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I am now teaching Calculus of several variables this semester. In apllications of integrals, the problem of finding the work done in moving an object under a force $F$ is one of the most common problems.

Let's assume that an object is moving from a point A to a point B (on a horizontal line) under a constant force $F$ pointing an angle $\alpha$ (w.r.t. the horizontal line). Then in most textbooks (Calculus), the work done in this case is defined by $W=|F||AB|\cos(\alpha)$. By accepting this definition, then one can meaningfully find a way to evaluate the work done in moving an object along a curve (in 2D or 3D) and under a non-constant force (this leads to line integrals).

When teaching this, the most uncomfortable fact is "why do we have such a definition of the work" in the first place. Since the word "work" should say something, it might give confusion.

My question: how do we give intuition to the definition of work above? For example, how do we give an explanation for the fact that the work is negative if $\alpha$ is greater than 90 degrees?

Thanks so much for any hints.

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    $\begingroup$ This is a standard definition of work: it characterizes a force applied to a body, causing movement. It is proportional to force and displacement. Definitions are something people come up with, this should not be "the most uncomfortable fact". With the angle between 90 and 270 degrees the force pushes the body against the chosen direction, thus its projection onto the chosen direction is negative. This can be easily seen from the unit circle. If one moves a body forward, then moves it backward, total work is zero, agreeing with the amount of mechanical energy of the system, it has not changed. $\endgroup$ – Rusty Core Jan 14 at 22:06
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    $\begingroup$ I think it may be useful to relate work to potential energy. (i.e. in a conservative system) With that in mind, you can see that this formula comes from integrating a gradient. The negative values come from moving against the gradient. See en.wikipedia.org/wiki/… $\endgroup$ – Adam Jan 15 at 0:23
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The best way to illustrate the sense of $W=|F||AB|\cos(\alpha)$ is to show that it makes sense that each of the factors is proportional to the work done.

  • If you double the amount of force to an object to move it a certain distance, it doubles the amount of work needed.

  • If you double the distance to move an object by applying a certain force, it doubles the amount of work needed.

  • If you apply the force in the same direction as the direction the object moves, you are being completely efficient in your work. The more your work is applied downward, the more effort you are wasting pushing the object into the ground instead of in the direction you want. At the end, if you are pointing straight down, you are doing no work no matter how much force you expend. And you if your force is in the opposite direction, you are doing "negative work" in the sense of moving the object further from your goal. These intuitive facts turn out to be modeled precisely by the cosine of the angle between the direction of your work and the direction the object is to move.

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  • $\begingroup$ This isn't really right physically. Work is not need in order to "move it a certain distance." An object in motion remains in motion even if you don't exert a force. No force is required in order to move an object. A force is required in order to change an object's state of motion. $\endgroup$ – Ben Crowell Jan 15 at 1:58
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I mainly teach physics, only a little math on the side. If I was teaching this math topic and wanted to spend 3 minutes giving the appropriate physical motivation, I would do something like the following.

There is something called energy. It comes in various forms. Food has energy, which is what we're talking about when we talk about calories. Heat, sound, and light all have energy. When a material object like this bowling ball is moving, then it has energy because of its motion.

Now I have this bowling ball, which I borrowed from the physics department's stockroom, and I'm going to roll it down the aisle of the classroom. Please pick up your legs and backpacks in case I don't do it perfectly straight. The ball started from rest, and then I made a force on it, which transferred energy into it. The amount of this energy transfer is called work. When the force is constant, we have $W=Fd$.

Now, Sally, the ball is near your desk. Could you please roll it back up the aisle toward me at a speed similar to the one I used? Now I slow the ball to a stop with my foot. This time the work I did was negative: I took energy out of the ball. This makes sense because if we pick some direction to call positive, then the force and the distance will have opposite signs, and their product will be negative.

Now when the force is not constant, we can't just multiply the distance by "the" force. This is a pattern that you'll see over and over again in applications of calculus. When we need to generalize a relationship involving multiplication to the case where one of the factors is varying, we do that using an integral.

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  • $\begingroup$ [1] I am not sure that "something called energy" is better than "this is the definition", or that "I made a force on it, which transferred energy into it. The amount of this energy transfer is called work" is better than simply stating that we are going to define work such and such, and it makes sense because it is proportional to force and to displacement. Richard Feynman wrote in his memoirs, "There was a book that started out with four pictures: first there was a wind-up toy; then there was an automobile; then there was a boy riding a bicycle; then there was something else. $\endgroup$ – Rusty Core Jan 15 at 6:20
  • $\begingroup$ [2] Quotation cont'd: "And underneath each picture it said, "What makes it go?" I thought, "I know what it is: They're going to talk about mechanics, how the springs work inside the toy; about chemistry, how the engine of the automobile works; and biology, about how the muscles work." I turned the page. For everything the answer was "Energy makes it go." Now that doesn't mean anything. Suppose it's "Wakalixes." That's the general principle: "Wakalixes makes it go." There's no knowledge coming in. The child doesn't learn anything; it's just a word!" $\endgroup$ – Rusty Core Jan 15 at 6:20
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I would start with the scalar. Work = force times distance. Starting with the scalar makes things easier, more intuitive. (Not a math logic point, just a psychological one.) You can then easily generalize, perhaps without showing a derivation and say "well this is what work is in these trickier situations (vectors, non constant force, more dimensions, etc.) At least they see the similarity of the look of the equation to what they know from high school basic physics.

As far as why W=F*d, you can devolve to "this is the definition". Or probably make some argument of how work is just another name for energy. (The simplest and most intuitive example of work done is moving an object up vertically in a gravity field. The work done is equal to the potential energy created.)

All that said, I would really try to avoid getting too tied up in the physics of the thing, especially when you are a little uncomfortable or unfamiliar with it yourself. I suspect a lot of what bothers you, here, is your own questions, not those of the students. But you are going to transfer that to them and bog the class down with a bunch of physics talk when they really ought to be practicing calculations. (This is frequent the commenters/questioners here want to explain to the class something that really bugs them from not being explained versus something that bugs the class from the omission.)

Remember, they will have other classes (most will) that go over this from more of a physics background and less of a math one. For a student taking both (at or near same time) it can be useful to see the same concept from two slants. One more physical and one more calculational. The physics class makes sense when you say "I've seen that math before, can concentrate on physical meaning of what is happening" (and conversely in math class). But I would try to keep math class as more calcuational versus deep digressions to physics. This supports the physics class better.

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  • $\begingroup$ If school students took physics in high school they would not need to get "comfortable" with the definition of work. Just sayin'. This site says that in 2005 only one quarter of high schoolers took physics. There is a version called conceptual physics, and of those who took physics at all one third took this non-math version, and this is not an elementary school course! $\endgroup$ – Rusty Core Jan 14 at 23:47
  • $\begingroup$ Yeah...you really need to take physics if you are going into pure math. Even if you are the purest of the purity, pure, pure. Because sooner or later you will end up teaching calculus or diffyQs. And they have a pretty intimate connection to physics. Going back to this guy called Newton. And the reason why math departments are not as sparse for funds as classics or philosophy departments is because of service course teaching requirements. Mostly to support hordes of engineers. $\endgroup$ – guest Jan 15 at 0:59
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    $\begingroup$ "This is the definition" is not an explanation. $\endgroup$ – Ben Crowell Jan 15 at 1:56
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    $\begingroup$ It's good enough for a math class, when you don't need to also be teaching a physics class and are just doing different applications. $\endgroup$ – guest Jan 15 at 5:32
  • $\begingroup$ And I think it actually is definitional. $\endgroup$ – guest Jan 15 at 5:34

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