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About this Lesson
- Type: Video Tutorial
- Length: 11:32
- Media: Video/mp4
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- Posted: 07/01/2009
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This lesson was selected from a broader, comprehensive course, Physics I. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/physics. The full course covers kinematics, dynamics, energy, momentum, the physics of extended objects, gravity, fluids, relativity, oscillatory motion, waves, and more. The course features two renowned professors: Steven Pollock, an associate professor of Physics at he University of Colorado at Boulder and Ephraim Fischbach, a professor of physics at Purdue University.
Steven Pollock earned a Bachelor of Science in physics from the Massachusetts Institute of Technology and a Ph.D. from Stanford University. Prof. Pollock wears two research hats: he studies theoretical nuclear physics, and does physics education research. Currently, his research activities focus on questions of replication and sustainability of reformed teaching techniques in (very) large introductory courses. He received an Alfred P. Sloan Research Fellowship in 1994 and a Boulder Faculty Assembly (CU campus-wide) Teaching Excellence Award in 1998. He is the author of two Teaching Company video courses: “Particle Physics for Non-Physicists: a Tour of the Microcosmos” and “The Great Ideas of Classical Physics”. Prof. Pollock regularly gives public presentations in which he brings physics alive at conferences, seminars, colloquia, and for community audiences.
Ephraim Fischbach earned a B.A. in physics from Columbia University and a Ph.D. from the University of Pennsylvania. In Thinkwell Physics I, he delivers the "Physics in Action" video lectures and demonstrates numerous laboratory techniques and real-world applications. As part of his mission to encourage an interest in physics wherever he goes, Prof. Fischbach coordinates Physics on the Road, an Outreach/Funfest program. He is the author or coauthor of more than 180 publications including a recent book, “The Search for Non-Newtonian Gravity”, and was made a Fellow of the American Physical Society in 2001. He also serves as a referee for a number of journals including “Physical Review” and “Physical Review Letters”.
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We've defined a new quantity in physics, work. Work is force times distance. In one dimension it's simple to compute. It contains some physical information. What happens if we move to two dimensions? We have the formula work is force times distance in the x direction. If you want to move to two dimensions, the story is really essentially the same. What I really want, if you look at this, it kind of looks like the force vector dotted with the displacement vector. Remember what the dot product does. It tells you force magnitude times delta r magnitude times the cosine of the angle between them.
Let me draw you a picture. Here's a little sketch. Here's an object. It starts here and it slides across this black path. And the force vector, let's assume that it is momentarily--well it's the whole time a constant. So here's the motion. I've got an object that's sliding horizontally, but I'm pulling up at some angle. And I'm pulling with a constant force. This force vector is staying in the same direction and the same magnitude. And what do I want when I think about work? I want the product of force times displacement. But I don't care about the up piece of the force. It's the piece of the force in the direction of motion that seems to me to be the part that is what I'm calling work. That would be F cosine theta times displacement, delta r. And that's exactly what our formula said. Force cosine theta times delta r. So the formula is exactly what seems reasonable, and after all, this is our definition of work. We'll see whether it turns out to be a useful quantity. It certainly is.
In one dimension, this is the same formula. Force dotted with delta r is just F[x] delta x. It's a dot product. When you have a force vector and you have a displacement vector, and you dot them, each is a number, and it's got a sign. It's the magnitude of this arrow times the magnitude of that arrow. That's always going to be positive, times the cosine of the angle between them. So if the angle is small, it's going to be positive, which makes sense. The force is essentially lining up with the displacement. You are doing positive work. If the force is exactly perpendicular, no work is being done. Force times displacement times the cosine of 90 degrees is zero. It doesn't make sense physically.
Imagine that you had an air track. And you put a cart on it, and you slid it off to the side. And now you just watch. This little cart is cruising sideways, and of course there are vertical forces. There is gravity pulling it down. And there is a normal force pushing it up. There are just no forces in the horizontal direction. So you ask yourself is the normal force, the force of the air track on the little object, doing any work? And I think a reasonable answer is no. If that thing extended forever, it would just happily continue to cruise along by Newton's first law. And it seems reasonable that the air track is not doing any work. It's just sitting there, and this thing is cruising along. No work being done when the normal force, or any force, is perpendicular to the displacement.
How about when the force is at an angle that is larger than 90 degrees? The cosine of angles larger than 90 is negative. This is a negative work being done. This is the extreme case. We talked about this in one dimension. The object is moving this way. The force is slowing it down, or trying to, and it's doing negative work. You might think of something skidding to a halt, and friction is doing negative work on it. So the sign makes some physical sense. And let's look at some examples.
