Why is it said that no work is done when I lift a box and walk?
Because the force that you apply to support the box as you walk is vertical but the motion of the box as you walk is horizontal. Direction matters.
But I am also applying force diagonally to the ground to move me and the box forward.
This is true only when you accelerate. If you are walking at constant speed, the only force you apply to the box is vertical.
Now, in a walk, the box might go up and down a bit, resulting in work actually being done alternatingly on the box and by the box - netting zero work done on the box.
This does not mean that your body is not doing work to generate the required force. However, this work ends up as internal heat rather than providing work on the box. Muscles are not the most efficient source of static forces ...
You just need to realise that "Work done on" (an object) is very strictly defined as the Force exerts on the object times the distance the object moves in the direction of the Force. It will always be different from "Work done by" (you), because practical things are not 100% efficient. If you stand and hold a massive weight for a day and it doesn't move up or down, you will have used a lot of energy but your efficiency will have been 0%.
There is no paradox and no one got anything wrong - it's just where real life and theoretical definitions do not happen to coincide - at first sight.
No, every time I walk I have to apply some force.
When I am moving, I am being displaced. So, the box is also displaced.
You apply force to the ground, not to the box. (Other than supporting its weight.)
But you are not exerting a force on the box in the direction of its displacement.
But friction is.
Friction is not acting on the box. Friction acts on the ground, which isn't moving.
Let's break the transport of the box into 5 phases:
1) lift the box
2) accelerate the box
3) move the box at uniform velocity
4) decelerate the box
5) set the box down
In step 1, work is done on the box to increase its PE.
In step 2, work is done on the box to increase its KE.
In step 3, neither the PE nor the KE is changing so there is no work.
In step 4, work is done by the box as its KE decreases.
In step 5, work is done by the box as its PE decreases.
The work done by the box in steps 4 and 5 are exactly equal to the work done on the box in steps 1 and 2, so no work is done over the whole period.
Friction acts on me. That is why I can move.
You are not getting the point that the definition of Work is very strict. You are applying it outside of its proper context. If you move the box up and down, the total displacement is still zero and the Work is zero. If you accelerate it and then slow it down, there is still no net work done 'on' it because it will return any energy it has. And remember - the displacement has to be in the direction that any force is applied, for work to be 'done on'.
You (and many other people, aamof) seem to want to disprove something or find a loophole in this work thing. There are no loopholes - if you stick to the strict definitions. OK, that may not make sense to you but, if you follow Physics further along the line, you will see the point of doing things the Physics way. When you think you have found something which doesn't quite add up - then you need to assume you are probably wrong and find out how, rather than getting cross about silly Physics.
I am not trying ti disprove anything. I am just trying to understand what I am unable to.
OK, that's fair enough. Let's say that you are struggling to 'accept' what is generally accepted. No one is saying that no work gets done, remember.
So you just need to apply the definition strictly without trying to put your own interpretation on it. Ask yourself where the forces (with directions) are and what the displacements are (in all directions). You will find each step in that process is understandable / acceptable, I think.
And it isn't a matter of "understanding" a definition. You can only apply it and see how it turns out. If Work were to have been defined differently then it would lead to different conclusions. Your problem may well be that you are not actually applying the definition rigorously enough. There are many instances of where people get confused when they do not stick strictly to a definition.
Yes, friction acts on you and, yes, that is why you can move. But that friction does no work on you, because the point of application is stationary. And you do no work on the box, as you exert no force in the direction of its displacement (when walking at a uniform velocity).
That of course does not mean that you are not expending energy, but that's internal to your body.
Once you are walking at a steady pace, the box will continue to move since it requires a force to stop it (we're ignoring air friction). At this point you are only applying a force upwards against the box to keep it in the air. Since the upwards force you provide is exactly equal to the force of gravity, the box stays at the same height. Since work is force acting through a distance, no work is done in either case.
W = F x D
Forward motion: Zero force applied in the direction of motion, so F x D equals zero, no matter how far the box moves as long as it moves at a steady velocity.
