How can we say we do work in walking up stairs or similar?

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In summary, the conversation discusses the concept of work in physics, specifically in the context of climbing stairs or lifting oneself. It is noted that there is confusion regarding whether a person is doing work on themselves or not, as well as the definition of "useful work." The concept of work is simplified as a transfer of energy, and the confusion is attributed to the use of language in describing systems in contrast to the simplicity of equations.
  • #1
alkaspeltzar
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I see physics problems asking one to calculate the work done in climbing stairs. However, isn't the system the person, and isn't the person not having any external work done? Then how can we say they did work?

Now I know when climbing or raising an object, work=mgh=potential energy.

But if the person is lifting themselves, then how can we say work was done? I thought it was noted that an object cannot change its own energy.

Another example would be ice skater, pushing off a wall, increasing kinetic energy, where the wall does no work.
 
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  • #2
alkaspeltzar said:
I see physics problems asking one to calculate the work done in climbing stairs. However, isn't the system the person, and isn't the person not having any external work done? Then how can we say they did work?
Not having external work done to them and doing external work themselves are exactly opposite things.
 
  • #3
russ_watters said:
Not having external work done to them and doing external work themselves are exactly opposite things.

Uh? I am not following. To do work you need to exert a force thru a distance. Work external on a system is equal to change in KE or PE or combination of both..

In the case of the person, are they doing work on themselves? I thought that was a no no.
 
  • #4
I found this, is this why we say work is done by a person when walking or climbing stairs?
Work done by a person is sometimes called useful work, which is work done on the outside world, such as lifting weights. Useful work requires a force exerted through a distance on the outside world, and so it excludes internal work, such as that done by the heart when pumping blood. Useful work does include that done in climbing stairs or accelerating to a full run, because these are accomplished by exerting forces on the outside world. Forces exerted by the body are nonconservative, so that they can change the mechanical energy (
\text{KE}+\text{PE}
) of the system worked upon, and this is often the goal. A baseball player throwing a ball, for example, increases both the ball’s kinetic and potential energy.
 
  • #5
alkaspeltzar said:
Uh? I am not following. To do work you need to exert a force thru a distance. Work external on a system is equal to change in KE or PE or combination of both..

In the case of the person, are they doing work on themselves? I thought that was a no no.
You are doing work against Earth's gravitational field on the way up an it is doing work on you on the way down.
 
  • #6
alkaspeltzar said:
I see physics problems asking one to calculate the work done in climbing stairs. However, isn't the system the person, and isn't the person not having any external work done? Then how can we say they did work?

You get hot climbing stairs, at least I do, that's doing work I think.

Cheers
 
  • #7
alkaspeltzar said:
I found this, is this why we say work is done by a person when walking or climbing stairs?
Work done by a person is sometimes called useful work, which is work done on the outside world, such as lifting weights. Useful work requires a force exerted through a distance on the outside world, and so it excludes internal work, such as that done by the heart when pumping blood. Useful work does include that done in climbing stairs or accelerating to a full run, because these are accomplished by exerting forces on the outside world. Forces exerted by the body are nonconservative, so that they can change the mechanical energy (
\text{KE}+\text{PE}
) of the system worked upon, and this is often the goal. A baseball player throwing a ball, for example, increases both the ball’s kinetic and potential energy.
Yes, that's fine.
 
  • #8
russ_watters said:
Yes, that's fine.
Thank you..

I guess I was confused becuase isn't it if we are doing work on ourself..most of the time work is done to another object, but in this case, doing work equals the change in PE but it is on the same body. Can you clarify why it is so?
 
  • #9
Let's say your mass is 70 kilograms.

If the problem was posed as this: "You carry a sack of concrete mix (mass=70 kg) up a flight of stairs 4 meters in height. How much work did you do on the sack?"

Now ask "you (70 kg) walk up a flight of stairs 4 meters in height, how much work has been done on your body?"

Same answer both times, right? 70 kilos mass has been raised 4 meters against gravity.

Whever the question is "how much work" always think "work on what"
cosmik debris said:
You get hot climbing stairs, at least I do, that's doing work I think.
This is just a distraction. Efficiency of the human physiology is not the issue here.
 
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  • #10
I think there is a lot of inherent confusion in the descriptions we used when work is "done". Langauge is (necessarily) sloppy in clearly defining when, where, and how energy is transferred, or what object is doing work upon which other object.
The physics expressed in equations is pretty simple. The confusion comes when people describe systems with words.
 
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  • #11
sorry @DaveE I was editing my response as you posted yours.
 
  • #12
alkaspeltzar said:
Thank you..

I guess I was confused becuase isn't it if we are doing work on ourself..most of the time work is done to another object, but in this case, doing work equals the change in PE but it is on the same body. Can you clarify why it is so?
Let me retry my first statement with different phrasing: These are the same thing with opposite directions. For example, if you push a box across the floor you are doing work on the box and it is doing negative work on you in the opposite direction.
 
  • #13
That's not my question. My question is how can I do work onto myself?

