# Conceptual question on work and potential energy

I'm confused by classic description of work and negative work. If someone pulls (slowly) on a rope to lift a bucket in a well, I understand that the person is doing work on the bucket and gravity is doing negative work on the bucket. So, the net work is zero.

But if the net work is zero, where is the energy coming from that increases the bucket's potential energy?

• possible duplicate of If Earth starts moving when attracted by an object, where does the energy come from?
– user65081
Oct 24, 2014 at 16:09
• I think the weight of bucket cancel the tension in the rope and hence they both cancel the effect of each other but the energy which is used to move the bucket is your force which you are exerting on it Hence the net work can't be zero Oct 5, 2021 at 21:36

Here are a few points to keep in mind:

1. Potential energy is always described as the potential energy of the system. For example, the gravitational potential energy of the Earth-Moon system, belongs to the system as a whole, not the Earth or the Moon individually. So for your example, if you are for instance throwing a brick upwards, it would be the potential of the brick-Earth system.
2. The Work-Energy Theorem can be written (in terms of conservative, non-conservative and other, external forces) as:

$$W_{tot}=W_{cons}+W_{non-cons}+W_{ext}=\Delta K = K_{f}-K_{i}$$

But for a conservative force, the definition of the associated potential energy is

$$W_{cons} = -\Delta U = -(U_{f}-U_{i})$$

and so our previous equation becomes: \begin{align*} W_{tot}=W_{cons}+W_{non-cons}+W_{ext}&=\Delta K = K_{f}-K_{i}\\ -(U_{f}-U_{i})+W_{non-cons}+W_{ext}&=\Delta K = K_{f}-K_{i}\\ K_{i}+U_{i}+W_{non-cons}+W_{ext}&= K_{f}+U_{f} \end{align*} If there are no external forces or non-conservative forces, then: $$K_{i}+U_{i} = K_{f}+U_{f}$$ So we see that we can either use the concept of the work done by gravity, OR we can use the concept of gravitational potential energy. But we don't want to do both at the same time, as then we would count the influence of gravity twice.

• Thanks. Just one follow up question, in the situation of the bucket in a well, I'm assuming the only time you have a conservative force at play would be if the bucket was in free-fall, in which case gravity would be doing positive work, which corresponds to reduction in the change in internal energy. Is that correct, or am I "double-counting" gravity in that case? Oct 25, 2014 at 13:21
• Right, if your bucket is in free-fall, gravity is doing positive work on the bucket, increasing its kinetic energy. Or you can say that gravitational potential energy is being converted into potential energy. Or that the positive work done by gravity corresponds to a decrease in gravitational potential energy (in others, is equivalent to, but should not be used at the same time when looking at total energy). And just to be clear, if you're pulling the bucket out of the well, gravity is doing negative work, but tension (an external force) is doing positive work. Oct 27, 2014 at 20:13

The energy comes from the chemical energy stored in the muscles of your arms. But ...

You have to be careful when you say that net work is zero. You have to make sure that your system is clearly defined so that you know what's in your system, and what's outside. Then you have to identify what work is internal, and what work is external, and which if any is zero.

One way of defining the system in your case is to say that the Earth and bucket are inside the system, everything else is outside. Then the rope (not you) does work on the system. This work is external work ... it come into the system from the outside.

So the work done on the system in this picture is not zero.

You can also analyze a system comprising the Earth, the bucket, the rope, and you. This case is more complicated because 1.) you now have an energy storage device (you) in the system, and 2.) you are deformable and 3.) The force produced by your muscles is not conservative. Nonetheless, in this case the total internal work is indeed zero. The energy comes from chemical energy in your muscles.

The person is doing work on the bucket by raising it against the gravitational force. ($W = mgh$).

Consider the definition of work $W = Fx$ (constant force). Since $F_{weight} = mg$, then $W = mgh$.

Whether you use a lot of force to quickly jerk the bucket up, or you use little force to slowly raise the bucket, as long as its initial and final positions and velocities are the same, the work done to the bucket would be the same. It does not matter how much force you exert in lifting the mass, all that matters is the distance the mass is lifted, as well as the gravitational force, mg. The work done is simply $W=mgh$. More explanation could be found here. https://physics.stackexchange.com/a/135194/59969