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I know that $$W_{net}=\Delta K\tag{1}$$

However, I don't know if this is only true in a system where only kinetic energy is present, or if it's also true for a system in which there's potential energy? And if it's the second case, can the equation become:

$$W_{net}=\Delta K+\Delta U\tag{2}?$$

Are there any cases in which this last equation holds true?

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  • $\begingroup$ For a system that does not have any losses, but I have not found one of those yet... $\endgroup$
    – user207455
    Jun 29 '19 at 5:50
  • $\begingroup$ One simple case I can think of is if you push a planet into another stable orbit. In that case both the Kinetic energy and potential energy will change. $\endgroup$ Jun 29 '19 at 6:53
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    $\begingroup$ I think your question arises from not yet having fully grasped that potential energy is a mathematical device for keeping track of field energy in suitable cases, whereas kinetic energy is energy actually transferred to a body. $\endgroup$ Jun 29 '19 at 9:52
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    $\begingroup$ @AndrewSteane is exactly right. In other words, even though potential energy is extremely useful (and necessary in other areas of physics), in classical mechanics potential energy is just a helpful energy-bookkeeping device. $\endgroup$ Jun 29 '19 at 12:17
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    $\begingroup$ @AndrewSteane potential energy is not necessarily a device to keep track of field energy, but rather to formulate conservation of energy in mechanics. In non-relativistic mechanics, it does not matter where in space that non-kinetic energy is, it belongs to the whole system. Field energy is a more special concept, connected with spatial localization of energy. $\endgroup$ Jun 30 '19 at 0:36
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In mechanics there is a general theorem, called work-energy theorem, or "theorem about work". It says that net work of all forces that act in the system (internal, external, conservative, nonconservative, whatever) equals increase of kinetic energy of the system.

If we denote position of $k-$th particle $\mathbf r_k$, velocity $\mathbf v_k$ and net force it experiences $\mathbf F_{-k}$, and if we watch process from time $t_1$ to time $t_2$, the theorem can be written in this way: $$ \sum_k \int_{t_1}^{t_2} \mathbf F_{-k}\cdot \frac{d\mathbf r_k }{dt}dt = \sum_k \frac{1}{2}m_kv_k^2(t_2) - \sum_k\frac{1}{2}m_kv_k^2(t_1) $$ or in short, net work on the system equals increase in its kinetic energy: $$ \Delta W = \sum_k \Delta E_k. $$

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  • $\begingroup$ I wish I had included this in my answer. I do acknowledge it. $\endgroup$
    – garyp
    Jun 30 '19 at 14:00
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Are there any cases in which this last equation holds true?

No, because the sum of the change in mechanical kinetic and potential energy of a system equals the total change in mechanical energy of the system . The term on the left side of the equation should be $\Delta ME$ where $\Delta ME$ is the change in the mechanical energy of the system. So the equation should be

$$\Delta ME=\Delta KE + \Delta PE$$

Work is not a part of the energy of a system. It is energy transfer between objects. For an isolated system, it is energy transfer between components of the system. For an isolated system subject only to conservative forces (like gravity) the change in mechanical energy of the system is zero, or from the first equation

$$\Delta KE=-\Delta PE$$

The work energy theorem states that the net work done on an object equals the change in kinetic energy of the object, or

$$W_{net}=\frac{mv_{f}^2}{2}-\frac{mv_{i}^2}{2}$$

Where $f$ and $i$ indicated the final and initial velocities. Emphasis is on the net work done.

Now let's look at these equations with respect to changes in gravitational potential energy. We define our "system" to be comprised of an object of mass $m$ and the earth. We further stipulate that our system is isolated, that is there are no external forces acting on the system. The first equation tells us, for our isolated system, if there is a change in the potential energy of the system there is also a change in its kinetic energy, and vice-versa.

Consider the following example.

An object of mass $m$ rests upon the edge of the surface of a table where the surface is a height $h$ from the floor. The object falls off the table. The only force acting on the object is the downward force of gravity acting in the same direction as the displacement of the object. Just before impacting the floor, the net positive work done by gravity equals its change in kinetic energy of $\frac{mv^2}{2}$. Per the work energy theorem:

$$W_{net}=W_{grav}=\frac{mv^2}{2}$$

But there is also a change in the gravitational potential energy of the system, therefore

$$\Delta PE=-mgh$$

Therefore from the first equation for conservation of mechanical energy we have

$$\Delta KE=\frac{mv^2}{2}=-\Delta PE=-(-mgh)=mgh$$

or

$$\frac{mv^2}{2}=mgh$$

Now let's consider how the object got on the table in the first place. That required a force external to the object/earth system, say me. In this case the object/earth system is no longer isolated since I interacted with the system to raise the object.

I pick up the object at rest on the floor and place it on the surface of a table a height $h$ above the floor. Since it starts at rest and ends at rest its change in kinetic energy is zero. According to the work energy theorem the net work done on the object has to be zero. This doesn’t mean no work was done, just that the net work done is zero. Here’s why.

