Work done running on an inclined treadmill

[harrylentil]

Why does it cost roughly the same effort to run against an inclined treadmill as up a hill of the same inclination? That is neglecting the movement of the legs and the bobbing up and down as we run and the wind resistance. I remember being told in school physics that there is no work done unless a mass is accelerated or raised. To a first approximation, on a treadmill we do not ascend and we do not accelerate.
When running up a hill you are going up. So the treadmill should be nearly effortless as you are not raising a weight hence not doing work. Of course it is not effortless so where is my error?

 

[PeroK]

Why does it cost roughly the same effort to run against an inclined treadmill as up a hill of the same inclination? That is neglecting the movement of the legs and the bobbing up and down as we run and the wind resistance. I remember being told in school physics that there is no work done unless a mass is accelerated or raised. To a first approximation, on a treadmill we do not ascend and we do not accelerate.
When running up a hill you are going up. So the treadmill should be nearly effortless as you are not raising a weight hence not doing work. Of course it is not effortless so where is my error?

It's not as hard to move on a treadmill as uphill on the ground at the same angle, but it is harder than a treadmill on the flat. Basically, the treadmill is moving part of your body (the foot and leg) downhill at each step - so, to some extent, your centre of mass. To stay at the same point on the treadmil, you have to keep moving at least some of your body upwards.

On a real hill, you have to move all your body up the whole time. And that, I think you'll find, is a lot harder than uphill on a treadmill.

 

[harrylentil]

It's not as hard to move on a treadmill as uphill on the ground at the same angle, but it is harder than a treadmill on the flat. Basically, the treadmill is moving part of your body (the foot and leg) downhill at each step - so, to some extent, your centre of mass. To stay at the same point on the treadmil, you have to keep moving at least some of your body upwards.

On a real hill, you have to move all your body up the whole time. And that, I think you'll find, is a lot harder than uphill on a treadmill.

Is it not the case that you have to keep moving your whole body upwards on a treadmill to counter the tread's movement? Say you weigh 60kg. Is the inclined treadmill not somehow forcing you to move all 60kg constantly upward (in a way I don't understand) and not just your legs?

 

[PeroK]

Is it not the case that you have to keep moving your whole body upwards on a treadmill to counter the tread's movement? Say you weigh 60kg. Is the inclined treadmill not somehow forcing you to move all 60kg constantly upward (in a way I don't understand) and not just your legs?

No. Your head and upper body remain essentially at the same height.

 

[harrylentil]

No. Your head and upper body remain essentially at the same height.

I'm having difficulty with this. I can see that your body mostly does not ascend on the treadmill at any time and wholly does not ascend on average (at the end of the run you are in the same place) and you seem to be saying that you do not have to work against the backward movement of the tread to stay put, except for a small component of work involving your lower body. This is not an answer I was expecting because I basically think it is false. I think you do have to work to keep your whole weight stationary against the tread. I asked the original question to find out how.

 

[PeroK]

I'm having difficulty with this. I can see that your body mostly does not ascend on the treadmill at any time and wholly does not ascend on average (at the end of the run you are in the same place) and you seem to be saying that you do not have to work against the backward movement of the tread to stay put, except for a small component of work involving your lower body. This is not an answer I was expecting because I basically think it is false. I think you do have to work to keep your whole weight stationary against the tread. I asked the original question to find out how.

Suppose you attached a monitor to your upper body. How would it detect that you are on a inclined treadmill? If your upper body is not moving, it's not moving.

 

[harrylentil]

Suppose you attached a monitor to your upper body. How would it detect that you are on a inclined treadmill? If your upper body is not moving, it's not moving.

If you stop working then your whole body shoots downward. To keep your whole body stationary you have to work.

 

[PeroK]

If you stop working then your whole body shoots downward. To keep your whole body stationary you have to work.

I can do about 6.5 km/h on a 15° treadmill at the gym. On a 15° hill, I can do about 4 km/h. There is no comparison in the effort involved. You have to do no work to stop your body falling. That is no different from standing or walking on the flat.

 

[PeroK]

I can do about 6.5 km/h on a 15° treadmill at the gym. On a 15° hill, I can do about 4 km/h. There is no comparison in the effort involved. You have to do no work to stop your body falling. That is no different from standing or walking on the flat.

