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Toffo
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The new question isn't fundamentally different from the one you already started a thread about, where you're asking about where another property comes from on an accelerating train. In anticipation of further such threads, I've merged them into this one. - gm

I'm in an accelerating train. I'm walking to the direction of the acceleration, which is hard, like walking uphill.

Where's the energy going? I mean the work I'm doing, where's that energy stored?

Zamfir
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### Re: Energy of transition in accelerating transportation

Kinetic energy of your body. Example:

100kg body, walking at 1m/s, unmoving train. KE of body: 1/2 mass times velocity squared = 50J
Train moving at 10m/s, so body moving at 11m/s. KE of body.6050J.
Gain in KE: 6000J

Same situation, body at rest relative to train.
Start KE: 0J
Train at 10 m/s, so body also at 10m/s. KE of body: 5000J
Gain in KE: 5000J

The train delivers that 5000J to your body in both cases , but you have to put in the rest. If you didn't, you would just slow down relative to the train.

Suppose the train accelerates at 2m/s2. So it takes 5 seconds to get up to speed. The walker walks 5m relative to the train in that time, all the time pushing applying a force of ( 2m/s2 times their body mass of 100kg )=200 N
Energy expended by that force: 200N times 5m, equals 1000J. That's the energy that went to the extra KE

Toffo
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### Re: Energy of transition in accelerating transportation

Ok so the energy becomes kinetic energy of my body, which energy becomes kinetic energy of the train when I stop walking.

Toffo
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### Re: Energy of motion in accelerating vehicles and relativity

Let's consider light in an accelerating spaceship, light is traveling from the rear to the nose of the spaceship.

Does that light give energy to the spaceship?

speising
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### Re: Energy of transition in accelerating transportation

consider that in the equations above, there is a term for mass.
set that to the mass of light (0) and you get zero kinetic energy.

thoughtfully
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### Re: Energy of transition in accelerating transportation

Another way to look at this is to use the Equivalence Principle. You and the train are accelerating, and this is equalent to an increase (and change in direction, since it isn't inline with existing gravity) of your weight. It's like pulling yourself up a ladder (or climbing stairs, whatever) while carrying a heavy load. That's pulling off to some oblique angle. Umm, hope that helps?

Perfection is achieved, not when there is nothing more to add, but when there is nothing left to take away.
-- Antoine de Saint-Exupery

Toffo
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### Re: Energy of transition in accelerating transportation

Zamfir wrote:Suppose the train accelerates at 2m/s2. So it takes 5 seconds to get up to speed. The walker walks 5m relative to the train in that time, all the time pushing applying a force of ( 2m/s2 times their body mass of 100kg )=200 N
Energy expended by that force: 200N times 5m, equals 1000J. That's the energy that went to the extra KE

... And when I stop walking, I give the 1000J to the train. My walking did not cause the train to lose any energy, so I donated a net amount of 1000J to the train.

But I immediateletely start walking again. I accelerate to 1 m/s relative to the train in 0.01 s, pushing with force 10000N, so train pushes me with force 10000N, a distance 10m/s*0.01s, doing work 1000J. After 5 m I stop and give 1000J to the train. My net energy contibution is 1000J - 1000J = 0J. <--- Wrong

So I guess it must be like this:
After 5 m I stop and give 2000J to the train. My net energy contibution is 2000J - 1000J = 1000J.

Toffo
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### Re: Energy of transition in accelerating transportation

speising wrote:consider that in the equations above, there is a term for mass.
set that to the mass of light (0) and you get zero kinetic energy.

Yes, but is that correct?

p1t1o
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### Re: Energy of transition in accelerating transportation

I'm guessing that light hitting the inside surface of the front of the train will in fact, donate some kinetic energy, as it does in the operation of a solar sail.

Photons do not have mass, but they do carry momentum (the maths and physics of this is slighty above my pay grade) and therefore can impart kinetic energy.