I did positive work external to the system equal to $mgh$ on the object since my average upward force of $mg$ is in the same direction as the displacement of the object. However, at the same time the downward force of gravity does an equal amount of negative work taking the energy I gave the object and storing it as gravitational potential energy equal to $mgh$ In the object-earth system. Therefore the net work done on the object is zero and the change in kinetic energy is zero.

$$W_{net}=\Delta KE=0$$

The work I did winds up as gravitational potential energy of the earth/object system and not a change in kinetic energy of the object. Note that now the mechanical energy of the system (earth/object) is not conserved because of an external (to the system) force (me). There has been an increase in its potential energy of $mgh$.

Hope this helps.

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Wnet = Δ K is sufficient, Δ U term is not required. ΔU is defined as the negative of the work done by conservative force. Thus in the equation if you have written Wnet then it means that you have already added this W done by conservative force along with work done by other forces, and hence does not need to be added as ΔU on other side of '=' sign.

Conside an example in which a ball is dropped from a tall building. It covers a distance h while free fall in some time say t. Now we want to calculate its K at that point. We can do this by 2 methods

  1. Using, W by other forces = ΔK + ΔU now W in this case would be = 0 (gravity will do work but here we will count it in potential energy) ΔU = -mgh ........1 therefore 0 = ΔK - mgh =》ΔK = mgh

  2. Using Wnet = ΔK ......2 now in this case gravity is doing work = mgh (we will not count here the potential energy change due to gavity as we have counted it in work done) =》 ΔK = mgh

You can see that the value of ΔK is same by both the methods. Which equation (1 or 2 ) is to be used depends on the choice of including W done by conservative force in Wnet or not.

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    $\begingroup$ Please use MathJax to format variables, equations, etc. $\endgroup$ Jun 29 '19 at 12:14
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The two answers that have been posted at the time that I write this are correct. So why do we see written $$W_\mathrm{net} = \Delta K + \Delta U\,\mathrm{?}$$

Well, these two answers are correct as far as they go, but their application is limited to single particle systems. There's more to the story. There exists in the world multi-particle deformable systems, and for those we need to take another look.

What does $W_\mathrm{net}$ mean in this case? We need to keep in mind the distinction between what is inside the system, and what is outside. We can draw an imaginary boundary separating what's inside and what's outside. Note that we are free to draw that boundary however we like, including or excluding whatever particles we choose. The answer we get will be the same in all cases, but the analysis that gets to that answer will be different.

The object of our attention in this case is the system, that is, the particles inside the boundary. $W_\mathrm{net}$ is the net force on the system due to interactions with objects outside of the system. We call these external forces, and we should write this as $W_\mathrm{external, net}$, but for simplicity we write $W_\mathrm{ext}$. Forces due to conservative forces between objects inside the boundary are internal forces, $W_\mathrm{int}$. Now if we apply an external force, the change in kinetic energy of the particles is not equal to the net (external) work done. The internal forces can, for example, slow them down. To do the energy accounting (and that's all it is, energy accounting) in this case, we define potential energy: $$\Delta U = -W_\mathrm{int}$$ and then $$W_\mathrm{net} = W_\mathrm{ext} + W_\mathrm{int} = \Delta K$$ $$ W_\mathrm{ext} = \Delta K - W_\mathrm{int}$$ $$ W_\mathrm{ext} = \Delta K + \Delta U.$$

As I said, this is really only bookkeeping. Notice that my analysis starts with the work-energy theorem. The work-energy theorem is ok, as long as you understand that it does not apply to multi-particle deformable systems. If you want to use the work energy theorem, you have to treat each individual particle separately. [Edit: The answer by Jan Lalinsky demonstrates this beautifully and explicitly, and should be read and understood.] The introduction of potential energy is a very useful bookkeeping device. And as others have commented, there's even more to it than that, because we can introduce the concept of field energy, which in fact becomes necessary when considering magnetic forces. But for the basics of classical mechanics, we don't have to go down that road quit yet.

And to answer your question, net work on a system is not always equal to the change in kinetic energy. But as always in science, and in any other field for that matter, you must clearly define the meanings of the words you use.

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  • $\begingroup$ So, to be sure, when there are both internal and external forces acting the net work is no longer equal to change in kinetic energy, but rather Wnet = ΔK+ΔU? This must be what I was missing, then, because I had tried to solve a problem that included both external and internal forces using only change in kinetic energy, and I couldn't get the right answer. I had to include potential energy. $\endgroup$
    – Agus
    Jun 30 '19 at 18:12
  • $\begingroup$ You should get the same answer either way. It would be instructive for you to get the two approaches to agree with each other. $\endgroup$
    – garyp
    Jun 30 '19 at 18:26
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The second equation is true only in the cases of work done by conservative forces like gravitational force. In the case of a body projected vertically upwards, at any point of its journey the work done by the gravitational force is equal to sum of change in KE and change in PE. This is true only if the system is an isolated system.

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