PS I was at the gym today - honestly. I did 20 minutes at 15° at an average of 6 km/h. That is about 500m of ascent in 20 mins. My normal outdoor speed on a mountain is about 600m of ascent an hour. The reason is clear: on the treadmill my upper body is not moving; on a hill, my whole body is moving upwards.

 

[harrylentil]

PS I was at the gym today - honestly. I did 20 minutes at 15° at an average of 6 km/h. That is about 500m of ascent in 20 mins. My normal outdoor speed on a mountain is about 600m of ascent an hour. The reason is clear: on the treadmill my upper body is not moving; on a hill, my whole body is moving upwards.

I admit I'm surprised.

 

[PeroK]

I admit I'm surprised.

It's quite different muscles as well. The uphill treadmill mostly works the calves and not much on the quads. In the mountains, it's both calves and quads that do the work. You really have to push hard to move your body up a mountain. It's much harder than the same speed on a treadmill. If I set the treadmill at my normal outdoor uphill speed of about 3 km/h it would hardly be a workout. That's just easy.

 

[jbriggs444]

No. Your head and upper body remain essentially at the same height.

Completely irrelevant.

The work that is not done raising your body is done moving the treadmill belt downward. Exactly the same net work done running uphill as on an upward-slanting treadmill.

 

[PeroK]

Completely irrelevant.

The work that is not done raising your body is done moving the treadmill belt downward. Exactly the same net work done running uphill as on an upward-slanting treadmill.

The treadmill moves itself. There is no need to do any work to move the treadmill. To make it go faster, you press a button!

In any case, it is far, far easier to walk or run uphill on a treadmill than on a real hill. For example, at 15° I can comfortably do 6 km/h, which equates to 1,500m of ascent an hour. On a real hill, I would max out at 750m of ascent an hour.

1,500m of ascent an hour would make me a world-class mountain runner!

 

[PeroK]

PS There is a record for the "vertical kilometer": 1,000m of ascent in about 30 minutes at an average angle of 27.5°. I could just about do that on a treadmill.

https://www.runnersworld.com/sweat-science/the-optimal-slope-for-running-uphill

 

[phinds]

The work that is not done raising your body is done moving the treadmill belt downward. Exactly the same net work done running uphill as on an upward-slanting treadmill.

jbriggs, I usually find your comments spot on but I think you seriously missed the boat on this one. (1) The treadmill moves exactly the same whether you are on it or not and (2) yes it DOES matter that you are not raising your center of mass.

 

[jbriggs444]

jbriggs, I usually find your comments spot on but I think you seriously missed the boat on this one. (1) The treadmill moves exactly the same whether you are on it or not and (2) yes it DOES matter that you are not raising your center of mass.

Sorry, you are wrong in this case. The force of the feet on the treadmill is not irrelevant.

 

[jbriggs444]

The treadmill moves itself. There is no need to do any work to move the treadmill. To make it go faster, you press a button!

Completely irrelevant. The work done by feet on belt is the same regardless of what other forces act on the belt.

 

[PeroK]

Completely irrelevant.

What isn't irrelevant is that my gentle warm-up on the treadmill is 4 km/h at 15°, which equates to 1,000m of ascent an hour and is much faster than I could do on a real hill.

There is simply no comparison. If you climbed mountains and went to the gym, you would know this for yourself.

 

[jbriggs444]

What isn't irrelevant is that my gentle warm-up on the treadmill is 4 km/h at 15°, which equates to 1,000m of ascent an hour and is much faster than I could do on a real hill.

Irrelevant. The physics of the situation is what it is.

 

[PeroK]

Irrelevant. The physics of the situation is what it is.

... says he from the comfort of his armchair!

 

[jbriggs444]

Generate the free body diagram. Calculate the work done by the body on the treadmill. Pick up a pencil and calculate.

 

[PeroK]

Generate the free body diagram. Calculate the work done by the body on the treadmill. Pick up a pencil and calculate.

By that method, the work is approximately zero, as the body is not moving. On the treadmill, your centre of gravity is (approx) stationary. As one leg moves dowwards, the other moves upwards, maintaining your centre of gravity at a fixed point.

The only work is moving your limbs.

 

[jbriggs444]

By that method, the work is approximately zero, as the body is not moving.

Where is that free body diagram? I did not ask for the work of the treadmill on the body. I asked for the work of the body on the treadmill.

Edit: I am interested in the real work done at the foot-belt interface rather than "center-of-mass" work done on the treadmill as a whole. Neither the runner nor the treadmill are rigid bodies. Attempting to apply conservation-of-energy arguments based on center-of-mass work will lead to erroneous conclusions -- this thread being a case in point.