One example, in the core of a hydrogen bomb, the secondary (basically a block of lithium deuteride and uranium) is compressed to fusion density/pressure by heat ablation of the external surface but also the X-ray environment is such that radiation pressure alone is on the order of Tera-Pascals.

stoppedcaring
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### Re: Energy of transition in accelerating transportation

p1t1o wrote:I'm guessing that light hitting the inside surface of the front of the train will in fact, donate some kinetic energy, as it does in the operation of a solar sail.

Photons do not have mass, but they do carry momentum (the maths and physics of this is slighty above my pay grade)...

Photons do have mass. E = mc2, so an x-ray photon with a wavelength of 0.01 nm has an energy of 100 keV and thus an associated mass of 0.18e-12 femtograms, about 1/5 the mass of an electron. Photons simply don't have rest mass.

PM 2Ring
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### Re: Energy of transition in accelerating transportation

Toffo wrote:
speising wrote:consider that in the equations above, there is a term for mass.
set that to the mass of light (0) and you get zero kinetic energy.

Yes, but is that correct?

Not really.

It is true that light has zero rest mass, and under Newtonian physics that'd imply it has zero momentum and zero kinetic energy. However, we know that light carries energy, traditionally referred to as radiant energy, and that it has non-zero momentum. That radiant energy is essentially kinetic energy.

The energy E, momentum p, and frequency f, of a photon are related by
E = hf = pc
where h is Planck's constant, and c is the speed of light.

See Energy–momentum relation.

p1t1o
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### Re: Energy of transition in accelerating transportation

stoppedcaring wrote:
p1t1o wrote:I'm guessing that light hitting the inside surface of the front of the train will in fact, donate some kinetic energy, as it does in the operation of a solar sail.

Photons do not have mass, but they do carry momentum (the maths and physics of this is slighty above my pay grade)...

Photons do have mass. E = mc2, so an x-ray photon with a wavelength of 0.01 nm has an energy of 100 keV and thus an associated mass of 0.18e-12 femtograms, about 1/5 the mass of an electron. Photons simply don't have rest mass.

Simply? Meh, I think that is just semantics. Rest mass = mass. Photons have none. What they do have is relativistic mass, your 0.18e-12fg, this is not the same property used in the classical kinetic energy equation.

stoppedcaring
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### Re: Energy of transition in accelerating transportation

p1t1o wrote:
stoppedcaring wrote:
p1t1o wrote:I'm guessing that light hitting the inside surface of the front of the train will in fact, donate some kinetic energy, as it does in the operation of a solar sail.

Photons do not have mass, but they do carry momentum (the maths and physics of this is slighty above my pay grade)...

Photons do have mass. E = mc2, so an x-ray photon with a wavelength of 0.01 nm has an energy of 100 keV and thus an associated mass of 0.18e-12 femtograms, about 1/5 the mass of an electron. Photons simply don't have rest mass.

Simply? Meh, I think that is just semantics. Rest mass = mass. Photons have none. What they do have is relativistic mass, your 0.18e-12fg, this is not the same property used in the classical kinetic energy equation.

Well, (Rest mass) = (mass) - (kinetic mass-energy). When we're dealing with particle physics, you kind of have to take it all into account.

Toffo
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### Re: Energy of transition in accelerating transportation

thoughtfully wrote:Another way to look at this is to use the Equivalence Principle. You and the train are accelerating, and this is equalent to an increase (and change in direction, since it isn't inline with existing gravity) of your weight. It's like pulling yourself up a ladder (or climbing stairs, whatever) while carrying a heavy load. That's pulling off to some oblique angle. Umm, hope that helps?

Let's say I'm again walking in the accelerating train, and I'm carrying a heavy box, which contains light, light has mass, I have been informed.

Am I increasing the kinetic energy of the light? Do I feel the box becoming more heavy?

p1t1o
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### Re: Energy of transition in accelerating transportation

Toffo wrote:
thoughtfully wrote:Another way to look at this is to use the Equivalence Principle. You and the train are accelerating, and this is equalent to an increase (and change in direction, since it isn't inline with existing gravity) of your weight. It's like pulling yourself up a ladder (or climbing stairs, whatever) while carrying a heavy load. That's pulling off to some oblique angle. Umm, hope that helps?

