Work done when lifting an object at constant speed

Time to jump into the fray. This equation here

$$W=intmathbf Fcdottext dmathbf x$$

is just the definition of the work $W$ done by a force $mathbf F$ along some path that you are performing the integral over. It is always applicable, as it is a definition. However this equation $$W=Delta K$$ is only valid when $W$ is the total work being performed on your object. If there are multiple forces acting on your object then, you would need to first add up all of the work done by each force, and then this total work will be the change in kinetic energy.

But if you imagine lifting up a rock from the ground at constant speed, am I not doing work on the rock by converting the chemical energy stored in my muscles into the potential energy of the rock? I am confused because the kinetic energy of the rock does not change and yet I am still converting energy from one form to another, which is the qualitative definition of work. What's the right way to think about this and the concept of work in general?

Your force is doing positive work on the rock. Gravity is doing negative work on the rock. The net work in this case is $0$, so the change in kinetic energy is $0$. The definition of work is all you need. Is $intmathbf Fcdottext dmathbf xneq0$? Then your force is doing work. You can try to follow where all of the energy is going, but that is just bookkeeping (which is essentially the purpose of energy in introductory Newtonian mechanics). All you need to do to determine if a force is doing work is look at the integral.

The problem with your qualitative definition of work is that it is hard to apply generally, as similar scenarios will need to be explained very differently. For example, in your case with lifting the rock, one could argue that the work you are doing is increasing the potential energy of the rock-Earth system. Ok, that is fine. But now let's say you, the rock, and myself are in space far away from any substantial gravitational influences. Let's say you also start pulling up on the rock (that already has some upward velocity) with a force equal to what its weight would be on Earth, but I pull down on the rock with that same amount of force. You are doing the exact same thing here, and you are doing the same amount of work as before. But now we cannot say that you are changing the potential energy of the rock, as it is not being acted upon by some gravitational force. Instead now you have to consider the rope / me / energy I am expending, etc. It becomes complicated and unnecessary if all you wanted to know is if you are doing work or not. Just look at the integral and you are good to go.

Work is always associated with some form of energy exchange, but I would highly recommend against trying to use energy exchange as a definition of work, as the application of this definition is usually much less practical than just doing the integral found in the definition of work.

But if you imagine lifting up a rock from the ground at a constant speed, am I not doing work on the rock by converting the chemical energy stored in my muscles into the potential energy of the rock?

Yes, you do. There are a few things to be noted here. First, as you lift it up, the rock has to move which serves as it's kinetic energy. As you lift the rock, you increase its potential energy. And you can observe muscles in contact with the rock gets compressed, meaning there is also work done and some potential energy is stored there!! Change of work need not be necessarily the change in kinetic energy but can also be in the form of potential energy.

I am confused because the kinetic energy of the rock does not change and yet I am still converting energy from one form to another, which is the qualitative definition of work. What's the right way to think about this and the concept of work in general?

The energy of the rock changes as I explained above. The right way to think about work is basically you do some changes in kinetic and potential energy by transforming them or add or remove energy from/to the system.

I think maybe you are confused by an example that is often given to emphasize that the physics definition of "work" doesn't always match how it is used in everyday speech. If you move a rock horizontally at constant speed, then you are doing no work on the rock. Its kinetic and potential energy are both the same through the whole process. This is a common example because it certainly feels like you are doing work, and that is because maintaining constant tension in your muscles takes energy, which is converted into heat. Individual cells contract and then get tired and loosen up; you can't just lock your muscle at a certain length.

But lifting a rock upward is different, even if the velocity is constant. Then you really are doing work on it in the physics sense, because you are converting chemical energy to gravitational potential energy.

The problem with work is the word we use for it. Since we work everyday, we're used to associate "physical work" and "effort", and that's confusing.

Physical work is a well defined quantity, but it needs 3 surnames, and this is usually omitted. You must specify these three parameters:

Work done [by some force] [on some system] [along this path]

Note how deep is this sentence. Work is done by forces, not by people. You don't do work, a force does work, not you.

Plus, work must be considered on a system. If you lift a rock, you are doing possitive work on the rock. However, gravity is doing negative work on the rock.

$W_{you}=mgcdot hcdot 1$; because $cos(0º)=1$

$W_{gravity}=mgcdot hcdot -1$; because $cos(180º)=-1$

So the total work on the rock is $0$ and that's why you are not increasing its kinetic energy. If you perform extra forces, then you will accelerate teh rock ($F=ma$, you know). So you'll accelerate the rock and hence the KE changes. It makes sense.

But you have to realize that you are considering the rock as a well delimited system. If you include the chemical energy stored in your arms, then your system under consideration is not the rock alone anymore. Your new system under study will be the rock-person system, and then you have to consider ALL forces involved in that system, not just your arm.

In conclusion: work needs 3 specifications. You seem to be needing to focus on the "on what system" part.

Hope this helped

I think perhaps the easiest (perhaps way too didactic) method to understand it is in terms of currencies. For example, think of dollars (for potential energy), euros (for kinetic energy) and Yens (in the case of muscles chemical energy).

In the example of lifting up a stone (supposing somehow it was originally in motion), you are exchanging Yens for dollars (meanwhile increasing the stone's potential energy), but you do nothing with the entire system's total energy (no one is actually changing the total amount of euros in your pocket). In this simple metaphor, your total capital is constant and therefore the Energy is conserved.

You can in fact think the total Work done as the exchange house total money exchange

Hope this is useful. Regards

Assume you have two opposite forces of the same magnitude acting on a particle. The total work is zero, and there is no change in kinetic energy. However one of the forces made positive work on the particle and the other negative work. Whatever did positive work lost some form of energy, and the one that did negative work won some energy. The net effect on the particle is zero, but there was an exchange of energy between two systems, the ones which generated the forces. Alternatively, if one of the forces is dissipative, like friction, some energy will be dissipated as heat.

My professor explained it to me by giving me the spring example.

Suppose there is a spring attached to a block of mass m at one end and wall at the other.Neglect friction everywhere.

If we now pull the mass with constant velocity. Let's us see what changes happens:

In the above example there are 3 objects that are me pulling the block , the block , and the spring.

Now since we pull block with constant velocity Net force on it 0. But since the spring is attached to the block it's potential energy increases. Now we generally treat mass and spring as a system hence potential energy of system increases.

In your case block and earth = spring and mass .In your case the potential energy of block is stored between mutual gravitational field of block and earth. You applying the force is a constant variable.

I hope it makes sense.