Upward force: The upwards force is counterbalanced by gravity, so the distance the force is applied over is zero, so F x D still equals zero. (Forward motion doesn't count here because the upwards force is perpendicular to the direction of motion)
The key here is to understand that we are talking about work done on the box, not on your body or any part of your body. Obviously it requires work to walk forward, but we don't care about that work because it isn't be done on the box, only on your body. We're also ignoring some real life issues such as not realistically being able to keep the box perfectly steady in your hands and the fact that your body isn't a perfectly efficient machine and has to expend energy (and thus perform work) just to keep the box in the air. Just remember that it's a learning tool, nothing more.
If it helps, replace the person with a frictionless surface. Once the box is accelerated, it requires no work for the box to remain in motion.
But consider this. As soon as I lift the box, the box and me act as a system and so if force is applied to my body, force is also applied to the box.
And also with the box in my hand I will have to apply a greater force to move forward.
All this has nothing to do with the definition of work done on the box. The box doesn't care how you are producing the forces on it or what forces happen to be exerted at any particular time. All that matters, for the purpose of calculating the work done on the box is F.Δx (dot product). If Δx is zero then no work has been done. Why do you keep arguing when all you need to do is to accept the definition and see how it applies here? You will not 'understand' how the definition applies if you do not apply it strictly.
'Work done on' is a tiny subset of the set 'Work'. Arguments involving Work may not apply to 'Work done on'.
I just don't understand the thing that how can force be applied just on my body and not on the box when the box and me are the same system?
The ground does no work on you and the box because it does not move and has no energy source. You and the box are a 'system', if you like, (in which some work is done) but what has that to do with work done on the box? You are mixing yourself (and the thread) up with irrelevancies. Can we just deal with one thing at a time, please?
NB Force and Work are two different things.
You can always consider you and the box to be the same system, even when you are not touching the box. You can always consider you and the box to be different systems, even when you are not touching the box. The boundary of your system is completely arbitrary. You can even consider half of the box to be a system and the other half of the box to be a different system.
The boundaries of the system are completely arbitrary, but if you don't make the box its own system then you cannot make any statements about the work done on the box. If you make the person and the box a system together then you would only be able to analyze work done on the person box system.
Using the same 5 steps, but considering the person-box as the system:
In step 1, the system converts chemical energy (CE) to potential energy (PE) and thermal energy (TE).
In step 2, the system converts CE to KE and TE.
In step 3, the system converts CE to TE.
In step 4, the system converts CE and KE to TE.
In step 5, the system converts CE and PE to TE.
Again, no work is done, but in the end you have converted a lot of CE to TE, which explains why the system gets tired.
And so is the choice of reference frame, which also affects the work done by a certain force. That's another reason why one shouldn't try to understand 'work done' as some objective statement about the situation.
You can consider you and the box as a single system, but you do not have to. I can easily consider you and the box to be different systems, and in this example it's a requirement if you want to understand. For example, if I drop a hat on your head as you are walking with the box, the hat applies a downward force on you, but not on the box.
When first accelerating, yes. Once you are walking you do not. Remember that we are assuming an idealized, non-realistic version where you can keep the box perfectly steady in your hands, regardless of what actually happens when you walk.
Honestly, you are missing the point here which is that once the box is moving at a steady velocity, no more work is performed on it. It doesn't matter how the box got up to speed or what is holding it up, it's moving and therefore will remain moving until a force is applied to stop it. Your body simply serves as an inefficient machine to keep the box lifted up off the ground.
Okay. Now, I get it. It was a moment for me when I made the silliest of mistakes. Sorry of all the trouble especially to @sophiecentaur who got a bit angry.
In order for us to help you, you may need to be more specific about what forces you are referring to because right now people are having to guess.
One possibility is that when you walk you do indeed apply forward force to the box....and then you apply backward force to stop it. Both with each step and for the initial acceleration and final deceleration. But these forces (or their works) sum to zero.
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