Let's say I'm again walking in the accelerating train, and I'm carrying a heavy box, which contains light, light has mass, I have been informed.

Am I increasing the kinetic energy of the light? Do I feel the box becoming more heavy?

The “box of light” is a tricky concept and may derail your analogy.

Light inside the box will hit one of the internal surfaces of the box, be absorbed and re-radiated. Light radiating from the inside-rear wall of the box (emitted forwards) will be slightly blue-shifted, your forward motion has increased the frequency of that photon, increasing its energy and its relativistic mass (as opposed to its rest-mass).
It will then hit the forward inside wall of the box, be absorbed and re-radiated in a rearwards direction. Your forwards motion will now slightly red-shift this photon, reducing its frequency, energy and relativistic mass.

I would imagine, then, that the photonic energy contained in a “box of light” will probably remain in equilibrium no matter how you move it.

It is important to remember that relativistic mass and rest-mass are not the same property, and the weight felt by your hands of say, a cannonball, will remain the same to you as you walk forwards, as you and the cannonball are travelling together, in the same reference frame – a stationary observer might see something else however, though I can’t claim to have a 100% handle on this.

Toffo
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### Re: Energy of transition in accelerating transportation

p1t1o wrote:and the weight felt by your hands of say, a cannonball, will remain the same to you as you walk forwards, as you and the cannonball are travelling together, in the same reference frame – a stationary observer might see something else however, though I can’t claim to have a 100% handle on this.

It's a quite well known fact that in an accelerating vehicle front clock ticks faster than rear clock.
Less well known fact is that in a accelerating vehicle front accelerometer reads less than rear accelerometer.
A human being is an accelerometer. So the cannonball becomes less massive according to you when you carry it towards the front.

stoppedcaring
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### Re: Energy of transition in accelerating transportation

Toffo wrote:It's a quite well known fact that in an accelerating vehicle front clock ticks faster than rear clock.

As this is your premise...are you sure?

Toffo
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### Re: Energy of transition in accelerating transportation

Yes, I'm sure. It can be derived from gravitational time dilation and equivalence principle.

I'm sure about the second fact too. It can be derived from ... Lorentz contraction ... and probably time dilation has some effect too.

Toffo
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### Re: Energy of transition in accelerating transportation

Toffo wrote:So the cannonball becomes less massive according to you when you carry it towards the front.

... because at the front the acceleration is smaller ... very simple.

If we ask the train for its opinion, it's the same: cannonball becomes less massive. (because of the smaller acceleration, although train does not necessarily know that its front has smaller acceleration than the rear)

Now we use the equivalence principle: A cannonball that is being raised into a flagpole is becoming lighter, although its energy is increasing.

Hey, maybe gravitatonal potential energy has negative weight.
Last edited by Toffo on Sat Jun 21, 2014 11:23 am UTC, edited 1 time in total.

Copper Bezel
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### Re: Energy of transition in accelerating transportation

Now we use the equivalence principle: A cannonball that is being raised into a flagpole is becoming lighter, although its energy is increasing.

You're confusing weight with mass.
So much depends upon a red wheel barrow (>= XXII) but it is not going to be installed.

she / her / her

Toffo
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### Re: Energy of transition in accelerating transportation

Copper Bezel wrote:
Now we use the equivalence principle: A cannonball that is being raised into a flagpole is becoming lighter, although its energy is increasing.

You're confusing weight with mass.

I don't think so. But some additional attacments may be in order. Let's mount the flagpole on a scale. And let's keep the flagpole short, so that gravity field is "homogeneous".

The reading on the scale goes down as the cannonball is winched up.

Zamfir
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### Re: Energy of transition in accelerating transportation

Toffo wrote:
Zamfir wrote:Suppose the train accelerates at 2m/s2. So it takes 5 seconds to get up to speed. The walker walks 5m relative to the train in that time, all the time pushing applying a force of ( 2m/s2 times their body mass of 100kg )=200 N
Energy expended by that force: 200N times 5m, equals 1000J. That's the energy that went to the extra KE

... And when I stop walking, I give the 1000J to the train. My walking did not cause the train to lose any energy, so I donated a net amount of 1000J to the train.

