[MUSIC] This week, we introduce work and energy. And also, power, which is the rate of doing work. These words, work, energy, power are used to mean many different things in everyday language, but sometimes they are used in the physical sense. [SOUND] We can both tell that I'm doing work here. Work done is proportional to the force that I apply. It's also proportional to the distance over which I apply the force. The trolley is supplying the force now, the normal force, but it's not doing work, because it hasn't moved. The angle is also important. No work's being done on the bags here, because the force and the displacement are at right angles. Here's a definition of work. If constant force F is applied over a displacement S with an angle theta between them, the work done, W, equals F S cos theta. You can think of this in two different ways. F times S cos theta, in other words, F times the component of S in the direction of F. Or, you could think of it as S times F cos theta. S times the component of F in the direction of S. This is what we call our scalar product of vectors. Although F and S are vectors, work is a scalar. It doesn't have a direction. Its units are joules. A newton times a meter is a joule. You know how big a meter is. And a Newton is a fairly small force to the weight of this mandarin. So a joule, a meter times a newton, is not much work. More about that in a moment. But first, something that requires care. We use the letter W for both work and weight. Usually we can tell them apart easily because weight is a vector and its units are newtons. Work is a scalar, and it's units are joules. Let's answer a question for practice. W equals F S cos theta. There're two easy special cases to remember. When theta is 0, work equals F S, and when theta equals 90 degrees, work is 0. Let's apply those cases, approximately, to my little movie. The weight of each of these bags is about 200 newtons down. So, because their acceleration is small, I must be applying a force of roughly 200 newtons up. In the vertical part of the lift, theta is approximately 0, so work equals F S equals MGS. So, estimating the distances, we can now be quantitative. One bag at 200 Newtons, 0.7 meters, so about 140 joules of work. Then another bag over a longer distance, more work. Two bags and a bigger distance, more and more work. But this shows it's possible to do hundreds of drills of work in a second or two, though you'd get tired pretty quickly doing that. Now remember that cos theta term. Note what happens when the trolley moves. The force and the displacement are at right angles. [LAUGH] because 90 degrees is 0. The trolley does no work at all. Here, I apply a force F to slide the mass a distance s. The force and the displacement are parallel. Figure is 0 so cos theta is 1. I do positive work Fs. But what about the frictional force? It's in the opposite direction to my applied force, so the angle between the frictional force and the displacement is 180 degrees. Cos theta equals minus one, friction does negative work. What does it mean to say friction does negative work? Well it means I have to do work against friction. Friction takes work and turns it into heat. Try this experiment. [SOUND] So, remember to think about that cause theta term when calculating work. Have a go now Question, is it easy to lift 20 kilograms up one meter, or to lift 2000 up a centimeter? And which of the two requires more work? Let's go back to this experiment. My weight is 700 newtons, but, because three ropes are pulling me upwards. The tension in each one is only about one-third that. I only need to pull with 230 newtons. Compare how far the rope moves, with the distance I go up. Because three ropes are shortening, I have to pull three times further. One third of the force, but three times the displacement. I do the same work as if I had climbed with my hands and no pulleys, but I assure you, it's a lot easier to pull 230 newtons over six meters than 700 newtons over two meters. Pulley systems and other devices that vary the applied force and the distance, are called simple machines. We see a similar effect with the lever. Here I can lift a mass with a force much smaller than its weight. But I have to apply that force over a larger distance, to do the same amount of work. Big force small distance, small force big distance. Or, using an inclined plane, smaller force, big distance. Bigger force, small distance. All familiar ideas but to make sure, let's do a quiz.
