My friend and I were arguing about this and I was wondering if someone out there could settle this for us.

Basically, he and I were walking to buy some stamps. When we were on our return trip he made the assertion that when I returned to my desk, which is where our trip began, I would have accomplished zero "net work". That is, utilizing our admittedly simple understanding of work from Wikipedia:

In physics, a force is said to do work when it acts on a body, and there is a displacement of the point of application in the direction of the force.

when I return to my departure point, that is my desk, the total displacement would be zero and using the definition from Wikipedia that $W = Fd$ and $d$ is displacement, no work was done. Intuitively, I suggested that there were two quantities of work done, one quantity of work from my desk to the shop and one quantity of work from the shop to my desk, but when I look at some decriptions of a displacement vector, I feel like the two displacements may also cancel each other out. Can someone help us sort this out?

Thanks in advance for the help!

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    $\begingroup$ Related, possible duplicate of physics.stackexchange.com/q/90947 $\endgroup$ – Brandon Enright Jan 7 '14 at 21:02
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    $\begingroup$ The short answer, which most of the posted answers dance around is: Zero work is done if you return to your starting point AND you are dealing with a conservative field. Gravity is a conservative field, but friction is not. Res ipsa loquitur. $\endgroup$ – Carl Witthoft Jan 8 '14 at 1:32

The work done on an object by a force is the distance traveled by the object multiplied by the magnitude of the force in the direction traveled. So if you move along some closed path (i.e. a path that gets you back to where you started) of length $d$, acted on by a constant magnitude force $F$ that is always parallel to your path and pointing backwards along it (like friction), then the force will have done work $-Fd$, not zero.

If instead your force was a constant in both magnitude and direction, then the force would do zero work over the journey. This is because on the way to the shop, the force would be (say) pointing forwards along your path and so doing positive work, whereas on the way back from the shop the force would be pointing backwards along your path and so doing negative work. These would cancel out.

In this particular example, the force (friction) is of the former kind --- it always points backwards along your path, trying to impede your motion. An example of a force of the latter kind is gravity --- it always point downwards (at least if we're talking about small terrestrial matters). Suppose you threw something vertically up into the air. To begin with the force of gravity would be pointing in the opposite direction to the path taken, doing negative work on the body. Then as the object turned around in the air, the force of gravity (whose direction remains the same) would start to act in the same direction to the path taken. Gravity would then do positive work, precisely equal and opposite to the work it did on the way up, such that when the object returned to its original position, it would have neither gained nor lost energy. The net work done would be zero.

To reiterate: for the case of friction, the force changes direction so as to be perpetually impeding your motion. This is why the work done by the friction builds up continually, rather than cancelling out.

This makes sense: the force of friction is dissipating energy in your legs for the entirety of the journey --- when you get back, you know you've used up some energy, since you might be a bit peckish! The friction hasn't at any point been putting energy back into your legs --- it's just been consistently sapping it --- and so it must have done some work (work being the amount of energy a force puts into an object).

  • $\begingroup$ Thanks for taking the time to respond. Your answer really helped me. In your example of the ball being thrown into the air, you say that the force of gravity does zero net work. That only holds for the force of gravity right? My arm has done work to toss the ball up, is that correct? $\endgroup$ – Purest Jan 8 '14 at 15:00
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    $\begingroup$ In short: yes. The ball was stationary to begin with, and when it left your hand it was moving. The ball has thus gained energy; the forces in your arm have done work on the ball. If you then caught the ball, the forces in your arm would do negative work on the ball in order to slow it down; the net work done by the forces in your arm would be zero. This can be appreciated since the ball started at rest and finished at rest; no net energy was transferred. Caveat: your arm is inefficient, and so in reality the forces in it will do a lot of work, wasted as heat. Result: playing catch is tiring. $\endgroup$ – gj255 Jan 8 '14 at 17:38

The problem is due to unclear meaning of the sentence

I would have accomplished zero "net work".

In a colloquial sense, you did not do any useful work, so someone, perhaps your boss, would evaluate your trip as "you did no work". But in the sense word "WORK" is used in physics, the answer may be different. This is because the sentence "Body A did non-zero work on body B" in physics means that the body $B$ moved while $A$ exerted force on it which had non-zero component in the direction of motion of $B$.

Now, if we take your body as the object $A$ and the ground as the object $B$, there is indeed work done, because the ground moves to your back while you're pushing it there with your feet (you have to push to overcome air resistance). So in this sense, you did work on the ground, and your friend was mistaken.

The total work being zero when object returns to its original position is a situation that occurs when the work is taken as special work due to one special force, for example the gravity force. Then, because for the gravity force $\mathbf F_G$ the integral over every closed path $\gamma$

$$ \oint_\gamma \mathbf F_G \cdot d\mathbf s = 0, $$

the gravity force did zero net work.


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