In walking up stairs with a sack, your doing work on the sack, that makes sense. But in the case when it is just I going up the stairs aren't I doing work on myself? I thought one couldn't change their own energy. Although here were are doing work, using our chemical energy and therefore increasing our PE. It's really like we are doing internal work.

So my question is how can we do work on ourselves. I thought work had to be external to a system
 
  • #14
alkaspeltzar said:
That's not my question. My question is how can I do work onto myself?
The way it is usually described (as I described above), you don't.
 
  • #15
russ_watters said:
The way it is usually described (as I described above), you don't.
What about a case where an ice skater pushes off a wall and increases their kinetic energy. Isn't the work done by the skater really internal work? The skater is doing work by using their internal chemcial energy which then becomes kinetic right?
 
  • #16
alkaspeltzar said:
What about a case where an ice skater pushes off a wall and increases their kinetic energy. Isn't the work done by the skater really internal work? The skater is doing work by using their internal chemcial energy which then becomes kinetic right?
Recognize that your question isn't really about the physics here. It's about how you define what the skater is and what the external environment is. If you would clearly define the pieces, as you did when you discussed the change in chemical potential energy, then the question becomes trivial.
You are asking questions about the definitions people use when they describe things with words. This is best dealt with by breaking the problem down into smaller, more easily defined pieces.
 
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  • #17
DaveE said:
Recognize that your question isn't really about the physics here. It's about how you define what the skater is and what the external environment is. If you would clearly define the pieces, as you did when you discussed the change in chemical potential energy, then the question becomes trivial.
You are asking questions about the definitions people use when they describe things with words. This is best dealt with by breaking the problem down into smaller, more easily defined pieces.
I don't know how to make it clearer. If you are a skater, and push against a wall, we accelerate away. Is this considered work? If so, how? As isn't work energy transfer and isn't the skater then performering work on themselves? Work is usually done on an object thru an external force. If you define the system as the person, how can this be?

If I am not clear, I'd love someone to clarify how it should be. I guess I'm confused as work definition is always based on external force.in the case of the skater, it's their internal muslces doing work
 
  • #18
alkaspeltzar said:
I guess I'm confused as work definition is always based on external force.
The work based on external forces tells you how much mechanical energy is transferred to the object from the outside.

alkaspeltzar said:
If you define the system as the person, how can this be? ... In the case of the skater, it's their internal muslces doing work
If you define the skater as one body, then there are no muscles in your analysis. We have been though this before many times with you:
https://www.physicsforums.com/threads/static-frictions-role-in-walking.977275/post-6233067
 
  • #19
alkaspeltzar said:
What about a case where an ice skater pushes off a wall and increases their kinetic energy. Isn't the work done by the skater really internal work? The skater is doing work by using their internal chemcial energy which then becomes kinetic right?
The conventions exist to keep the math consistent. w=fd. force is a vector. You have to pay attention to what is applying force to what, and in what direction.

If you want to break down the person into parts and model him as a complex machine you can do that too, but it is really way more work than is needed. You are getting too bogged down in details that are beyond the question you asked.

The wall is external to the person, so the work done is external. No matter what happens internal to the person, that's the end result and what the question is asking about.
 
  • #20
Perhaps I am getting bogged down. But then how do you look at the situation I have presented then, say the ice skater and wall?

Books have you calculate the force from the skater and distance applied to show W=ke, but this implies work is being done on an outside system. The system here is the skater, so how is this even correct?

I'm asking, everyone tells me I'm thinking too deeply, yet no one is giving me any definition of how they would break down the example.

Reason I bring up external versus internal work is it seems it would be work done internally, which transforms into kinetic energy. See linkhttps://physics.stackexchange.com/questions/132978/who-does-work-while-walking
 
  • #21
alkaspeltzar said:
Perhaps I am getting bogged down. But then how do you look at the situation I have presented then, say the ice skater and wall?

Books have you calculate the force from the skater and distance applied to show W=ke, but this implies work is being done on an outside system. The system here is the skater, so how is this even correct?
I don't see that the words "outside system" add anything here, since you have correctly applied the equation to the problem. Why are you trying to define systems at all? Where are you getting the idea you need to?
I'm asking, everyone tells me I'm thinking too deeply, yet no one is giving me any definition of how they would break down the example.
You've already successfully broken it down and gotten the right answer; there is nothing useful left to add!
Reason I bring up external versus internal work is it seems it would be work done internally, which transforms into kinetic energy. See linkhttps://physics.stackexchange.com/questions/132978/who-does-work-while-walking
Yes, in the case of a person walking at constant speed on flat ground, you can say the person is doing no external work, but is burning energy and doing internal work. But is that what this is really about? Human physiology?
 
  • #22
alkaspeltzar said:
Perhaps I am getting bogged down. But then how do you look at the situation I have presented then, say the ice skater and wall?

Books have you calculate the force from the skater and distance applied to show W=ke, but this implies work is being done on an outside system. The system here is the skater, so how is this even correct?

I'm asking, everyone tells me I'm thinking too deeply, yet no one is giving me any definition of how they would break down the example.

