Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: saspinski on 27/01/2015 22:24:48
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Two twin kids are making fun in a shopping centre. One of them is "going up" an escalator that is going down, staying in the same level. His brother takes another escalator that is not working and goes up to the next floor.
If the first kid plays on the escalator at the same time that his brother climbs to the next floor, do both expend the same amount of energy?
I've never thought of a satisfactory explanation for the objection to the positive answer: Where that energy goes? That means E=mgh is clear for the second kid because he increased its potential energy. But not for the first one.
Now I believe to have an answer: the motor that moves the escalator needs less power while the boy is pushing the steps downwards. If we could measure the motor power during the 'action", comparing to when he stops and come down, the difference would be: mgh/t.
There are a lot of situations like that. Is it the right explanation or there is another one?
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Hmmm...
Sounds like an inclined treadmill problem.
The kid on the down escalator has to expend energy to counteract gravity. Good point that the energy would be going into the escalator. In fact, there are many self-powered treadmills, powered by the movement force the user puts into them.
Now, if both kids are taking the same number of steps/frequency of steps, then then one that is climbing would actually be expending more energy (I think) due to the gain in elevation.
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Wouldn't both spend about the same amount of energy, although the second kid gaining more 'potential energy' due to his elevation? you can exchange the first kids elevator to him stepping up and down a stone on the floor instead I think? It's not the exact same as the elevator step is moving against him, each one shrinking the step he start to take, but assuming that he by each step lifts himself up, maybe?
The potential energy of a 1 kg stone laying on a ladder change with the ladders elevation, but the energy content in the stone (restmass) doesn't. Here we find 'identical twins' :) not stones, but the energy they spend doesn't have to do with the elevation per se? That is if I read you right?
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you can exchange the first kids elevator to him stepping up and down a stone on the floor instead I think? It's not the exact same as the elevator step is moving against him, each one shrinking the step he start to take, but assuming that he by each step lifts himself up, maybe?
If it was a perfect elastic kid, in a perfect elastic floor, I think he would need only to spend energy for the first step. Later on, he would keep bouncing without any effort. But in the actual world all the potential energy "recovered" after going down the stone is lost. Transformed in heat, mainly in his body I suppose.
About the stone, it is necessary more power for the motor to move the escalator with the stone than without it. That is, the increase in its potential energy is supplied by the motor.
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Two twin kids are making fun in a shopping centre. One of them is "going up" an escalator that is going down, staying in the same level. His brother takes another escalator that is not working and goes up to the next floor.
If the first kid plays on the escalator at the same time that his brother climbs to the next floor, do both expend the same amount of energy?
If they spend the same amount of time on the escalator then both kids expend the same amount of energy.
I've never thought of a satisfactory explanation for the objection to the positive answer: Where that energy goes?
Your confusion lies in what's doing the work. In the one where the escalator is not working then its not doing any work. All the energy spent is biochemical energy of muscle motion. The boy's muscles are doing work raising him a distance h. In the other case when the boy is walking up the escalator which is in motion then the boy's muscles are working and the escalator is also doing work.
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In the other case when the boy is walking up the escalator which is in motion then the boy's muscles are working and the escalator is also doing work.
The escalator is doing work: E1, and the boy is doing work: E2. If E1 is the same for a given period of time, with or without the kid "walking up", then all the muscular energy are transformed in heat.
But both kids spent the same amount of energy as you said, and in the first case not all of that energy became heat, because the potential energy of the first kid increased.
Both kids are thermal machines very similar because they are twins. Because all the physical movements are the same for them, the portion of energy not going to heat should be the same.
That's why I thought of a modification in the escalator work when the kid is acting, E1' < E1.
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I look at this way:
Ignoring the energy involved in getting to the steady-state condition:
- You have two boys of identical mass (say 20kg), both walking upwards for (say) 10 seconds
- One escalator is stationary, and the boy is gaining altitude at 0.5 m/s.
- The other escalator is moving, and the boy is not gaining any altitude, but his feet are still moving vertically at 0.5m/s
- The average force on the escalator is the same for each boy - about 200 Newtons, averaged over this 10s period.