 

[jbriggs444]

That's still approximately zero, as your feet are stationary with respect to the treadmill's surface.

Pick a frame of reference and stick with it.

From the ground frame, the treadmill's surface is moving. But it seems that you wish to adopt the belt frame. That's fine. But in that frame, the runner's legs are lifting his body exactly as if he were running up a fixed incline.

 

[jbriggs444]

Let me try this thought experiment on you. It takes the form of a series of scenarios which I claim are all equivalent to one another from the runner's point of view. I want you to identify which scenario in the series first breaks the equivalence.

1. The runner is climbing a track on a hill with a fixed incline. The track is made of a rubbery material and extends for several hundred yards up-slope.

2. The hill is mounted on an aircraft carrier steaming slowly west on calm waters as the runner runs east. The aircraft carrier maintains a speed so that the runner makes no progress compared to the water beneath. The runner does, however end his run higher than he started.

3. The aircraft carrier is still steaming west, but now it is sinking as it moves so that the runner's height above water remains constant. [Assume that the carrier maintains speed and remains upright as it sinks]. Now the runner ends his run exactly where he started.

4. Replace the long track that was fixed to the carrier with a treadmill on an island, arranging for the surface where the runner runs to be moving west and down exactly like the track surface that was on the aircraft carrier in scenario 3.

 

[lewando]

Since the human body in motion is a complex system, why not simplify the matter. Consider using a Segway on the treadmill next time you are at the gym. Now it becomes a rotational work problem.

 

[harrylentil]

I didn't think my query would be controversial. I was expecting someone to explain where and how my assumption of zero work was wrong, because I thought it was wrong. I am glad to see that it might clarify a more widespread misunderstanding. I haven't seen a clear and simple explanation yet, though.

 

[jbriggs444]

I didn't think my query would be controversial. I was expecting someone to explain where and how my assumption of zero work was wrong, because I thought it was wrong. I am glad to see that it might clarify a more widespread misunderstanding. I haven't seen a clear and simple explanation yet, though.

Let's see you justify the claim of zero work. What force are you considering acting on what body?

 

[PeroK]

Where is that free body diagram? I did not ask for the work of the treadmill on the body. I asked for the work of the body on the treadmill.

Edit: I am interested in the real work done at the foot-belt interface rather than "center-of-mass" work done on the treadmill as a whole. Neither the runner nor the treadmill are rigid bodies. Attempting to apply conservation-of-energy arguments based on center-of-mass work will lead to erroneous conclusions -- this thread being a case in point.

We have:

a) a theoretical argument that shows that the work done on a treadmill is the same as on a real hill.

b) the empirical evidence that this is not the case.

Edit: I'll leave this post in, but I don't think we need any such complicated explanation. See my later posts about gravity providing the force in the case of a treadmill, but not in the case of moving uphill.

I think I have the explanation. It's certainly more complicated than it first appeared. When you walk (on the flat or uphill) you tend to bounce up. On a inclined treadmill, it's clear that the force is not constant. You give a very definite push with each foot for a short time and then ease off. This raises your centre of gravity for a short time. The push is vertical, rather than tangential to the treadmill, but it's probably a bit of both. The greater the vertical force, the longer the weight is off your feet and the longer you do not need to apply a force against the treadmill.

By this mechanism, you avoid having to apply the tangential force against the treadmill for all the distance.

On the flat or at slow speeds, this bouncing would be inefficient. But, as the treadmill speed increases, it becomes more economical to adopt this bouncing gait. Certainly, this is what happens with me as the treadmill speeds up. It's a very definite sequence of short sharp pushes with each foot.

It's too late today to look at this, but I'll try to do some basic calculations tomorrow. It's not clear to me whether you do save energy this way, but the evidence of walking uphill suggests it must be so.

 

[harrylentil]

Let's see you justify the claim of zero work. What force are you considering acting on what body?

Slow down. I was expecting to be contradicted, and rightly. Read my sentence again. I am stating, clearly I think, that my zero-work assumption is wrong.

 

[jbriggs444]

Slow down. I was expecting to be contradicted, and rightly. Read my sentence again. I am stating, clearly I think, that my zero-work assumption is wrong.

Fair enough. The zero-work assumption is wrong.

