Khan!
December 6, 2012 at 8:36 PM by Dr. Drang
I’d like to like Khan Academy. I like its goals. I like how its website works (not surprising, given that John Resig works there). And I like how it makes its videos available in different ways: streaming on its site, streaming on YouTube, and downloadable for playback anytime. But every time I’ve watched a Khan Academy video on a topic I know something about, I’ve been disappointed.
It’s not the roughhewn nature of the videos—that makes them more personal, more accessible. No, it’s that they’re wrong. Sometimes they’re just factually wrong, with minor mistakes of the kind that are easy to make when teaching live but which shouldn’t be left in a permanent record. More often, though, they’re pedagogically wrong, taking an approach that won’t serve the student well.
Here’s the video on the behavior of springs and Hooke’s Law, a topic I’ve been dealing with as a student, a teacher, and a practitioner for about 35 years. Certainly, it’s worth 10 minutes of your time. Give it a look, and then we’ll discuss it.
Let’s start with the numerical examples. They are bizarre and don’t come close to describing the kinds of springs the viewer is likely to have experience with. A 10meter compression under a 5newton load? It may not seem weird if you don’t have a good feeling for what a newton is, but 5 newtons is the weight of roughly half a kilogram,^{1} or a bit more than a pound. What kind of spring deflects 33 feet under a onepound load?
You might think this doesn’t make any difference, that the numbers are just there for illustration, not meant to represent any real spring. But students tend to learn better when examples are concrete and realistic. The deflection in the example could just as easily be 10 millimeters, and then you’d have a spring that makes sense.
And while we’re talking about units, you must have noticed that the video commits the cardinal sin of physics and engineering problems: it gives answers without units. Is the spring constant in the example ½, as the narrator says and writes? No. The spring constant is ½ N/m. You might say I’m being picky here, and you’d be right, but I’m not being any pickier than every science and engineering instructor I’ve ever run across. If you’re a student—and I have to believe many of Khan Academy’s users are students looking for a little help—this video is going to give you the impression that units aren’t that important. Tell that to the designers of the Mars Climate Orbiter.
And then there’s the sign convention. If you ask any engineer to write down Hooke’s Law, they’ll write
\[F = kx\]not
\[F =  kx\]I recognize, and appreciate, that the narrator wants to emphasize that the \(F\) in the equation is the restoring force of the spring rather than the applied force, but that minus sign isn’t the right way to do it. All it does is add something to the equation that’s easy to get wrong. How easy? In the video’s second example, which starts at about the 8:30 mark, the narrator himself forgets his sign convention and ends up calculating a spring constant of 2, an absurd answer (and not just because he forgot the units).
I’m not blaming the narrator for messing up his signs.^{2} I’ve made countless sign errors; it’s the easiest kind of mistake to make. Which is why you shouldn’t make life more difficult by forcing yourself to keep track of more signs than is absolutely necessary.
Here’s why Hooke’s Law shouldn’t be written with a minus sign. There is no single restoring force—which is better referred to as the internal force—in a spring. In the video examples, the spring is not just exerting a force at the right end, it’s also exerting a force on the wall at the left end. And that force is in the opposite direction of the one at the right end. How does the minus sign handle that situation?
In fact, if we go by what the video says, the force at the wall is a mystery. That end of the spring doesn’t move at all, \(x = 0\), and yet we know there’s a force there—it’s necessary for equilibrium.
Which leads to my main criticism: Hooke’s law should not be taught as a vector equation. The \(F\) and the \(x\) in the equation should be treated as signed scalars, not as vectors. The \(x\) represents the change in length of the spring from its natural (unloaded) length, with positive values for stretching and negative values for shortening. The \(F\) represents the internal force in the spring, with positive values for tension and negative values for compression. With these definitions, the equation \(F = kx\) doesn’t need a negative sign.
Defining and writing Hooke’s Law this way makes it easy to apply in every case. It doesn’t matter which end of the spring moves, or if both ends move, the force in the spring is determined entirely by its change in length. And the directions of the forces exerted by the spring on whatever it’s touching are simple: a spring in compression pushes out; a spring in tension pulls back in. Apply those ideas to the examples in the video and you’ll see how straightforward these problems can be.
More important, this view of Hooke’s Law is the one that will serve the student better when she moves on from elementary problems. The way it’s taught in the video will have to be unlearned at some point in the future.
As I said at the top, I don’t want to be mean to Khan Academy. It’s heart is in the right place. But I’m not convinced poor teaching is better than no teaching.

“The weight of half a kilogram” is an awkward construction, but I’m trying to be colloquial and scientific at the same time. The kilogram is commonly thought of as a unit of weight, which is a force, but it is, strictly speaking, a unit of mass. ↩

I am, however, blaming him for not immediately realizing that he’d made a mistake. Spring constants aren’t negative. ↩