# Homework Help: A conceptual problem regarding work

1. May 20, 2005

### misogynisticfeminist

Work is the force multiplied by displacement. If i carried my sofa to the moon and back, at the exact same spot. My displacement would be 0, does this mean that i have done no work?

2. May 20, 2005

### Justin Lazear

Work, in general, is not force times displacement. However, the result is still the same. If you moved your sofa to the moon and back (presuming you never gave away any energy in the process), you would have done no net work. The work you do to move the sofa off the earth is given back to you when you move the sofa back onto it. You don't even have to move the sofa to the same spot on the earth.

--J

3. May 20, 2005

### primarygun

Not displacement but the distance in the same direction travelled.

4. May 20, 2005

### misogynisticfeminist

hmmm in what sense is it given back to me? are you talking about gravity?

5. May 20, 2005

### SpaceTiger

Staff Emeritus
Hmm, I'm not sure I agree with this. Work, in general, is given by:

$$W=\int \vec{F}\bullet d\vec{s}$$

If you're pushing the sofa in its direction of motion for the entire trip, then the dot product will always return the same sign and the net work will be positive. Even if you just push it up and let gravity do the job of bringing it back down, you're still doing net positive work because the work you do on the return trip is zero (with the force zero). Physically, this makes sense if you consider that the sofa will return to earth with a larger velocity than it started with (implying that you gave it energy).

It is true that the gravitational field does no net work. This is because the gravitational field is a conservative force field, meaning we can define a potential energy at each position and the work done is indendent of path. However, the force imposed by our "pushing" is not conservative by any stretch of the imagination.

It's also true that if you carried the sofa back and brought it to rest that the net work would be zero (since on the return trip your force would have to oppose the direction of motion). Perhaps this is what the OP meant, in which case you're absolutely right.

Last edited: May 20, 2005
6. May 20, 2005

### Justin Lazear

You're right. The work done on the sofa is zero, but the work that you do on the sofa is not. I'm not (and wasn't) in much condition to post anything very sensible, so I'll just shut up until I have more rest. Until then, probably best to ignore me.

--J

7. May 20, 2005

### SpaceTiger

Staff Emeritus
No no, even your work done is zero if the sofa is brought back to rest (since the gravitational field's net work is zero, yours must also be if it is not moving at the end of the trip). Given that assumption, you were right.

The OP didn't specify a final velocity, though, so I wanted to make sure that everyone was on the same page. In retrospect, the stationary interpretation of the question probably was more reasonable.

8. May 20, 2005

### Justin Lazear

Actually, I was thinking more along the lines of pushing the sofa out of the earth's field, letting the moon do a bit of stopping via the crater method, then pushing the sofa out of the moon's field and then letting the earth do the stopping. And then you have done work (No guarantees! All work and no sleep makes Justin a dull boy. Haha, get it? Work?).

So I guess the moral of the story is to be specific in describing your scenario.

--J

9. May 20, 2005

### SpaceTiger

Staff Emeritus
Heh, yeah, that would seem reasonable. I think you're right that if you let the ground dissipate the energy of the impact, your net work would be positive (and the ground's negative to compensate).

10. May 20, 2005

### misogynisticfeminist

ahh alright, maybe the moon was a bad example because there's stuff about the gravitational field. what if i carried the sofa around the world, at constant velocity?

11. May 21, 2005

### Staff: Mentor

Forget about going around the world, just think of carrying the sofa across the room. Ignoring any work done lifting the sofa, are you doing any work on the sofa by just carrying it parallel to the ground?

When you are carrying the sofa, what force must you exert? And what direction are you moving? Are they in the same direction? Are you doing any work on the sofa?

(Another way to think of it: Work is a means of transferring energy. Does the energy of the sofa increase as you carry it?)

12. May 21, 2005

### SpaceTiger

Staff Emeritus
Let's take this one step at a time. First, we'll simplify a bit and say you pushed it the same distance as the circumference of the world (i.e. a straight line instead of a circle) at a constant velocity. If we neglected friction, you would only need:

$$W=\frac{1}{2}mv^2$$

in order to get the sofa up to that velocity. Notice there's no depedence on distance. Why? Well, remember Newton's First Law. An object that's in motion stays in motion unless acted upon by an unbalanced force. All you'd have to do is get it up to that velocity and it would keep going all by itself. The above energy is just the kinetic energy of an object with mass m and velocity v. If it started out with no energy (v=0), then the above must equivalent to the work you did on it (where else would it get the energy).

Next, let's add some friction. For friction on a horizontal surface, the first thing you have to worry about is static friction. There is a nice discussion of that here (jump to mine or Doc Al's post). The basic idea is that it resists your force up until a breaking point, after which the friction gives and you can freely move the object. This means, to get the sofa started, you need:

$$F > \mu_smg$$

where $$\mu_s$$ is the coefficient of static friction. This alone wouldn't change the work required to move it the distance you requested, it would only mean that you'd have to use a larger force to do it. However, there's also kinetic friction. What this does is produce a constant force in opposition to your force once you already have the object moving. The magnitude of this opposing force is given by

$$F_f = \mu_kmg$$

where $$\mu_k$$ is the coefficient of kinetic friction. This means that the total force on the object once you've passed the static friction barrier is

$$F_{tot}=F_0 - F_f$$

where F0 is the force that you are a applying. If we use Newton's second law and say that the object's acceleration is the force divided by its mass, this equation becomes

$$a=\frac{F_0}{m}-\mu_kg$$

Remember that a constant velocity corresponds to zero acceleration, so let's solve for the force we would need to apply to keep it at a constant velocity for the whole trip:

$$F_0=\mu_kmg$$

Notice that this is just the force of kinetic friction. In order to have zero net force on the object, your applied force must equal the resisting force.

So let's calculate the work that comes out of all this, remembering that

$$W=Fd$$

Since static friction is usually stronger than kinetic friction, we're going to have to apply two forces, one to get it up to speed (F1) and another to keep it going against friction (F0). The total energy at the end of the application of F1 is going to be the same, but you'll have to do more work because friction will be working against you all along the way. Thus, the work that you do to get it up to speed is

$$W=F_1(\frac{E_f}{F_{tot}})$$

where the term in parentheses is just the total distance travelled by the sofa in accelerating from 0 to v. This can be expanded to

$$W_1=F_1\frac{\frac{1}{2}mv^2}{F_1-F_f}=\frac{1}{1-\frac{F_f}{F_1}}\frac{1}{2}mv^2$$

Notice that this is the same as before, except with an extra term out front that takes into account the work you do against friction. This is not it, however. We also must do work to keep the sofa going at a constant speed. This is a bit easier and is simply given

$$W_2=F_0d=F_fd$$

where d is the distance you push the sofa. We'll assume that the total distance is much larger than the distance you push it in accelerating to v. So now, finally, we have our expression for total work:

$$W=W_1+W_2=\epsilon E_k+F_fd$$

where

$$F_f=\mu_kmg$$

and

$$E_k=\frac{1}{2}mv^2$$

and

$$\epsilon=\frac{1}{1-\frac{F_f}{F_1}}$$

Notice I've still neglected air drag, circular motion, and complications brought in by carrying the sofa. I had hoped to address those, but this took longer than I thought, so maybe I'll come back to it later. :tongue2:

13. May 23, 2005

### misogynisticfeminist

hey spacetiger, i really liked your explanation, thanks alot.....