- #1

Guy135

Depending on when I make the trip the amount of work I do varies but I have the same amount of energy (none) at the end of the trip

How does the conservation of energy apply here?

Thanks

Guy

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- #1

Guy135

Depending on when I make the trip the amount of work I do varies but I have the same amount of energy (none) at the end of the trip

How does the conservation of energy apply here?

Thanks

Guy

- #2

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- #3

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I think there is a bit more to it than that. You're actually doing real work when moving forward in an accelerating train, wich goes into raising your own kinetic energy.

Suppose you are in a train that accelerates from 0 to 100m/s at 1m/s

The moment you stop walking however, this extra kinetic energy will be given back to the train. Suppose it takes you 1s to stop. That will need an acceleration of 1/m/s

So, if you have to walk in a train, you should be doing it while it is decelerating. (unless you want to lose weight)

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I did not want to get into this in a B-level thread, but this depends on the frame you are considering. Energy isYou're actually doing real work when moving forward in an accelerating train, wich goes into raising your own kinetic energy.

The bottom line is that you are at rest when you start and at rest when you end and therefore the total work done on you is zero in all cases. There really is nothing more to it. Of course, you can play around with what forces have performed what work and what effort you have exerted, but this is not what the question was about (the question was about conservation of energy). Again, note that the work done on or by you is frame dependent, just as the energy itself.

This is very oversimplified. It depends on when you accelerated in comparison to when the train accelerated. Also "came from you" is not very well defined. The energy you gain comes from the work done by the train on you in all cases. Again, please separate work (which has a precise physical definition) from effort. For example, as long as you are not walking while the train is accelerating, you will spend exactly the same effort walking from the back to the front when the train is standing still as when it is moving relative to the ground.Suppose you are in a train that accelerates from 0 to 100m/s at 1m/s-2. You weigh 100 kg. If you don't do anything you'll get (1/2)mv2 = 500 kJ of Kinetic energy, all delivered by the train. If you walk forward at 1m/s all this time, you'll get (1/2)*1012 = 510 kJ of kinetic energy, of wich 10kJ came from you.

- #5

russ_watters

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In one case you are doing some work and in the other case you aren't, but in both cases the train does the same amount of work. When you are walking while the train is accelerating, the train is doing more work to enable that acceleration.How does the conservation of energy apply here?

- #6

russ_watters

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I agree with @wilhen here: B-thread or not, the fact that energy is not frame dependent is exactly what this question is about. The OP is trying to account for the real work being done by him and the train. The choice to walk when the train is accelerating vs standing still when it is accelerating means you are taking the force of the acceleration while stationary on the train (you are doing no work) vs while moving (you are doing work). But vs the ground, the train is doing more work to accelerate you when you are moving too. That extra kinetic energy is re-absorbed when you stop walking.I did not want to get into this in a B-level thread, but this depends on the frame you are considering. Energy isnotframe independent.

For a train, all of the work done is during the accelerations: positive work during the positive acceleration, negative work during the negative acceleration (deceleration). The catch is that when you are walking in the train, you add an acceleration or deceleration the train was not expecting to do. It's easier to see when the train is moving at constant speed: when you start walking, the train applies an extra force on you to make you accelerate and when you stop walking the train applies a force to decelerate you.

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If the train is doing work on you, you are doing work on the train. By definition, these works have the same magnitude and opposite sign.In one case you are doing some work and in the other case you aren't, but in both cases the train does the same amount of work.

I disagree, the OP clearly states that he is puzzled by the fact that regardless of how he moves, he ends up with the same velocity. It is perfectly well suited to analyse the situation in an inertial frame and in any inertial frame the work done by the train on you is going to be equal to zero, just as your work on the train is going to be.the fact that energy is not frame dependent is exactly what this question is about.

- #8

russ_watters

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Yes, I know -- but I may have been unclear: I meant in total during the trip. In one case you do positive work and then a different amount of negative work and in the other case you a very small amount of both and of equal magnitude. The amount of workIf the train is doing work on you, you are doing work on the train. By definition, these works have the same magnitude and opposite sign.