- The vertical distance moved by their shoes is 5m in this 10s period
- The energy expended is Force x Distance = 200 N x 5m = 1000 J
Including the effort required to get on and off the escalators is left as an exercise for the reader... (Read: "I've had a long day, and I'm going to bed".)
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I look at this way:
Ignoring the energy involved in getting to the steady-state condition:
- You have two boys of identical mass (say 20kg), both walking upwards for (say) 10 seconds
- One escalator is stationary, and the boy is gaining altitude at 0.5 m/s.
- The other escalator is moving, and the boy is not gaining any altitude, but his feet are still moving vertically at 0.5m/s
His center of mass is not moving upwards, it may be oscillating a little bit.
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[li]The average force on the escalator is the same for each boy - about 200 Newtons, averaged over this 10s period.
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[li]The energy expended is Force x Distance = 200 N x 5m = 1000 J [/li]
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Including the effort required to get on and off the escalators is left as an exercise for the reader... (Read: "I've had a long day, and I'm going to bed".)
If we disregard the energy lost as heat, I agree on the energy spent. But in the case of the moving escalator, its motor spent 1000 J less during this period, that is my point.
- The vertical distance moved by their shoes is 5m in this 10s period
It is the case for the stationary escalator and also for a different situation: if he just goes down from the upper floor, in a downward moving escalator, but being at rest on the steps. In my opinion, the motor would be relieved by the same 1000 J for this last case.
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"If we disregard the energy lost as heat, I agree on the energy spent. But in the case of the moving escalator, its motor spent 1000 J less during this period, that is my point. "
Yes, the guy 'helping' the elevator going up against it going down do help that motor, saving some energy. But that's not how I read you? I was thinking of whether one guy would spend more energy than the other. And there I can't see it matter where the energy 'spent' goes? The energy the boy at a higher elevation gains is a result of him being inside a gravitational 'field' (gravity well), not a result of him suddenly gaining some sort of 'mass'. If you weighted him he should weight a minuscule amount less than the twin left down at the floor. Although their 'rest mass' is the same naturally. as he falls down he will gain that 'weight' again, and in a relation with the ground also transform whatever potential energy he gained into kinetic energy. It's perfectly transformable to the ground rushing to hit him.
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You can define it two ways. If you define the universe as a 'container' then it's a result of there being a given amount of 'energy' inside it. That 'energy' is also part of 'gravity'. But it is not measurable in this case, there is no way I know of to measure 'potential energy' directly. It's a definition between two 'objects', and you're free to pick any relation you like, simultaneously. This 'elevated boy' will have a simultaneous different 'potential energy' relative the moon for example, as will the boy on the elevator.
The other way to define it is relative conservation laws, ignoring any idea(s) of a 'container'.
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But it is not measurable in this case, there is no way I know of to measure 'potential energy' directly.
I think there is a way to measure the increased potential energy of the boy going up to the second floor. As soon as he is there, his weight is added to the weight of the building structure. All that structure deflects a little bit under the increased weight. The increase in the elastic energy stored in the building is a measure of his gain in potential energy. If he goes up to the third floor, the force (his weight) is the same, but the deflection increases, and so the elastic energy.
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The guy above would weight less than when was 'down under', so to speak :) So adding that decreased weight to the building should have a opposite effect, Don't you agree? But I think you may considering it in a way similar, not the same though, as GR does? In where you find the stress energy tensor.
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The loss of weight is too small for a 3 meters higher floor.
If we simplify the building structure to a column, the compressive deflection ΔL/L is (almost) the same under the same boy weight, if it follows Hooke's law, regardless of the position of the load (the kid).
So, the absolute deflection ΔL is greater for greater L, (how high is the boy). And the elastic energy increases when ΔL increases for the same load.
But going back to the subject: I think that both kids spend the same energy. You could think of a 2 floors airplane, losing altitude. One person going up to the 2nd floor, could be seen as keeping the same altitude by an external observer, but his effort to climb can not be different if he does the same movement before the plane descend phase.