But I immediateletely start walking again. I accelerate to 1 m/s relative to the train in 0.01 s, pushing with force 10000N, so train pushes me with force 10000N, a distance 10m/s*0.01s, doing work 1000J. After 5 m I stop and give 1000J to the train. My net energy contibution is 1000J - 1000J = 0J. <--- Wrong
.

Nope, not wrong. When the train stays moving at constant velocity, those energies do sum to zero. You slow down, you lose 1050 Joule. 1000 goes to the train, 50 is lost as braking heat. You start moving again, you gain 1050 J. 1000 is supplied by the train, 50 by you. You brake again, you lose another 50.

So from the walker's perspective, walking and stopping on a constant-velocity train is just like walking and stopping on the ground. Takes 50 J effort to start, you have to dissipate 50J to stop. The other 1000J is hidden, taken care of by the train's machinery.

gmalivuk
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### Re: Energy of transition in accelerating transportation

Toffo wrote:
Copper Bezel wrote:
Now we use the equivalence principle: A cannonball that is being raised into a flagpole is becoming lighter, although its energy is increasing.

You're confusing weight with mass.
I don't think so. But some additional attacments may be in order. Let's mount the flagpole on a scale. And let's keep the flagpole short, so that gravity field is "homogeneous".

The reading on the scale goes down as the cannonball is winched up.
Which is still a matter of weight, not mass.
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Toffo
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### Re: Energy of transition in accelerating transportation

Zamfir wrote:When the train stays moving at constant velocity, those energies do sum to zero.

And when the train does not stay moving at constant velocity, those energies do not sum to zero.

(If the passanger feels like he has climbed a hill, then the energies sum to a positive number)

Toffo
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### Re: Energy of transition in accelerating transportation

gmalivuk wrote:Which is still a matter of weight, not mass.

Yes, weight. Weight is the force needed to keep a flagpole-cannonball system from falling. If that force decreases we say weight decreases.

Force is for example a force needed to accelerate a flagpole-cannonball system. If that force decreases we say mass decreases. (Well maybe we could say that artifical weight decreases.)

gmalivuk
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### Re: Energy of transition in accelerating transportation

No, we most definitely do not say mass decreases in that case. You're correct that strictly speaking "weight" may not be the best word either, since the change isn't due to gravity, but when force changes because acceleration changes, we don't just lie about it and say mass is what changed instead. (Remember F=ma, so different acceleration means different force directly, without anything happening to the mass.)
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Zamfir
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### Re: Energy of transition in accelerating transportation

Toffo wrote:
Zamfir wrote:When the train stays moving at constant velocity, those energies do sum to zero.

And when the train does not stay moving at constant velocity, those energies do not sum to zero.

(If the passanger feels like he has climbed a hill, then the energies sum to a positive number)

Yes, when you are walking at constant velocity with respect to the train while the train is accelerating, then you expend quite some energy to keep that constant velocity differential. That's the situation in your original post.

In the other post, you wrote:

... And when I stop walking, I give the 1000J to the train. My walking did not cause the train to lose any energy, so I donated a net amount of 1000J to the train.

But I immediateletely start walking again. I accelerate to 1 m/s relative to the train in 0.01 s, pushing with force 10000N, so train pushes me with force 10000N, a distance 10m/s*0.01s, doing work 1000J. After 5 m I stop and give 1000J to the train. My net energy contibution is 1000J - 1000J = 0J. <--- Wrong

So I guess it must be like this:
After 5 m I stop and give 2000J to the train. My net energy contibution is 2000J - 1000J = 1000J.

That's another situation: the walker changes velocity with respect to the train, while the train stays at constant velocity. In these cases,most of the energy transfers go unnoticed to the walker. When they slow down, the train engines get a temporary reduction in load to the tune of 1000J. When they start walking again, the train engines have to put in 1000J extra to stay at constant velocity.The walker only experiences 50J of change either way.