In fact, let's start off with this example. I'm tugging this thing along, and let's draw a force diagram. Before we can really calculate the work, we need to know what are all the forces acting on it. I know that it's moving in the horizontal direction. And I know that there's a tension in the string, which is pulling it up and right. I'm not going to draw an arrow on this diagram, which represents its motion, because this is a force diagram. I just remember that it's moving to the right. There is mg down. It's sitting on the ground, so the ground will exert a normal force up on it. Then there will be a friction, and let's just think about something which is being scraped for simplicity. So I've got this string, and I am pulling the string at an angle, so I really don't want to have to worry about the issue of wheels. So let me just drag it along the ground so that this is kinetic friction. And that's all of the physical forces.
And now I can ask various questions like what is the work done by the string on the object? So all you need to know is the force of the string times delta r, or in this case it's moving in the x direction, delta x times the cosine of the angle between them. I think I can safely draw this angle into this force diagram. So the work done by the string is going to equal to T times the displacement, how ever far I drag it, times the cosine of theta. And can I figure out what T is? Well I might just measure it. If I was dragging something along with constant speed, I could actually work it out. If it's moving with constant speed, it's not accelerating. So the net force in the x direction would have to be zero. So T cos theta would have to equal to force of kinetic friction. And remember kinetic friction is just mu[K]N. So you think I'm done. , And I can solve for tension. You've got to be careful. What's N? Is N equal to mg in this picture? Warning, it's not. People get used to setting N equal to mg when something is moving flat, but look at Newton's law in the vertical direction. I've got this up arrow, and I've got this up component. So N plus T sine theta is equal to mg. So I really have two equations and two unknowns, if I want to solve for T and N simultaneously. And then I know everything and I can figure out the work done by the string on the car. The work done by friction on the car would be . The work done by the normal force would be zero, because it's at 90 degrees, gravity does no work. So it's straightforward, but requires a little thought, and sometimes a force diagram, to really get everything right.
Let's do one more example. It's a nice example, very useful in a lot of problems. I'm sliding a block up a hill. The hill has an angle theta. Let me push it. So this is my force. It's an external force like me pushing this thing straight up the hill. And are there any other forces acting on it? Well there is going to be a normal force, and there's going to be a weight force, which is of course mg. And let's assume no friction for now. I can accelerate this thing up the hill, but I can also pick F external so that the some of all these three forces is zero. I can always choose to slide this object up the hill with constant speed. It's my choice. If I do that, you could ask how much work am I doing? What's the work done by the external force? So the work done by the external force is going to be the following. It's F external times the distance traveled. So let's suppose that this thing is a grand total length of L. So it's times L times the cosine of the angle between the external force and the displacement. Well think about that for a second. The displacement is now in two dimensions. It's up at this angle. But so is my force. So the angle between them is zero. So that's it. That's the answer, F external times L. I could work out F external by looking at this force diagram. It's the usual story of an incline plane. If I know that there is no acceleration, then I know that the sum of the forces in the x direction is equal to zero, and the sum of the forces in the y direction is equal to zero, two equations and two unknowns. So I can certainly figure out F external and the normal force. And then I would have the work done by the external force.
Let's look at the work done by gravity in this problem. I've got a gravity force, which is down, and this is not a force diagram. I'm just drawing a sketch to remind myself that I had a displacement vector of length L. So what I want is mg, that's the magnitude of the force of gravity, times L times the cosine of the angle between them. That's this angle. What is that angle? This was theta, and this is 90 degrees. So it's the cosine of 90 plus theta. So you have to use a trig identity. This is cos theta minus sine 90 sine theta. So that's minus sine theta. It's . The work done by gravity is negative. Just like when I lifted something up with constant speed gravity was doing negative work, when I push something up a hill with constant speed gravity is doing negative work. And in fact, if you look at the picture, L sine theta is just the height of the hill. So it's just negative mgh. It's the same old formula as I got when I lifted something up a height h. It turns out that the work done by gravity is always negative mgh, if you lift it up a height h, no matter what path you take. It could be a diagonal line, or a twisty line. I haven't proven that, but it's a true statement.
In two dimensions, calculating work is a little bit trickier, because you really have to figure out the direction of the force and the direction of the displacement. They're not always the same. There might be lots of forces in the problem. It might have had some initial velocity. You've just got to look at the problem, and figure out force times displacement times the cosine of the angle between them. To figure out the force sometimes you need to write down and solve Newton's second law. Sometimes they'll be given to you.
I still haven't told you why we really want to know this number. We've got some intuition that work is certainly a physically meaningful quantity. It's telling you something about the force and the motion together. And we're getting there. We still need to work out a few more technical details. How do you compute work? And then we'll start to use it.
The Work Done by a Constant Force in Two Dimensions Page [2 of 2]
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