In order for all the "works" to sum to zero, the train's work against the ground must also be imbalanced: when you walk while the train is accelerating it does more work in getting up to speed than it does to stop at the end of the trip. It re-absorbs that extra kinetic energy when you stop walking, preserving the balance. And because of the power of the square function of kinetic energy (puns intended), the train does a surprising amount of negative work on you when you stop walking while it is in motion. That's the real work you got to feel when you were walking while it was accelerating.

I'll say it in a different way: when you start walking while the train is accelerating, you get to do work in Earth's frame, not just in the train's frame. The amount of work you do against the train and the amount of work *it* does against the Earth to accelerate you are different.

Here's the premise again:I disagree, the OP clearly states that he is puzzled by the fact that regardless of how he moves, he ends up with the same velocity. It is perfectly well suited to analyse the situation in an inertial frame and in any inertial frame the work done by the train on you is going to be equal to zero, just as your work on the train is going to be.

That's not an "effort" thing; when you walk while the train is accelerating, you do real work, applying a force over a distance.Guy135 said:If I walk when the train is accelerating forwards it is harder for me to get to the front and I do more work. If I make the trip when the train is stopped or at a constant speed I do less work.

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What do you mean by "work you do" here? The total work you do on your environment is zero in both cases. This follows directly from the work-energy theorem.The amount of workyoudo on the total trip is different for the two cases.

I think this is very vague. Work is not something that is done in a particular frame. If you apply a force there is going to be some work done regardless of the frame you are using for computations unless it so happens that the velocity of the point of application is zero in the particular frame.I'll say it in a different way: when you start walking while the train is accelerating, you get to do work in Earth's frame, not just in the train's frame.

Of course, you could consider the work done in an accelerated frame of reference, but there is no guarantee that conservation of energy will hold then as you are mixing the energy concepts of different inertial frames with each other when you accelerate. Put in other terms, there is no time-translation invariance in a general accelerating frame and so there is no conserved quantity associated to it. This is why analysing the situation in an inertial frame is much preferable as it is where you actually have access to energy conservation. What I am trying to dispel here is the notion of there being a frame-independent amount of work that you do. There is not. There is no objective amount of "work" that can be defined without reference to a particular frame. I think that it is quite clear that the OP (at B-level) is making the typical layman mistake of confusing work with perceived effort. Depending on the frame, the work done when walking while the train is accelerating can be positive, negative, or zero, in all cases. It only depends on the applied force and the instantaneous velocity of the point of application.

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Whenever the guy's kinetic energy is decreasing, then he might be doing work. In post #3, when the guy stopped walking, he lost 10 kJ of kinetic energy, that energy is equal to falling from 10 meters, he did not absorb that amount of energy into himself, instead he did work on something else.

- #11

russ_watters

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Let me restate my critique and then expand on my answer:

Usually when the issue of "effort" vs "work" is brought up it is in regard to

That issue does not apply here because there is no part of this problem where physiological effort and physics work are out of sync. The only time the person is applying extra static force is when standing or sitting still while the train accelerates. There's (practically) no extra physiological work involved in remaining in your seat or leaning forward/backwards a little bit.

The OP is asking about the *real*/physics work involved in walking forward on an accelerating train: You do more physiological effort because there is real physics work being done.

The questions are:

Why is it so large?

Where does it go?

The answer is in the surprising amount of kinetic energy involved and the fact that when you walk while the train is accelerating, you get to contribute more to it (or, rather, experience it). Let's put some real numbers to it:

Your mass: 50 kg

Your walking speed: 1 m/s

Your acceleration time: 1s

Train's speed: 20 m/s

Train's acceleration time: 20s

Consider the accelerations to be constant (constant force).

Base case: You are sitting on the train while it accelerates. The train does 10,000 Joules of work on you, for an average of 500 Watts. But note: the wattage isn't constant, it starts small and grows as the train accelerates.

Next: You start walking when the train is stationary or moving at constant speed. By the kinetic energy equation, *you* do 25 Joules of work in one second or 25 Watts (average) to get moving. Once moving, you expend no more [physics] energy keeping moving.

Now, when you walk while the train is accelerating: Let's say you start walking before the train starts accelerating (1s, 25J). Then you walk for 20s and expend an additional 1000 J (20W, constant this time) in the effort because of the train's acceleration. Big difference! Where does this come from/go?