@lewando is trying to direct us on a productive course. But rather than replace legs with wheels, I would prefer to idealize things differently and assume for the moment that we are dealing with massless legs. The legs have two external interfaces. One at the hips where they push upward. One at the feet where they push downward. If we wish to compute the net work done by the legs, we should examine both interfaces.

We can make the further simplifying assumption that there are, on average, no horizontal forces. Realistically there would be wind resistance, but we can reasonably neglect this (or put a fan in front of the treadmill to generate a headwind). Realistically, there is energy dissipation in the strike of foot on track. But I do not think that is the effect that that we are trying to concentrate on. So let us ignore that as well. In the absence of horizontal forces, we are concerned entirely with the upward force of leg on hip and the downward force of leg/foot on track. And we are concerned entirely with the vertical motion at those two interfaces.

We have two scenarios to compare.

In one case (climbing a hill), the hips are moving upward at a vertical velocity of +v and the feet, when planted on the ground, are not moving at all. The force at the hip is equal to the weight of the runner's body. Call that mg. The force at the feet is, on average, equal and opposite (recall that we are still under the simplifying assumption of massless legs). Call this scenario A.

In the other case (climbing a treadmill), the hips are stationary and the feet, when planted on the track, are moving downward. Their vertical velocity is -v. The force at the hip is still equal to the weight of the runner's body. The force at the feet is, on average, equal and opposite. Call this scenario B.

Let us tot up the work being done by the legs in scenario A. That's $mg \cdot v + (-mg) \cdot 0 = mgv$
Let us tot up the work being done by the legs in scenario B. That's $mg \cdot 0 + (-mg) \cdot (-v) = mgv$

Edit: A concern had been raised earlier that the motor on the treadmill was somehow contributing to the energy balance. However, that concern is misplaced. The work and energy associated with the motor and any friction in the bearings are not relevant. Those forces act between the treadmill's mechanism and its belt. They do not enter into a calculation of work done by the runner.

 

[PeroK]

Pick a frame of reference and stick with it.

From the ground frame, the treadmill's surface is moving. But it seems that you wish to adopt the belt frame. That's fine. But in that frame, the runner's legs are lifting his body exactly as if he were running up a fixed incline.

No. Because when you are on the treadmill it's gravity that is providing the force. The force on the treadmill is not being generated by the use of your muscles, but by gravity pushing your weight into the treadmill.

When you genuinely move uphill, your weight is supported, but each step you must move your entire mass against gravity by using your muscles.

 

[PeroK]

We can make the further simplifying assumption that there are, on average, no horizontal forces. Realistically there would be wind resistance, but we can reasonably neglect this (or put a fan in front of the treadmill to generate a headwind). Realistically, there is energy dissipation in the strike of foot on track. But I do not think that is the effect that that we are trying to concentrate on. So let us ignore that as well. In the absence of horizontal forces, we are concerned entirely with the upward force of leg on hip and the downward force of leg/foot on track. And we are concerned entirely with the vertical motion at those two interfaces.

We have two scenarios to compare.

In one case (climbing a hill), the hips are moving upward at a vertical velocity of +v and the feet, when planted on the ground, are not moving at all. The force at the hip is equal to the weight of the runner's body. Call that mg. The force at the feet is, on average, equal and opposite (recall that we are still under the simplifying assumption of massless legs). Call this scenario A.

In the other case (climbing a treadmill), the hips are stationary and the feet, when planted on the track, are moving downward. Their vertical velocity is -v. The force at the hip is still equal to the weight of the runner's body. The force at the feet is, on average, equal and opposite. Call this scenario B.

Let us tot up the work being done by the legs in scenario A. That's $mg \cdot v + (-mg) \cdot 0 = mgv$
Let us tot up the work being done by the legs in scenario B. That's $mg \cdot 0 + (-mg) \cdot (-v) = mgv$

Edit: A concern had been raised earlier that the motor on the treadmill was somehow contributing to the energy balance. However, that concern is misplaced. The work and energy associated with the motor and any friction in the bearings are not relevant. Those forces act between the treadmill's mechanism and its belt. They do not enter into a calculation of work done by the runner.

Let's take scenario C: you are standing on a treadmill that is moving down.

The work being done by the legs is still $mgv$. It's the same forces as before. You can see the error in this case. The leg muscles are doing no external work, as it's gravity that is providing the force. Your body is simply transmitting that force as it does when you are standing.

