The Work–Energy Theorem

Transcript

0:00Introduction

Hello everyone! One small puzzle first. You throw a one kilogram stone straight up at twenty metres per second. Because of air resistance, it rises only fifteen metres. Now tell me: how much work did the air do on the stone? Hmm. Nobody told you the force of the air. Looks impossible, right? Yet by the end of this lesson, you will solve it in a few short lines. The tool is the work energy theorem, and it comes up often in JEE problems. So, let's begin.

0:34The work–energy theorem

So, here is the big idea. Forces do work on a particle, and its kinetic energy changes. This theorem connects the two: W net equals K f minus K i. W, with the small word net below it, is the net work. K with a small f is the final kinetic energy, and K with a small i is the initial kinetic energy. Okay? Now, read W net very carefully. It is the work done by all the forces acting on the particle. Not one force: all of them, added up. Remember this word, all. Right? Now think. Suppose the net work comes out negative. What happens to the kinetic energy? It decreases! Yes, exactly! The right side, K f minus K i, is the change in kinetic energy. Negative net work makes K f less than K i, so the particle slows down. Positive net work, and it speeds up. Clear?

1:31Why it is true: the proof

Now, why is this true? The proof is short. Take motion along the x axis, and start with Newton's second law. F, the net force, equals m times d v over d t, the rate of change of velocity. Okay? But work needs a distance: d x, a tiny step along x. So how can we bring d x into this equation? The chain rule! d v over d t equals d v over d x, times d x over d t. And d x over d t is just the velocity, v. So F equals m times v times d v over d x. With me? Now multiply both sides by d x and integrate; the tall S is the integral sign. On the left: the integral of F d x, from x one to x two, the start and the end, with small subscripts. That is exactly the work done, right? On the right: the integral of m v d v, from v one to v two, the speeds at those two points. And the integral of v d v is v squared over two. So we get half m v two squared, minus half m v one squared: the final kinetic energy minus the initial one. That's it, the theorem is proved! Got it?

2:48When does the theorem hold?

Now, does this theorem need a special kind of force? What about friction? Or a force that keeps changing? It works for all of them! Look back at the proof: we never asked what kind of force F was. So it holds for conservative forces, like gravity, and for non-conservative forces, like friction and air resistance. Okay? And it holds for constant forces and for variable forces. We integrated, so the force was free to change along the way. Right? Now, one condition, and it is about the frame. The theorem holds in any inertial frame, a frame that is not accelerating. In an accelerating frame, you must include the work of the pseudo-force as well. Clear? And for a system of particles, include the work of the internal forces too. Careful: don't throw them away just because they are internal! Got it?

3:42Topper's trick: when to use it

Now a topper's trick. Suppose a question gives you forces and distances, and does not ask for time. Which method should you pick? The work energy theorem, exactly! Work needs only forces and distances, and kinetic energy needs only speeds. Time never appears. Right? But when time is given, or time is asked, switch tools. Use impulse and momentum, or the equations of motion. With me?

4:11Example 1: average braking force on a car

Now, let's use it: a one thousand kilogram car, moving at seventy two kilometres per hour, is braked to rest in forty metres. On screen, k g is kilograms, k m h to the minus one is kilometres per hour, and m is metres. Find the average braking force. Forces and distances, no time: you know which tool! Pause the video, and try it first. Done? Okay, let's check. The speed is seventy two, so we put v equals seventy two in half m v squared, right? No, no, wait! Seventy two is in kilometres per hour. First change it to metres per second: multiply by five over eighteen. So v i, the initial speed, is twenty metres per second. Okay? Now the kinetic energies. K i equals one half times one thousand times twenty squared: half of one thousand times four hundred, that is two times ten to the power five joules, capital J for joule. And the car comes to rest, so K f is zero. Right? Now the work. The braking force stops the car; it acts against the motion. So its work is negative: minus F times forty. And gravity and the normal force? They are vertical, and the car moves horizontally, so their work is zero. Clear? Put it into the theorem: minus F times forty equals zero minus two times ten to the power five. The minus signs on both sides cancel. Got it? So, F equals two times ten to the power five, over forty: five thousand newtons, with the capital N on screen for newton. See? No time needed, just energy. That's it!

6:00Example 2: work done by air on a stone thrown up

Now, remember our puzzle from the start? Let's solve it. So: a one kilogram stone is thrown straight up at twenty metres per second. Because of air resistance, it rises only fifteen metres. Find the work done by the air. Take g, the acceleration due to gravity, as ten. Pause and try. Hint: count every force. Done? Okay, let's check. The stone loses all its kinetic energy on the way up. So the air did minus two hundred joules of work, right? No, no, careful! This is the classic trap. W net means the work of all the forces. While the stone rises, two forces act on it: gravity and the air. So W g, W with a small g for gravity, plus W air, with the small word air, equals K f minus K i. Okay? Now the energies. K i equals one half times one times twenty squared: two hundred joules. And at the top the stone stops for a moment, so K f is zero. Right? Next, gravity. It pulls down, but the stone moves up. So is the work done by gravity positive, or negative? Negative, correct! The force and the motion are opposite. W g equals minus m g h, where m is the mass and h is the height risen: minus one times ten times fifteen, that is minus one hundred and fifty joules. Clear? Now put everything into the theorem. Minus one hundred and fifty, plus W air, equals zero minus two hundred. With me? So W air equals minus two hundred plus one hundred and fifty: minus fifty joules. Negative, because the air opposes the motion. Got it? And look, puzzle solved! We never knew the force of the air, and it did not matter. The theorem works for any force, if you count them all.

8:00Quick revision: net work, when it holds, when to use

So, quick revision. W net equals K f minus K i, and W net is the work of all the forces, okay? It holds for any forces, constant or variable; in an accelerating frame, add the work of the pseudo-force. And when forces and distances are given and time is not asked, use the work energy theorem. Now go and try the practice problems on this page. All the best!