Reason I bring up external versus internal work is it seems it would be work done internally, which transforms into kinetic energy. See linkhttps://physics.stackexchange.com/questions/132978/who-does-work-while-walking

The work-energy theorem applies to rigid bodies. A rigid body cannot walk, ice skate or push itself away from a wall. For example:

"The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work done on the body by the resultant force acting on that body. Conversely, a decrease in kinetic energy is caused by an equal amount of negative work done by the resultant force."

If you consider a system that is a) not rigid; and b) has an internal store of chemical energy at its disposal, then your analysis based on "all work is external" fails.
 
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  • #23
How can we say we do work in walking up stairs?

By deciding that a person does not own "his" potential energy.

Now a cyclist climbing uphill is losing energy, the lost energy goes somewhere else than the cyclist, and a cyclist coasting downhill is not doing any work, the energy to stir the surrounding air comes from somewhere else than the cyclist.
 
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  • #24
PeroK said:
The work-energy theorem applies to rigid bodies. A rigid body cannot walk, ice skate or push itself away from a wall. For example:

"The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work done on the body by the resultant force acting on that body. Conversely, a decrease in kinetic energy is caused by an equal amount of negative work done by the resultant force."

If you consider a system that is a) not rigid; and b) has an internal store of chemical energy at its disposal, then your analysis based on "all work is external" fails.

Perok, so is that my mistake? Work energy theorem applies to rigid bodies in the sense I'm thinking about but not to a person or machine etc. In those cases, they do work which can change their kinetic or potential energy as they have their own internal energy to use/transform.

So in the case of walking up stairs or pushing away from a wall, we are doing work, using our energy to gain kinetic or potential and that's why we don't worry about any external work. Please let me know if my understanding is correct
 
  • #25
alkaspeltzar said:
Perok, so is that my mistake? Work energy theorem applies to rigid bodies in the sense I'm thinking about but not to a person or machine etc. In those cases, they do work which can change their kinetic or potential energy as they have their own internal energy to use/transform.

So in the case of walking up stairs or pushing away from a wall, we are doing work, using our energy to gain kinetic or potential and that's why we don't worry about any external work. Please let me know if my understanding is correct

Yes. Here's a simple model for "motion".

An object of mass ##2m## splits itself (by release of chemical energy) into two equal parts of mass ##m##, moving in opposite directions with equal and opposite momenta. Energy has been released, but total momentum of the system is zero.

One part hits a wall and rebounds, so that both parts are now moving in the same direction. Assume an elastic collision with no change in energy.

The result is that now the two masses have achieved a non-zero momentum.

The energy came from the internal chemical store. The non-zero momentum came from interaction with fixed external components, which themselves did no work.

That's a crude model for the ice-skater scenario of pushing against a wall.
 
  • #26
PeroK said:
Yes. Here's a simple model for "motion".

An object of mass ##2m## splits itself (by release of chemical energy) into two equal parts of mass ##m##, moving in opposite directions with equal and opposite momenta. Energy has been released, but total momentum of the system is zero.

One part hits a wall and rebounds, so that both parts are now moving in the same direction. Assume an elastic collision with no change in energy.

The result is that now the two masses have achieved a non-zero momentum.

The energy came from the internal chemical store. The non-zero momentum came from interaction with fixed external components, which themselves did no work.

That's a crude model for the ice-skater scenario of pushing against a wall.
I think that makes sense. Like I said then, really when looking at the work of the skater, that would be internal work technically that they are doing to create kinetic energy right? Their muscles create a force against the wall to accelerate away. This force from the body and the distance is the work from the skater. The wall provided a reaction force but really did no work
 
  • #27
alkaspeltzar said:
I think that makes sense. Like I said then, really when looking at the work of the skater, that would be internal work technically that they are doing to create kinetic energy right? Their muscles create a force against the wall to accelerate away. This force from the body and the distance is the work from the skater. The wall provided a reaction force but really did no work
Perok can you let me know if that makes sense. Thank you
 
  • #28
alkaspeltzar said:
Perok can you let me know if that makes sense. Thank you
It makes sense to me.
 

Related to How can we say we do work in walking up stairs or similar?

1. How does walking up stairs count as doing work?

Walking up stairs requires the use of muscles in our legs and feet, which exert force to move our bodies against gravity. This force is considered work, as it results in a change in position or motion.

2. Is walking up stairs considered a form of exercise?

Yes, walking up stairs is a form of physical exercise as it involves the use of muscles and increases our heart rate and breathing. It can also help improve cardiovascular health and strengthen leg muscles.

3. How is the amount of work done while walking up stairs calculated?

The amount of work done while walking up stairs is calculated by multiplying the force exerted (weight of the body) by the distance traveled (height of the stairs). This is known as the work-energy principle.

4. Can walking up stairs be more efficient than using an elevator?

In most cases, walking up stairs can be more efficient than using an elevator as it burns more calories and does not require electricity. However, for individuals with physical limitations or when carrying heavy objects, an elevator may be more efficient.

5. How does walking up stairs affect our energy levels?

Walking up stairs can actually increase our energy levels by triggering the release of endorphins, which can improve mood and reduce fatigue. Regular exercise, including walking up stairs, can also increase overall energy levels and improve overall health.

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