 

[jbriggs444]

No. Because when you are on the treadmill it's gravity that is providing the force. The force on the treadmill is not being generated by the use of your muscles, but by gravity pushing your weight into the treadmill.

Again, this is completely irrelevant. The work performed by your muscles is the work performed by your muscles.

 

[jbriggs444]

Let's take scenario C: you are standing on a treadmill that is moving down.

The work being done by the legs is still $mgv$. It's the same forces as before. You can see the error in this case. The leg muscles are doing no external work, as it's gravity that is providing the force. Your body is simply transmitting that force as it does when you are standing.

Do the math.

 

[PeroK]

Do the math.

Do the exercise!

 

[jbriggs444]

Do the exercise!

Do the math. The work being done by the legs is indeed zero and not mgv. See if you can find out why.

Hint: There are two interfaces.

 

[PeroK]

Again, this is completely irrelevant. The work performed by your muscles is the work performed by your muscles.

Your muscles do little or no work to stand up. Your body can produce a force of $mg$ downwards without any effort. It's the force over and above $mg$ that is provided by your muscles.

On a treadmill (or step machine), the additional force is to keep lifting your legs.

On a real hill the additional force is to lift your whole body.

If you go to the gym and experiment, you will see this immediately. The difference is fundamental.

 

[jbriggs444]

Your muscles do little or no work to stand up. Your body can produce a force of $mg$ downwards without any effort. It's the force over and above $mg$ that is provided by your muscles.

On a treadmill (or step machine), the additional force is to keep lifting your legs.

On a real hill the additional force is to lift your whole body.

If you go to the gym and experiment, you will see this immediately. The difference is fundamental.

Do the math. Here. I'll do it for you. There are two interfaces. Hips and feet. You are standing on a treadmill in motion.

Hips: $(mg) \cdot -v$ [Upward force on downward-moving hips]
Feet: $(-mg) \cdot -v$ [Downward force on downward-moving track]
Total: $0$

 

[russ_watters]

Completely irrelevant. The work done by feet on belt is the same regardless of what other forces act on the belt.

I'll add a +1, but elaborate:

Consider a zero friction inclined plane. It requires a constant force to hold stationary against gravity's tendency to accelerate the object (you). Now move your legs against that force, creating a distance moved(the moving treadmill). That's the work, and it is exactly the same as on a hill (neglecting wind).

You guys are letting the fact that the treadmill moves itself confuse you. In fact, the fact that the treadmill maintains a constant speed against your pushing on it means it is absorbing power from you, not powering itself when you are pushing on it. We recently had a thread about non-powered treadmills, where we discussed the fact that they required brakes.

 

[russ_watters]

The treadmill moves itself. There is no need to do any work to move the treadmill. To make it go faster, you press a button!

Let me try a different way: since the treadmill moves itself, you have to apply a force to it to keep from falling off the back of it. That force is larger the higher the slope. And that force, times the speed of treadmill, is the power you apply, and it is exactly the same as climbing a hill.

 

[PeroK]

Let me try a different way: since the treadmill moves itself, you have to apply a force to it to keep from falling off the back of it. That force is larger the higher the slope. And that force, times the speed of treadmill, is the power you apply, and it is exactly the same as climbing a hill.

Imagine a flat treadmill. You need to walk to stop going off the end of it, but once you have attained the required speed, you only need the biomechanical energy to keep walking.

On an inclined but stationary treadmill, friction keeps you from sliding down.

On an inclined moving treadmill, you need a combination of these: You need to walk. Friction still holds your weight in place, but your legs are pulled down and you need extra energy to keep lifting them up.

The evidence, however, is clear: you can do speeds and "height" gains on a treadmill that are impossible on a real hill.

 

[jbriggs444]

Yes, but most of that force is provided by gravity, not by your muscles. That's the difference.

The muscles go through the same motion either way. The notion that the force is "provided by gravity" is not well posed and is irrelevant.

Anyway, a force is only needed to accelerate; not to stay stationary.

A force is needed to resist gravity. If you prefer we can treat gravity as a fictitious force and adopt a free fall inertial frame, but that seems pointless.

Imagine a flat treadmill. The same applies. You need to walk to stop going off the end of it, but once you have attained the required speed, you only need the biomechanical energy to keep walking.

Yes. Walking or running on a flat treadmill is equivalent to walking or running on a springy sidewalk. However, walking up an incline requires force to maintain one's state of motion in spite of gravity.

On an inclined but stationary treadmill, friction keeps you from sliding down.

