuase you don't know. and my dear m not interested in the answer .

 

[JaredJames]

its simple u can't tell the answer becuase you don't know. and my dear m not interested in the answer .

I've just derived the answer on a pad in front of me for you.

I can tell you it's relating to the KE work done and velocity equations.

Drop the attitude or no one here will help you. If you don't care about the answer I recommend this thread be locked right now as any further help will be pointless if you don't care about it.

I again refer you to the rules you agreed to when you signed up relating to you must show some attempt before we can help.

 

[pankazmaurya]

I've just derived the answer on a pad in front of me for you.

I can tell you it's relating to the KE work done and velocity equations.

Drop the attitude or no one here will help you. If you don't care about the answer I recommend this thread be locked right now as any further help will be pointless if you don't care about it.

I again refer you to the rules you agreed to when you signed up relating to you must show some attempt before we can help.

would you elaborate on your bakwas...what kind of efforts do you want

 

[JaredJames]

would you elaborate on your bakwas...

Last chance or I'll report the thread and stop offering help. Drop the attitude.

what kind of efforts do you want

EDIT: Forget that. The answer is more obvious than all my workings here.

Look at the values for speed quoted. What is wrong with Pauls speed values?

 

[pankazmaurya]

Last chance or I'll report the thread and stop offering help. Drop the attitude.


You indicated you're not sure where to begin, so I pointed you in the direction by asking you for some equations:

stop being bossy...the energy of the fuel will be converted to kinetic energy of the car according to half my square

 

[JaredJames]

stop being bossy...

You asked for help, I'm giving it to you and for that all I ask is you converse in a respectful manner and follow the rules of the forum. So far you have not done that and you won't get much help if you keep it up.

the energy of the fuel will be converted to kinetic energy of the car according to half my square

I edited the previous post last minute.

Look at the speeds quoted in the question and where Pauls reference point is. What is wrong with his speeds.

 

[lalbatros]

This homework has already been discussed once of this forum.
No need to do it twice.

 

[JaredJames]

This homework has already been discussed once of this forum.
No need to do it twice.

Has it really? Could you link to it please? Save this having to continue for no reason.

 

[lalbatros]

I could not give you the link, since this would amount to giving you an answer to an homework.

 

[JaredJames]

I could not give you the link, since this would amount to giving you an answer to an homework.

In which case we need to discuss it twice, making your first post pointless.

 

[JaredJames]

they are right according o me...

So Peter is right saying it will take 3x litres of fuel to go from 100 to 200 and Paul is right saying that it will take (5/3)x litres of fuel to go from 200 to 300?

Are you telling me it would take less fuel to go from 200 to 300 than it would 100 to 200?

Again, look at the values for the speeds - there's something wrong with one set of speeds.

 

[pankazmaurya]

So Peter is right saying it will take 3x litres of fuel to go from 100 to 200 and Paul is right saying that it will take (5/3)x litres of fuel to go from 200 to 300?

Are you telling me it would take less fuel to go from 200 to 300 than it would 100 to 200?

Again, look at the values for the speeds - there's something wrong with one set of speeds.

the train in moving in opposite direction of the car and hence in train frame of refrence corrolay pauls frame of refrence the initial speed of car is 100 and after consuming x amount of fuel it will be 500

 

[JaredJames]

the train in moving in opposite direction of the car and hence in train frame of refrence corrolay pauls frame of refrence the initial speed of car is 100 and after consuming x amount of fuel it will be 500

Where did you get 500 from?

The kinetic energy to go from 100 to 200 for the car will be the same, regardless of frame of reference. The train is independent and Paul is making a mistake by adding the trains velocity into the equation.

The car goes from 0 to 100 and then 100 to 200. The fuel to go from 0 to 100 is not the same as the fuel to go from 100 to 200. The fuel to go from 100 to 200 is not the same as the fuel to go from 200 to 300. So with Paul using these values he is incorrect.

You can't add the trains motion into the equation because it is independent of the car. The kinetic energy for acceleration of the car isn't affected by the train.

 

[pankazmaurya]

Where did you get 500 from?

The kinetic energy to go from 100 to 200 for the car will be the same, regardless of frame of reference. The train is independent and Paul is making a mistake by adding the trains velocity into the equation.