By stopping with respect to the train, 1000J of energy is transferred to the train. By walking again, that 1000J comes back. You end up in the original situation energy-wise, without noticing the effect on the engines either way. When you walk up and down a moving train, you do not gain or spend that 1000J yourself, it feels just like moving on still ground. That 1000J of effort that you put in originally, will only return to you when the train decelerates. Or when you jump out of the train. Then you can retrieve the whole 6050J of kinetic energy, most likely in the form of broken bones.

Toffo
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### Re: Energy of transition in accelerating transportation

Well, I was trying to calculate what happens in a accelerating train. That's what I was trying to do, in that calculation.

... And in the first post I was trying to say this:
While a train accelerates: you start walking, you walk, you stop walking. Where is the energy you used.

Let me try again that calculation:

impulse needed to change the speed of the 100 kg person by 1 m/s: 100 Ns

energy loss of train that absorbs that impulse when speed of train is 10 m/s: 100 Ns * 10 m/s = 1000 J
energy gain of train that absorbs the opposite impulse when speed of train is 20 m/s: 100 Ns * 20 m/s = 2000 J

Net energy change of train = 2000 J - 1000 J = 1000 J

drachefly
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### Re: Energy of transition in accelerating transportation

stoppedcaring wrote:
p1t1o wrote:I'm guessing that light hitting the inside surface of the front of the train will in fact, donate some kinetic energy, as it does in the operation of a solar sail.

Photons do not have mass, but they do carry momentum (the maths and physics of this is slighty above my pay grade)...

Photons do have mass. E = mc2, so an x-ray photon with a wavelength of 0.01 nm has an energy of 100 keV and thus an associated mass of 0.18e-12 femtograms, about 1/5 the mass of an electron. Photons simply don't have rest mass.

You can't apply the rest mass equation to photons and expect to have the result make sense. Let's use the total energy equation:

E^2 = (mc^2)^2 + (pc)^2

For photons, that simplifies to E = pc

Toffo wrote:... because at the front the acceleration is smaller ... very simple.

So is the train getting shorter and shorter over time? If not, then the acceleration is not less in the front than in the rear.

gmalivuk
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### Re: Energy of transition in accelerating transportation

Yes, it is.

To maintain the same proper distance, the front of a long ship and the back of the same long ship must undergo different proper acceleration. If the ship is 1ly long and the rear is accelerating at 1g, then the front has to accelerate at only 0.5g in order for the ship to maintain its 1ly length (as measured on the ship itself). If they both undergo 1g of proper acceleration in the same direction, then the relativity of simultaneity will mean that from both frames (front and rear of ship), the front will appear to have started accelerating earlier than the rear and they will drift apart as measured in both frames. This is of course bad for the whole middle of the ship, as it will be ripped apart in the process.

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Hypnosifl
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### Re: Energy of transition in accelerating transportation

Toffo wrote:Well, I was trying to calculate what happens in a accelerating train. That's what I was trying to do, in that calculation.

... And in the first post I was trying to say this:
While a train accelerates: you start walking, you walk, you stop walking. Where is the energy you used.

Let me try again that calculation:

impulse needed to change the speed of the 100 kg person by 1 m/s: 100 Ns

energy loss of train that absorbs that impulse when speed of train is 10 m/s: 100 Ns * 10 m/s = 1000 J
energy gain of train that absorbs the opposite impulse when speed of train is 20 m/s: 100 Ns * 20 m/s = 2000 J

Net energy change of train = 2000 J - 1000 J = 1000 J

I think that would work only if both interactions between the person and the train were elastic collisions where 100% of the energy goes to changing each object's linear velocity. In an inelastic collision, some of the energy gets lost to heat (kinetic energy of molecules in random directions) or radiation (kinetic energy of light waves in random directions)...see elastic and inelastic collisions. Also, you can simplify the analysis by considering a scenario where the person is in a spaceship, so their only interactions are initially pushing off the back wall to get up to 11 m/s (while the ship's velocity will decrease from 10 m/s to some slightly lower velocity V), coasting for a while at 11 m/s without touching the ship, and then letting some part of the ship hit them again later to bring their velocity from 11 m/s to 20 m/s, after the ship itself has jettisoned some mass to change its velocity from V to 20 m/s. By choosing the right initial and final mass for the ship it might be possible to find a solution where both collisions between the person and the ship were elastic ones, but I'm not sure.