Yes, friction provides a force. But only as long as it is resisted.

On an inclined but stationary treadmill I would think that we are all in violent agreement that the leg muscles require no power. One could sit on a chair and achieve a similar effect.

On an inclined treadmill, you need to walk. Friction still holds your weight in place, but your legs are pulled down and you need extra enrgy to keep lifting them up.

Yes indeed. You need to power your legs in order to maintain the strain that allows the friction to exist. That's power required on the downstroke.

The evidence, however, is clear: you can do speeds and "height" gains on a treadmill that are impossible on a real hill.

Unfortunately, the proffered explanation for this "evidence" is just plain wrong.

 

[PeroK]

Unfortunately, the proffered explanation for this "evidence" is just plain wrong.

You dismiss the evidence too easily, by the way. Why not do the experiment yourself or get a friend to do it? You ought to be willing to test your theory by experiment.

a) How fast can you run up a hill?

b) How fast can you run up a moving treadmill of the same gradient?

You may deduce theoretically that a) and b) are equal, but you shouldn't be unwilling to put this to the test. You will, in fact, undoubtedly find that b) is far greater than a).

It may turn out that my evidence is "plain wrong", but you shouldn't dismiss evidence simply because it doesn't concur with your theorectical conclusions.

I can simplify my explanation (in the ground frame):

When the treadmill is stationary, there is a force of $F = mg \sin \theta$ acting down the treadmill. This is balanced by an equal and opposite friction force. Normal forces are of course balanced. There is no motion.

When the treadmill is moving, there is an additonal tangential force generated by the person's leg muscles of $f$ and the friction force increases to balance this. Again, the centre of gravity of the person remains constant. $f$ is required for walking at the required speed and to lift the legs/moving parts against gravity.

The work done/power by the person's muscles is $fv$. Where $v$ is the speed of the treadmill. There is no need for the muscles to assume the additional burden of providing the force $F$ when the treadmill is moving. $F$ continues to be provided by gravity.

 

[jbriggs444]

It may turn out that my evidence is "plain wrong"

I made no judgement about the evidence. It is the explanation that is incorrect.

 

[jbriggs444]

The work done/power by the person's muscles is $fv$. Where $v$ is the speed of the treadmill. There is no need for the muscles to assume the additional burden of providing the force $F$ when the treadmill is moving. $F$ continues to be provided by gravity.

This notion that a force of one object on another being "provided" by some other source is decidedly unphysical.

 

[FactChecker]

CORRECTION: I have been corrected about this. I no longer believe this is a good way to analyse work. Why? At the point of contact between foot and treadmill, work is being done (force times distance). If the man is not walking, gravity is doing the work and his mass loses altitude. If the man is walking, the man is doing the work and his mass altitude remains constant.

Using anything except the official definition of "work" opens a can of worms. But you can still talk about expending energy while no "work" is officially being done. Your zero-energy path on an inclined treadmill is the downward trajectory you would take while standing still without walking on the moving treadmill. By walking, you expend the same amount of energy to stay in place as if you were climbing a hill. No work is being done, but energy is being expended.

The calculation of the difference in potential energy between the zero-energy trajectory and the "walk to stay in place" is easy to do.

 

[A.T.]

Your muscles do little or no work to stand up. Your body can produce a force of $mg$ downwards without any effort. It's the force over and above $mg$ that is provided by your muscles.

On a treadmill (or step machine), the additional force is to keep lifting your legs.

On a real hill the additional force is to lift your whole body.

If you go to the gym and experiment, you will see this immediately. The difference is fundamental.

When the treadmill runs at constant speed, the only difference is air resistance. Just analyse the treadmill from the inertial frame, where the upper belt surface is at rest. Here you continuously move upwards, just as you would on a real hill.

 

[russ_watters]

On an inclined moving treadmill, you need a combination of these: You need to walk. Friction still holds your weight in place, but your legs are pulled down and you need extra energy to keep lifting them up.

Ok...so that's the work, right?

 

[FactChecker]

My two cents:
Stopping something from falling (either free fall or any other downward trajectory) can not be considered work. It would not be possible to say that a person on an inclined treadmill is doing work while at the same time saying that a table that keeps a book from falling is not doing work. The standard definition of work (force times distance) can not be changed without opening a can of worms.

But you can say that the man on the treadmill is expending energy and the table holding up a book is not. That can be done easily without any tricky definitions or controversy.