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- Thread starter pankazmaurya
- Start date

- #1

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- #2

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What do you think?

- #3

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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?

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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

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What do you think?

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

- #6

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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

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

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

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

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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?

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And what do you think the blunder is?

that is wat im not geting.now ur chance...

- #10

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that is wat im 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?

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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 cant tell the answer becuase you dont know. and my dear m not interested in the answer .

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its simple u cant tell the answer becuase you dont 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

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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 youmustshow some attempt before we can help.

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

- #14

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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?

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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

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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.

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This homework has already been discussed once of this forum.

No need to do it twice.

No need to do it twice.

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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.

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

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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.

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they are right according o me...

So Peter is right saying it will take 3x litres of fuel to go from 100 to 200

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.

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So Peter is right saying it will take 3x litres of fuel to go from 100 to 200andPaul 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

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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.

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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

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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.

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