Work does not require a change in potential energy. For example, moving a car on a flat and level surface requires energy and equates to work but it does not change the potential energy of the car.
In your scenario you and I each expended the same amount of energy and performed the same amount of work. Much like Sisyphus, however, your work was pointless.
This whole “potential energy” argument is irrelevant. If I walk a mile on flat ground, I gain no potential energy, yet I am doing work.
Not true, because you haven’t accounted for all the energy transfer in the system. I think the issue is earlier in your post, where you say this:
You do perform work on the escalator, whether it’s self-powered or not. A component of your weight will drive the escalator downwards.
If you don’t believe me, draw out a free body diagram. Whether you’re walking or standing on the escalator, your weight exerts a force parallel to the belt of the escalator. In a frictionless unpowered system, that force is unopposed and must result in acceleration of the escalator (an increase in its kinetic energy, in other words). In a powered system maintaining a constant velocity, the energy would be transformed into electric energy…but the point is, that energy must go somewhere.
To stay at the same height above the ground (maintaining your potential energy) you must add back energy into the system. THe energy comes from your legs.
I’ve looked over my posts and I can’t find any place where I said that work must be turned into potential energy. However, in this case the work done is climbing the stairs is turned into potential energy;
Why do you say that we both did the same amount of work?
True but the OP didn’t ask about people walking a mile on flat ground. It asked about people climbing stairs in which case potential energy changes are involved.
Good point and now that you mention it, I don’t think I thought the unpowered case through very thoroughly. If the system is frictionless there is no force opposing my weight component parallel to the stairs and there is no upper limit to the downward speed of the stair. I can’t run fast enough to keep up.
On the other hand, if there is friction in the system, how much work I do in overcoming that friction is unknown unless the friction is specified, which it is not.
In addition to the frictional work there is the kinetic energy of the rotating and translating parts of the stair. Those depend upon the moment of inertia of the rotating parts and the mass of those that are translating. None of those are specified either.
So the question, as posed in the OP can’t be answered without additional data.
In the case of a powered escalator, I believe that the force that I exert downward and the work that would put into the system is overcome by whatever powers the escalator and not by my legs. But, here again, I can be convinced otherwise.
David I think you are over thinking this.
You are gaining potential energy at the same rate I am, the difference is the movement of the escalator is taking yours away as fast as you are adding it.
You are also lifting your weight when you walk up the down escalator. If you do not raise your weight to the next step you will continue to the bottom of the escalator.
If you doubt this find a high end gym that has one of these, spend 5 minutes on it and tell me no work is being done. 
One more minor nit on which is more work, I submit that climbing an escalator seems like it is more work than stairs due to the design. A set of stairs usually has a rise and run that total 18 or 19 (10" rise + 9" run) Escalators have much higher steps, and much longer treads. If you have ever climbed a stopped escalator you are working your muscles through a range of motion they are not used to, so it seems like it is much harder to climb the escalator. Or maybe it is just my short little legs that make it seem that way.
I don’t think so. In the army I learned to walk without bouncing my CG up and down. In fact in Cadet College Tranining and Pre Flight school you got gigs (demerits) for bobbing up and down and had to walk punishment tours on weekends. So if I walk that way I’m not lifting my weight to the next step but am bending my knees so as to not lift my CG. With a little practice and the right incentive it becomes automatic.
And maybe I am overthinking it, but what else have I got to do. It’s raining outside so I can’t golf. 
That’s not a powered escalator. That’s like walking up stairs because it has an adjustable friction brake. I’ve already covered this in my last post. If the amount of friction isn’t specified, and it isn’t in the OP, then you don’t know how much work is being done on the unpowered escalator.
Go back and read the specs. It has a 110V motor that has a 1 year warranty.
Having used one, please believe me that it has a motor.
Why are you focused on potential energy? It is not required to define work. Why would the context of the OP affect whether work is being done? Either there is, or there isn’t. I still say potential energy is a red herring with no relation to the question.
All things being equal with respect to your movement relative to the stairs, you will expend approximately the same amount of force and energy going up a flight of stairs as going up an escalator.
From an external reference, you will do less work on the escalator.
In short, going up the down escalator is a highly inefficient way of going up.
All true, but…
…these conclusions don’t follow.
As long as the escalator is moving, your weight is transferring energy into the escalator. The different scenarios you list (frictionless, with friction, powered stair, etc) change how the energy is transferred from the escalator to the rest of the environment but it doesn’t change how the energy is transferred from you to the escalator.
Energy is force times distance. In this case, that’s (force parallel to the direction the stairs travel) times (distance the stairs travel in that direction). If the force exists (and it does, if you step on the escalator) and the distance exists (and it does, if the escalator is moving), then you must have a transfer of energy. That energy must come from somewhere, and it comes from some combination of your original potential energy and your legs, depending whether or not you’re walking.
Yes and no. Work is the rate of transfer of energy, so you have to have some form of energy to transfer to do work.
The energy being transferred doesn’t have to be potential energy, so in that sense it’s a red herring. However, many people make the first-blush assumption that the only forms of energy here are potential energy and “muscular” energy (for lack of a better term), and that if there’s no increase in potential energy, there’s no expediture of “muscular” energy (that’s essentially the OP’s question). From that perspective, potential energy is importnat, if only to contrast it with the other forms of energy in the system.
Again, yes and no. Force and energy is the same, yes. But work is also the same–the relative velocities and distances between you and the escalator are independent of reference frame.
The potential energy gained by the person walking up stairs results from the work done by that person and that is the standard for comparison. The question is does someone maintaining the same level on an excalator do as much, more or less than that amount?