Toffo
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### Re: Energy of transition in accelerating transportation

The 100 kg man was walking 1 m/s. So the kinetic energy was 50 J. Let's ignore that teensy 50 J.

Well ok let's not ignore:

The man uses 50 J to get going. Then he walks, from conservation of energy I guess 1000 J is used in walking. Then he stops, uses the 50 J of kinetic energy for his own purposes, pole-vaults to the top of a sill, for example.

Hypnosifl
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### Re: Energy of transition in accelerating transportation

Toffo wrote:The 100 kg man was walking 1 m/s. So the kinetic energy was 50 J. Let's ignore that teensy 50 J.

Well ok let's not ignore:

The man uses 50 J to get going. Then he walks, from conservation of energy I guess 1000 J is used in walking. Then he stops, uses the 50 J of kinetic energy for his own purposes, pole-vaults to the top of a sill, for example.

Why do you think 1000 J would be used in walking? 1000 J was the figure you had for the energy loss of the train in order for it to absorb an equal and opposite impulse to the man when he first pushed off with his foot to get to 1 m/s relative to the train, right? Each further step during walking would involve further energy transfers. That's why I suggested it's easier to analyze a situation in zero-G, where the person just pushes off once and then coasts inertially at 1 m/s relative to the train without touching it.

I also see another problem with your calculation--you assumed the train's change in energy would just be its velocity before the push-off times the impulse it received during the push-off, but this doesn't seem to be how it works even for purely elastic collisions. As an example, suppose you you have a ball #1 approaching another ball #2 that's at rest, with the first ball's mass as m1 = 200 kg and its initial velocity v1 = 0.75 m/s, and ball #2's mass as m2 = 100 kg. Then if the collision is elastic, the formulas on this page tell you that after the collision, the larger ball #2 is moving at v1' = 0.25 m/s, and the smaller ball #2 is moving at v2' = 1 m/s. Both then got an impulse of (mass)*(change in velocity) = 100 kg*m/s = 100 N*s, but the larger ball #1's change in energy is not just the impulse times the initial speed before the collision, since that would suggest a change of energy of 75 J, but you can see from these numbers that the change in kinetic energy of ball #1 was actually (1/2)*(200)*(0.75)^2 - (1/2)*(200)*(0.25)^2 = 100*(0.75^2 - 0.25^2) = 50 J. It does seem from the units that you should be able to calculate the change in energy by multiplying the impulse by some velocity (at least in an elastic collision where energy isn't lost to heat, sound waves, radiation etc.), but I think it should be the change in velocity of the train before and after the push-off, rather than just its initial velocity at the moment before the push-off. In the case of ball #1, its change of velocity was 0.5 m/s, which when multiplied by the impulse 100 N*s it received gives the correct change in kinetic energy of 50 N*m = 50 J.

Toffo
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### Re: Energy of transition in accelerating transportation

Hypnosifl wrote:Why do you think 1000 J would be used in walking? 1000 J was the figure you had for the energy loss of the train in order for it to absorb an equal and opposite impulse to the man when he first pushed off with his foot to get to 1 m/s relative to the train, right? Each further step during walking would involve further energy transfers. That's why I suggested it's easier to analyze a situation in zero-G, where the person just pushes off once and then coasts inertially at 1 m/s relative to the train without touching it.

The man is the load that the the train is carrying. When the train causes the man to accelerate, train is doing the work that the train is supposed to do. If said load decides to move ahead in the train, then the train motors don't have to work so hard to accelerate the load.

In the example, I calculated the energy saved to be 1000 J. Then I guessed that the man did that amount of work, 1000 J, when he moved ahead.

The train pushes the man's feet at the same force when the man is standing still and when the man is walking at constant speed.

1: Man is standing still, train is accelerating a normal load.
2: Man starts to walk, train is accelerating an extra heavy load.
3: Man is walking, train is accelerating a normal load.
4: Man stops to walk, train is accelerating an extra light load.
5: Man is standing still, train is accelerating a normal load.

