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- Thread starter misogynisticfeminist
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--J

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Not displacement but the distance in the same direction travelled.

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

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SpaceTiger

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Hmm, I'm not sure I agree with this. Work, in general, is given by:Justin Lazear said: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.

[tex]W=\int \vec{F}\bullet d\vec{s}[/tex]

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

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.

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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.SpaceTiger said:Hmm, I'm not sure I agree with this.

--J

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SpaceTiger

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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.Justin Lazear said:You're right. The work done on the sofa is zero, but the work that you do on the sofa is not.

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.

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

--J

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SpaceTiger

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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).Justin Lazear said:So I guess the moral of the story is to be specific in describing your scenario.

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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?Justin Lazear said:So I guess the moral of the story is to be specific in describing your scenario.

--J

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Doc Al

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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?misogynisticfeminist said:what if i carried the sofa around the world, at constant velocity?

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?)

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SpaceTiger

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Let's take this one step at a time. First, we'll simplify a bit and say youmisogynisticfeminist said: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?

[tex]W=\frac{1}{2}mv^2[/tex]

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

[tex]F > \mu_smg[/tex]

where [tex]\mu_s[/tex] 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

[tex]F_f = \mu_kmg[/tex]

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

[tex]F_{tot}=F_0 - F_f[/tex]

where F

[tex]a=\frac{F_0}{m}-\mu_kg[/tex]

Remember that a

[tex]F_0=\mu_kmg[/tex]

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

[tex]W=Fd[/tex]

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 (F

[tex]W=F_1(\frac{E_f}{F_{tot}})[/tex]

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

[tex]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[/tex]

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

[tex]W_2=F_0d=F_fd[/tex]

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:

[tex]W=W_1+W_2=\epsilon E_k+F_fd[/tex]

where

[tex]F_f=\mu_kmg[/tex]

and

[tex]E_k=\frac{1}{2}mv^2[/tex]

and

[tex]\epsilon=\frac{1}{1-\frac{F_f}{F_1}}[/tex]

Notice I've still neglected air drag, circular motion, and complications brought in by

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

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