# Homework Help: Understanding Work

1. Mar 23, 2014

### dragon-kazooie

Hello

I don't have one specific homework question, but a question about the concept of work. My book says that work is equal to the change in kinetic energy. But it also says work is equal to force x distance. So shouldn't work be equal to the change in all mechanical energy?

For example, if a crane is lifting a beam straight up, at a constant rate, it is moving the beam a distance, by using a force, so work is being done. But there is no change in velocity (once it starts moving), so there is no change in kinetic energy, just potential energy, so according to the work-energy theorem as my book states it, no work is being done.

Can anyone help me make sense of this?

Thank you!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 23, 2014

### Staff: Mentor

Yes. You are correct. Work is more general than just the change in kinetic energy.

Chet

3. Mar 23, 2014

### guitarphysics

Nice question, I wondered about this myself as well when I first started learning work. One thing that your book didn't mention is that work is also equal to minus the change in potential energy! So we have W=-∆U and W=∆K. From this, we get that -∆U=∆K, Ui-Uf=Kf-Ki and rearranging we find our equation for conservation of energy! Ui+Ki=Uf+Kf. Hope this helped a bit.

4. Mar 23, 2014

### BruceW

hey, welcome to physicsforums!
hmm. Yeah, it seems that your book was maybe not specific enough about that. It looks like you are fairly certain that total work should be equal to change in kinetic energy, plus change in potential energy. And you are correct! So when your book was saying work was equal to just change in kinetic energy, I think most likely it meant that this is true in the special case when the change in potential energy is zero. Probably a better way for your book to say it, is like this: "the work required to accelerate some mass is equal to the change in kinetic energy".

5. Mar 23, 2014

### dragon-kazooie

Hyper Physics is usually a helpful site but it also explains work as change in kinetic energy. Is there some detail here that I'm missing that makes it ok for them to define it that way?
http://hyperphysics.phy-astr.gsu.edu/hbase/work.html#wepr

6. Mar 23, 2014

### guitarphysics

Oh, net work :). That's the big thing here! So look, when you lift something upwards, you're doing work on it, right? BUT it's not changing in kinetic energy! How? Because gravity is doing work that is equal to yours but opposite in sign, so there is zero NET work done on the object, thus its kinetic energy won't change ;).

7. Mar 23, 2014

### Staff: Mentor

Hyperphysics is not defining work via the change in kinetic energy, they are describing the work-energy principle. And, as guitarphysics points out, it is the net work on a particle (including all forces acting) that gives the change in kinetic energy.

8. Mar 24, 2014

### BruceW

that's a good point. It is the net force on the particle (due to all forces acting on it), which is used to calculate the change in kinetic energy of the particle. Although, I don't really like the term 'net work'... does that just mean work done due to the net force on a particle? If this is true, then the total work done is equal to the 'net work', plus the work that is converted into potential energy. For example, when the particle is being pushed upwards through a gravitational field at constant velocity, the net force on the particle is zero, so the change in kinetic energy of the particle is zero. But work is still being done, which is converted into gravitational potential energy.

edit: The best way to explain is probably with an equation:
$$Work = \int_{t1}^{t2} \vec{F} \cdot \vec{v} \ dt + \Delta V$$
where $\Delta V$ is the change in potential energy. and $\vec{F}$ is the net force on the particle.

9. Mar 24, 2014

### BruceW

However, if we instead say that $\vec{F}$ is the just the force caused by the external thing (doing the work), then the work done is just:
$$Work = \int_{t1}^{t2} \vec{F} \cdot \vec{v} \ dt$$
But this is not equal to the change in kinetic energy of the particle, since $\vec{F}$ is just the external force under consideration, not the net force on the particle. Thinking back to the example of the particle being pushed by an external force, upwards at constant velocity through a gravitational field, the force $\vec{F}$ must be exactly opposite to the force of gravity. Therefore, the above equation for work done is simply the work stored as gravitational potential energy.

10. Mar 24, 2014

### nasu

What will be the work done by gravity (the only force, so the net force) on a falling body?
I mean, according to this formula.

11. Mar 24, 2014

### Staff: Mentor

If you use the net force on the particle (including all forces) then you cannot also have a gravitational potential energy term. The potential energy term already incorporates the work done by gravity--you'd be double counting the effect of gravity.

12. Mar 24, 2014

### nasu

This is what I thought.
There is no need to redefine the work. It works () just fine the way it is now.

13. Mar 24, 2014

### BruceW

when we talk about 'work', it must be a transfer of energy from one form to another. The work done by some force acting on a point particle is the work done by that force to change the KE of that particle, which is:
$$\int_{t1}^{t2}\vec{F} \cdot \vec{v} \ dt$$
Where here, $\vec{F}$ is the just the force under consideration (i.e. not necessarily the net force on the particle). Also, if our force under consideration is the net force, then the work done on the particle is equal to the total change in KE of the particle. Otherwise (if the force under consideration is not the net force), the work done by that force on the point particle is not equal to the total change of KE of the particle.