Hypnosifl
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### Re: Energy of transition in accelerating transportation

Toffo wrote:The man is the load that the the train is carrying. When the train causes the man to accelerate, train is doing the work that the train is supposed to do. If said load decides to move ahead in the train, then the train motors don't have to work so hard to accelerate the load.

Each time his feet push against the train, the train accelerates him again, otherwise he wouldn't be walking at a constant speed relative to the train from one step to the next. As I said, it would be simpler to just look at the situation where you have a spaceship (or train on a spaceship) in zero-G, so the man can just push off once and then coast inertially through the center of the room without touching it--in that case, while the man is out of contact, the engines do indeed have to do less work since the mass they're accelerating is reduced by the mass of the man.
Toffo wrote:In the example, I calculated the energy saved to be 1000 J.

Your calculation was for a single impulse, not for multiple impulses each time the man steps, so it doesn't make any sense to use it as the energy saved during walking. And in any case I already showed that the reasoning behind your calculation is incorrect, you can't multiply the impulse received by the train by its velocity immediately before the impulse to get the train's change in kinetic energy, rather you have to multiply the impulse received by the train's change in velocity immediately before and after the impulse. Do you think my calculation was incorrect?

Toffo
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### Re: Energy of transition in accelerating transportation

Hypnosifl wrote:Do you think my calculation was incorrect?

Yes.

Try calculating the energy change of a spacecraft moving 200000 km/s, when an apple is thrown out into the backwards direction with an impulse of 1 Ns. Let the mass of the spacecraft be million tons.

Use formula velocity change * impulse, velocity change must be the velocity change of the spacecraft.

Result should be a large number. But you will get a tiny number. Right? Relativity may be ignored.

gmalivuk
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### Re: Energy of transition in accelerating transportation

If you want to ignore relativity, maybe pick a thought experiment that isn't going 2/3 the speed of light...
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Hypnosifl
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### Re: Energy of transition in accelerating transportation

Toffo wrote:
Hypnosifl wrote:Do you think my calculation was incorrect?

Yes.

Can you point out the specific step of my calculation that you think was incorrect, rather than inventing a new problem? For example, if initially the 200 kg ball is moving at 0.75 m/s and the 100 kg ball is at rest, do you disagree that after an elastic collision the 200 kg ball is moving at 0.25 m/s, and the 100 kg ball is moving at 1 m/s? Do you disagree that both received an impulse of 100 N*s, and that each ball's change in kinetic energy was 50 J?
Toffo wrote:Try calculating the energy change of a spacecraft moving 200000 km/s, when an apple is thrown out into the backwards direction with an impulse of 1 Ns. Let the mass of the spacecraft be million tons.

Use formula velocity change * impulse, velocity change must be the velocity change of the spacecraft.

Result should be a large number. But you will get a tiny number. Right? Relativity may be ignored.

Why do you think the result "should" be a large number? For such a small impulse the change in velocity of the apple will also be small, and hence so will its change in kinetic energy. You didn't specify the apple's mass, but let's say it's 1 kg...then since impulse = mass*(change in velocity), its change in velocity would only be 1 m/s (so after being thrown out its new velocity is 199999.999 km/s), so its change in kinetic energy would only be 0.5 Joules. Why should it be surprising that the ship will only have a "tiny" change in kinetic energy after it supplies 0.5 Joules of kinetic energy to an apple?

Hypnosifl
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### Re: Energy of transition in accelerating transportation

gmalivuk wrote:If you want to ignore relativity, maybe pick a thought experiment that isn't going 2/3 the speed of light...

We can imagine as a thought-experiment that this is taking place in an alternate universe where Newtonian laws hold exactly, but Toffo is still wrong about the Newtonian analysis of these problems.

Toffo
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### Re: Energy of transition in accelerating transportation

Here is the energy change of the apple:
0.5*200000000^2 - 0.5* (200000000-1)^2) = 2.0e8 J

The formula is: kinetic energy = 1/2 m v^2