The car goes from 0 to 100 and then 100 to 200. The fuel to go from 0 to 100 is not the same as the fuel to go from 100 to 200. The fuel to go from 100 to 200 is not the same as the fuel to go from 200 to 300. So with Paul using these values he is incorrect.

the fuel required for 0 to 100 in peters frame is same as the fuel required 100 to 200 in pauls frame...and if u are trying to sat that paul is making mistake by adding trains velocity then how how can conclude that he is making a mistake...becuase he can never know whether he is moving or the car is moving

 

[JaredJames]

the fuel required for 0 to 100 in peters frame is same as the fuel required 100 to 200 in pauls frame

Yes it is, but in his equations he uses it wrongly and includes the trains velocity. If you include the trains velocity, you are including the kinetic energy of the train.

...and if u are trying to sat that paul is making mistake by adding trains velocity then how how can conclude that he is making a mistake...becuase he can never know whether he is moving or the car is moving

Irrelevant to the question.

 

[pankazmaurya]

Yes it is, but in his equations he uses it wrongly and includes the trains velocity. If you include the trains velocity, you are including the kinetic energy of the train.

but how will paul know that he is including trains velocity...
Irrelevant to the question.

m not discusing the exact question...i want to discuss the spirit of it

 

[JaredJames]

m not discusing the exact question...i want to discuss the spirit of it

What spirit? You specifically asked what's gone wrong in the question. I am trying to help you answer it - that will show what has gone wrong.

If you don't want to answer the question then you've completely wasted my time for the last two pages of posts. You should make that perfectly clear in the OP.

I will no longer respond to this thread. Between your attitude and this I'm just fed up.

 

[JaredJames]

Here is the final explanation:

The energy to go from 100 to 200 is not the same as 200 to 300. Therefore the fuel requirements are different.

If Paul ignores the trains speed and uses 200 to 300 for the second section his V² values for KE are going to be much higher.

V² = 200² = 40000
V² = 300² = 90000

That is a significant difference in energy values. You have to put in more than double the fuel to get to 300 than you do to get to 200. So if you don't compensate for the trains speed you end up with the incorrect energy values - because the car is not going to 300. Change in KE from 100 to 200 = 0.5m(200² - 100²) =/= change in KE from 200 to 300 = 0.5m(300² - 200²). So the readings gained by both people are not the same and hence cannot both be right for the second stage.

Pauls reference frame can be what ever he likes, it could show the car going from 1000 to 1100 but the energy used by the car will always be the amount to go from 100 to 200. You have to compensate for this based on your reference frame.

What ever speed Paul records must be altered to allow for the trains velocity.

 

[rcgldr]

If Paul ignores the trains speed and uses 200 to 300 for the second section.

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.

This same issue has been brought up a couple of times recently. If you include the energy change to the earth, and consider Earth and car as part of a closed system, then ignoring losses to heat, the total energy of the system is conserved, and the total kinetic energy gain is equal to the total chemical or electrical potential energy consumed by the engine, as observed from any inertial (non-accelerating) frame of reference.

If the effect on the Earth is going to be ignored (earth with infinite mass), then the frame of reference must correrspond to the point of application of force, which in this case, is the road, since that is where the force is being applied by the tires of the car, and since the assumption is the road is attached to an Earth with infinite mass (no energy gained by the earth).

 

[JaredJames]

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.

Correct, that is compensating for the trains motion. But, in the question Paul goes wrong and uses 100 to 200 and then 200 to 300 for the energy values.

 

[rcgldr]

In the question Paul goes wrong and uses 100 to 200 and then 200 to 300 for the energy values.

True, but I was ignoring that mistake in the the orignal post as well as the bad calculations of energy, and commenting on a corrected observation.

 

[JaredJames]

True, but I was ignoring that mistake in the the orignal post as well as the bad calculations of energy, and commenting on a corrected observation.

Precisely, your answer is exactly why Paul is incorrect in his calculation.

 

[Gokul43201]

Precisely, your answer is exactly why Paul is incorrect in his calculation.

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.

Nevertheless, pankaz needs to read the forum guidelines before continuing to post. The attitude displayed is not useful.

 

[JaredJames]

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.

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.

 

[pankazmaurya]

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..

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

 

[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.