So, continuing along these lines, let's say there are two forces on a point particle. The first force is a guy pushing the particle. The other force is gravity. The work done by just the guy is:
$$\int_{t1}^{t2} \vec{F_{guy}} \cdot \vec{v} \ dt$$
(where I've written subscript 'guy' to be clear). Also, we know the net force on the particle (let's call it $\vec{F}$) is equal to the force due to the guy, plus the force due to gravity:
$$\vec{F} = \vec{F_{guy}} + \vec{F_{grav}}$$
Or in other words,
$$\vec{F_{guy}} = \vec{F} - \vec{F_{grav}}$$
So if we now substitute this into our above equation for work done by the guy, we get:
$$\int_{t1}^{t2} \vec{F} \cdot \vec{v} \ dt - \int_{t1}^{t2} \vec{F_{grav}} \cdot \vec{v} \ dt$$
Here, the second term is just the negative of work done by gravity, on the particle. In other words, it is energy which is transferred from the KE of the particle into the potential energy of the gravitational field. Therefore,
$$\int_{t1}^{t2} \vec{F} \cdot \vec{v} \ dt + \Delta V$$
is the work done on the particle by the guy. (which is the force under consideration). And $\Delta V$ is the change in the gravitational potential energy, due to the movement of the particle. And $\vec{F}$ is the net force on the particle (due to the guy and gravity).

I think the important thing to remember is that when we talk about work, we need to state which force(s) are doing the work. Otherwise, there is ambiguity.

14. Mar 24, 2014

### nasu

Aha, so the expression is for the work done by the guy alone, right?

In the case of moving something up with constant speed (Fnet=0) it simply says that the work done by the guy is equal to the change in potential energy. Which is correct. But that is not total work, is it? Is just work done by the guy.

I thought that you said that this is an expression for "total work".

15. Mar 24, 2014

### guitarphysics

I think BruceW may be over-complicating things. Just use the expression for work that you know, and make sure that the force in that expression is the net force (if you're planning to equate that work with the change in kinetic energy of the particle).

16. Mar 24, 2014

### Staff: Mentor

A contrary opinion. I really like the way BruceW explained it, because it removes all ambiguity regarding what is doing work on what. Also, it lets the math do all the analysis for you. I guess what we're dealing with here is just a matter of taste.

Chet

17. Mar 24, 2014

### Staff: Mentor

Here's how I would simply express what BruceW is saying:

The work-energy theorem says that if you consider the net work done by all forces (including gravity) it will equal the change in the kinetic energy: ∫Fnet*dx = ΔKE

If you consider the work done by all forces except gravity, then the work done will equal ΔKE + ΔU. (The work done by gravity is already included in the potential energy term.)

18. Mar 24, 2014

### nasu

The work done by the guy can be written as
$$\int_{t1}^{t2} \vec{F}_{guy} \cdot \vec{v} \ dt$$

Why burying it in that formula with the net force and potential energy makes it clearer who does the work on what?

19. Mar 24, 2014

### guitarphysics

It doesn't make it clearer, but why do you need it to be? If you know which forces do what, and sum them all to find a net force (to then find the net work) it just makes everything simpler. You don't need separate terms for every force, that'll just complicate things too much (for my taste. Keep in mind that this is, of course, all personal).

20. Mar 24, 2014

### guitarphysics

Why is it necessary to take just the gravitational force out? You really simplified what BruceW was saying, thanks for that. But you also made it clear- it's just taking the gravitational potential energy out of it. It seems very arbitrary to me. (Although if other people prefer it that way, they're free to do it that way)

21. Mar 24, 2014

### BruceW

haha, yeah, I was saying 'total work' to mean 'total work done by the guy'. Sorry I wasn't clear enough about that.

yeah, in most cases we just want to calculate the change in kinetic energy of a particle, so the energy transfer under consideration is unambiguous. It is simply the transfer of any kind of energy into KE of the particle.

But there are cases when we want to talk about work as a transfer of energy more generally. For example in thermodynamics, we can have one system that does work on another system. Or we can have one system that is doing work on its surroundings. Also, in classical mechanics, if we have a dissipative system which loses energy, then we can talk about the work done on the environment (i.e. energy lost by the system). Also in classical mechanics, we can model a system where some external force is doing work on the system. In this case, we might only be interested in the work done by the external force on some particle in the system. We might not care about energy transferred between particles within the system itself. (This situation is the most similar to what I was talking about before, by considering the work done by just the guy).

22. Mar 24, 2014

### guitarphysics

But wouldn't the forces inside of the system cancel out, making it irrelevant to consider the internal energy transfers anyway? This specific example bothered me, but I do see your point in that being less ambiguous about which forces are doing the work on what/whom can be useful.

23. Mar 24, 2014

### nasu

I did not say that I need it.
But the claim seem to be that this formula makes the thinks clearer.

Please note that it is not me that started the discussion. I just added some comments about the formula proposed by Bruce. And I am OK with the work-energy theorem as it is. No need to convince me.

24. Mar 24, 2014

### guitarphysics

Sorry, didn't mean you specifically, I was saying "you" in general, to everyone.

25. Mar 24, 2014

### Staff: Mentor

Hi BruceW. I too had in mind the kind of application you are referring to for thermodynamics problems, particularly in situations where irreversability is involved. As you indicated, in thermo applications, one needs to focus specifically on the work done by the system on the surroundings. This is where your approach is often of great value.

Chet