Right. I believe you. I would suppose that the workout domes from running the motor fast enough that you must maintain a rapid pace to keep up. So I suppose one possible answer to the OP is that how much work you do keeping up with an escalator depends upon how fast the escalator is running. It does seem to me that if the escaltor runs at the same pace as the person walking up the stairs, then you would do less work on the escalator.
However, it does depend upon how you walk. If you step on the next step up and raise you weight you are correct, you are lifting your weight just as far on the escalator as you do on the stairs. I don’t think this is the way we would do it. I would put my foot on the step and as the step comes down I would tranfer my weight to that foot gradually so that I am standing with full weight on it when it reaches approximately the original level. Then put the other foot on the next step and repeat. I don’t think my weight would be lifted by anything like the full amount.
However, now we are in a matter of opinion as to just how we walk on an escalator so as to stay at the same approximate level.
It doesn seem to me that the force that your weight applies to the escalator comes from gravity and not your muscles.
Would someone please bring back that airplane?
You do, more or less, the same amount of work running on a treadmill as you do runnning on a road. Same thing for the stairs vs. escalator.
First I’ll run on the treadmill. Let’s look at the reference frame I am in. That being the one moving at the speed of the treadmills surface. Relative to the surface of the treadmmill I am moving forward with a certain velocity. I am also applying a force in the horizontal direction. Since I am moving a distance (velocityt), and applying a force I am doing work. That value being W=Vt*F.
I am also applying a force and moving in the vertical direction. In essence I am jumping up and down. I do work both propelling myself up into the air, and stopping myself as I come down.
So what happens to the energy? Well, the energy from going up and down essentially disappears into the ether. Some of it is converted to heat due to friction when my shoes compress, or when my joints rub together. Some of it goes to flexing or compacting the ground. Since the systems in question (the ground, and the atmosphere) are so huge, the amount of energy I am putting into it is negligible.
The energy that I am using to propel me forward goes into the turning of the treadmill. Nearly all of that is lost to friction in the gears, while a tinsy bit goes into the kinetic energy of the tread. If the tread were running on frictionless bearings, all my energy would go into the tread. The tread would speed up very fast until I find myself flying off the back of the readmill.
Same thing with the escalator.
[quot=brewha]Does my question make sense? Can someone explain to me why staying on an escalator is actually work aside from just counteracting friction?
[/quote]
I don’t think anyone can explain why it isn’t just counteracting friction, becuase that’s all it is. It’s hard to move the tread on treadmills. If you don’t turn on the motor it’s doubtful that you would be able to come close to running speed. There are unpowered treadmills but they don’t work quite as well as powered ones. It’s much more difficult run on an unpowered treadmill at a certain speed then it is to run on the road at that speed.
“Weight” comes from gravity by definition. The force equals m*g.
As Zut pointed out, potential energy is a red herring. Work is not defined by change in potential energy. If, however, you persist in wanting to know why the two scenarios do not end up with identical potential energy even though they produced the same amount of work, then Zut also explained that: the escalator is decreasing the poential energy of the person walking up the down escalator as fast as he is increasing it.
One last thought experiment: there two elevators with step stools on the floor. One is decending at a constant speed, the other is motionless. If a person in each elevator steps on to the stool do you think they do the same amount of work?
Look at it this way then: If I climb a set of stairs, I gain potential energy. But then I take an elevator back down to the ground level; what has happened to the potential energy? It was given back on the ride down, even though the elevator is powered. It is exactly the same climbing the down escalator, only you are giving back the energy one step at a time, instead of all at once as going down in the elevator.
David you are still missing it.
Let me restate the problem, and see if it makes sense this way.
You have an escalator and a staircase side by side. Same length. Furthermore they both have the same rise and run for the steps.
We will start with the escalator stopped.
Now if you walk from the first floor to the second floor you will increase your potential engery by X. It does not matter if you use the staircase or the stopped escalator your potential energy will increase by X.
Now you and I will both climb half way up the escalator / staircase. We both have increased our potential engery by 0.5X.
Now we will start the escalator. We will asume that the escalator moves downward at the exact same rate we have been climbing.
If you (on the stairs) keep climbing, you will go from 0.5X to X in time T. On the other hand, if I (on the escalator) keep climbing at the same rate, I will maintain a potential energy of 0.5X.
How is this possible? Simple, the escalator is subtracting potential engery at the exact same rate I am adding it, by my climbing. If I stop climbing, I will return to zero potential engery in time T (we started half way up remember).
Now if I climb faster, when I finally get to the top, I will have achieved potential engery X. It will have taken me longer than time T, but I have gotten there.
Now if you cimb the stairs for time T, and I maintain the same position in the middle of the escalator for the same amout of time, We have done equal work. how can it be otherwise?
David, David, David think for just a second. The only time you can do what you just described in at the bottom of the escalator. In the middle (in fact anywhere except the bottom) the treads are a constant distance apart. you can’t do what you are describing, because your lower leg is retreating at the exact rate the next stair is arriving. So going back to my example, we are starting in the middle of the escalator eliminating your arguement.
We have no argument that if you do raise you CG by one rise with each step when on the escalator the two energies will be the same. I deny that your CG is raised. You can’t step on a tread, transfer your weight and straighten your leg instantly so as to raise your CG by the length of the riser when the tread is descending.
I am confident that you walk in place with very little change in the location of your CG by bending and straightening of your legs and gradual transfer of weight from one foot to the other and you do this regardless of where on the escalator you are.
This being our difference, it appears unresolvable by discussion. Off to the lab. 
This particular argument is a red herring. It’s a red herring because of what Fear Itself says:
When you’re talking about reversible energy transfer (which this is, near enough), the final distribution of energy is independent of path. In other words, as you go from step to step, it doesn’t matter whether your CG moves smoothly, or bounces up-and-down, or even rockets to the top of the escalator and back. As long as you wind up in the same spot, the energy values are the same.