The answer lies in the train's motors and the fact that it has to accelerate you with respect to the ground. At 20 m/s, it gives you 10,000 J of kinetic energy, but at 21 m/s it gives you 11,025 J. You might recognize those numbers from above: 11,025 - 10,000 - 1,000 - 25 = 0. Conservation of energy has held.

Another way to look at it is to consider that you are making the train do a lot of work by walking when it is in motion. You only apply an average of 25 W to accelerate to 1m/s, but the train has to apply 1,025 W(!) to you if you do it while the train is moving. Fortunately the train gets that back when you stop walking.

- #12

russ_watters

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We're back on the inclined treadmill again: if you walk forward and then stop walking while the train is accelerating, you've expended actual, non-zero, physics work.What do you mean by "work you do" here? The total work you do on your environment is zero in both cases. This follows directly from the work-energy theorem.

Then please tell me how better to word the fact that you move different distances vs the train and the ground and account for the different work/energy involved. Should I use "versus" instead of "in"? "With respect to"? The point either way is that your legs do a different amount of real physics work when the train is accelerating with respect to the ground (vs not), and you and the train do different amounts of work in the same scenarios and I'm calculating those values.I think this is very vague. Work is not something that is done in a particular frame. If you apply a force there is going to be some work done regardless of the frame you are using for computations unless it so happens that the velocity of the point of application is zero in the particular frame.

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I disagree with this. Work is as arbitrary as your choice of reference frame. In this sense, they are always out of sync.That issue does not apply here because there is no part of this problem where physiological effort and physics work are out of sync.

This depends on the reference frame you are considering. Go to a reference frame that is moving sufficiently fast (faster than the train) and the work done will even change sign. Clearly, the physical effort you do is not going to depend on which frame I use to analyse the situation.You do more physiological effort because there is real physics work being done.

You have not stated what frame you are considering, but it seems to me that you are implicitly considering the ground rest frame. In a frame that is moving at 1 m/s relative to the train, the train is doing -10000 J of work on you. This underlines the fact that work is frame-dependent.Base case: You are sitting on the train while it accelerates. The train does 10,000 Joules of work on you, for an average of 500 Watts. But note: the wattage isn't constant, it starts small and grows as the train accelerates.

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Now, when you walk while the train is accelerating: Let's say you start walking before the train starts accelerating (1s, 25J). Then you walk for 20s and expend an additional 1000 J (20W, constant this time) in the effort because of the train's acceleration. Big difference! Where does this come from/go?

The answer lies in the train's motors and the fact that it has to accelerate you with respect to the ground. At 20 m/s, it gives you 10,000 J of kinetic energy, but at 21 m/s it gives you 11,025 J. You might recognize those numbers from above: 11,025 - 10,000 - 1,000 - 25 = 0. Conservation of energy has held.

In a fast moving train any extra force a passenger exerts on the train causes a large amount of energy to be exchanged between the train and the passenger. When one passenger has a different acceleration than all the other passenger that are sitting down calmly, that passenger is exerting extra forces on the train. But a person that is not accelerating away from the sitting passengers has the same acceleration as the sitting passengers. I mean the walking passenger, he does not have any extra acceleration.

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- #16

Guy135

- #17

jbriggs444

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One accounting:But why would it be easier for a train to accelerate whilst a passenger walks forwards at a constant (relative to the train) speed as opposed to the same passenger sitting still?

Because you started walking (exerting extra rearward force to accelerate) while the train was moving slowly relative to the tracks. The extra load on the train engines multiplied by a low speed relative to the tracks amounted to a tiny bit of extra power. You stopped walking (exerting forward force - or reduced rearward force) to decelerate while the train was moving rapidly. The reduced load on the train engines multiplied by a high speed relative to the tracks amounted to a larger bit of saved power.

If you had stayed in your seat for the entire trip you would have still been applying a rearward force on the train during its acceleration, but those two blips would have been absent.

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See post #3. Every single Joule that the walker uses to walk goes to increasing the kinetic energy of the walker, relative to the ground.

After million seconds of walking the walker has 100 megajoules of extra kinetic energy.

Now comes the important part: Walker stops walking. Let's say the process of stopping walking takes one millisecond, as the walker collides on the front wall of the train. Afterwards the former walker has no extra kinetic energy. 100 megajoules was lost in one millisecond, that is a power of 100 gigawatts.

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