Using the Theorem with Position, Velocity and Time Data

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Some force acts on a particle for three seconds, and at the end, the particle is moving at two metres per second. So, did the force do positive work on it, or negative work? Positive, most of you will say. It is moving, after all, right? Hmm, but how fast was it moving at the start? In this lesson, you will meet a particle exactly like this, with negative work done on it! Problems that give the motion as a formula come up often in JEE, so let's begin.

0:35Recipe 1: position given as a function of time

So, first, a quick recall: the work energy theorem. W net, that is W with the small word net below it, is the work done by all the forces together. It equals delta K, the change in kinetic energy; delta is the little triangle. And delta K is K f minus K i, where the small f and i mean final and initial. Okay? Now, suppose the question gives the position x as a formula in time. We write it as x of t. To find the work, do we need to find the force first? No, not at all! The theorem needs only kinetic energies, so all we need are the speeds. Differentiate x with respect to time: v equals d x over d t. Put in the two instants, and you have both speeds. Right? Then W equals half m v f squared, minus half m v i squared, with v f and v i the final and initial speeds. Two speeds and one subtraction. That is the whole recipe. Clear?

1:38Recipe 2: velocity given as a function of position

Next, suppose the velocity is given as a function of position, v of x. This is even easier. Put the two positions into v of x, and you get the kinetic energy, half m v squared, at each one directly. No time needed at all, okay? But what if they also ask for the force? Force is mass times acceleration, and acceleration is d v over d t. But here v is given in terms of x, not t. So how do we get the acceleration? The chain rule, yes! d v over d t equals d v over d x, times d x over d t; the dot means times. And d x over d t is just v. So the acceleration is v times d v over d x. Got it? Multiply by m, and F equals m v, d v over d x. Now look closely: m v d v over d x is exactly the derivative of half m v squared with respect to x, the bracket on screen. So the force is d K over d x, the rate at which kinetic energy changes with position. With me?

2:47Recipes 3 and 4: force or energy given in time

Now the third case: the force is given as a function of time, F of t, and the body starts from rest. Work is the integral of F d x. But this force depends on t, not on x. So, can we integrate F d x directly? Not easily, no! So we take a different road. Integrate the force over time instead. The integral of F d t is the impulse; the tall S shape is the integral sign. Since the body starts from rest, the impulse is exactly its momentum p, okay? Then K equals p squared over two m, the link from our kinetic energy lesson. It started with zero kinetic energy, so the work done is just K minus zero. Momentum first, then kinetic energy. Clear? And the last case: the kinetic energy is given as a function of time, K of t. The rate at which kinetic energy grows is the power of the net force, capital P, okay? So P equals d K over d t. And power is force times speed, so the force is P over v. Differentiate K to get P, then take P over v. Done!

3:59Example 1: work done on a body moving as x = 2t^2

Right? Alright, time for Example one. Here is a three kilogram body, k g on screen, moving as x equals two t squared metres; the small m is metres. Find the work done on it in the first three seconds, small s for seconds. Pause the video, and try it first. Done? Okay, let's check. Position is given in time, so which recipe? The first one! v equals d x over d t, and the derivative of two t squared is four t. Simple, right? The first three seconds means from t equals zero to t equals three. So, what is the speed at t equals zero? Zero. Correct! Four times zero is zero, so this body really does start from rest. See, we checked it; we did not just assume it. At t equals three seconds, v equals four times three, twelve metres per second. The m s to the minus one on screen means metres per second, okay? So W equals half times three times twelve squared, minus zero. Twelve squared is one hundred and forty four; times three is four hundred and thirty two; and half of that is two hundred and sixteen. Two hundred and sixteen joules, capital J for joule. That's it!

5:21Example 2: negative work when x = (t - 2)^2

Now a slightly harder one, and it settles our puzzle. Here is a one kilogram particle moving as x equals, t minus two, whole squared, metres. Find the work done on it in the first three seconds. Pause and try. Hint: look very carefully at the start. Done? Okay, let's check. Position in time again, so recipe one. Differentiate: v equals two times, t minus two. With me? At t equals three, v equals two times one, that is two metres per second. It starts from rest, so W equals half times one times two squared: two joules, right? No, no, careful! Who said it starts from rest? Put t equals zero: v equals two times minus two, that is minus four metres per second! At the start it is already moving at four metres per second, in the negative x direction. And at three seconds, v is plus two metres per second. Got it? Now, the kinetic energy at the start. Does that minus sign change anything? Not at all! Kinetic energy has v squared, and minus four, squared, is sixteen. So W equals half times one times, two squared minus the square of minus four, inside the big brackets. Half of four minus sixteen: minus six joules. Negative work! And there is our puzzle, solved: the particle ends at two metres per second, but it started faster, at four. It lost kinetic energy, so the net work on it is negative. Clear? So what is the force doing? The acceleration is the derivative of v, that is two. So F equals one times two, two newtons, capital N, and constant. It first slows the particle down, and then speeds it up the other way. Right? And here is a topper's quick check. For a constant force, the work is force times displacement. x goes from four metres to one metre, a displacement of minus three metres. Two times minus three: minus six joules, again. Perfect!

7:36Quick revision: recipes for x(t), v(x), F(t) and K(t)

So, quick revision. Position in time: differentiate to get the speed at both instants, then W equals delta K, and never assume it starts from rest. Velocity in position: the kinetic energy directly, and the force is d K over d x. Force in time, from rest: the impulse gives p, then K is p squared over two m. And kinetic energy in time: power is d K over d t, and the force is P over v. Now go and try the practice problems on this page. You will do them well. All the best!