Potential Energy and Conservative Forces

Transcript

0:00Introduction

Hello everyone! Let's start with a small argument. I lift a two kilogram book from the floor onto a shelf, one and a half metres high. Now, one student says: on the floor, the book's potential energy is zero. Another student says: no, on the floor it is minus thirty joules. Who is right? Keep your answer in mind; by the end of this lesson, it may surprise you. Potential energy comes up often in JEE energy problems, so let's build it properly. Let's go.

0:32Potential energy: minus the work done

So, a quick recall from Chapter one. We call a force conservative if its work between two points depends only on the end points, not on the path. With me? Now picture a ball falling. Gravity pulls it down, and it moves down, so gravity does positive work on it, right? So tell me: does its potential energy go up, or go down? It goes down. Yes, exactly! Positive work by the force, so the potential energy falls. That is the definition: delta U equals U sub B minus U sub A, equals minus W from A to B, the work done by the force. Delta, the little triangle, means change; capital U is the potential energy; the small B and A written below are subscripts naming the two points; and the arrow means going from one point to the other. Okay? Now, look carefully. Only differences of potential energy matter. So the zero, which we call the reference, can be put anywhere you like: the floor, the shelf, anywhere. Hold on to this; it will settle our book argument. Got it? One more question. Can we define a potential energy for friction? No, we cannot! Friction is not conservative; its work depends on the path. And potential energy is defined only for conservative forces. Clear?

1:53Force from potential energy

Alright, now let's turn it around. Suppose I give you the potential energy as a function of x, U of x. How will you get the force back from it? Take a very small step, d x. The work done by the force is F times d x, so the change in potential energy is d U equals minus F d x. Right? Now just divide both sides by d x. F equals minus d U over d x. Here d U over d x is the derivative of U with respect to x, the slope of the potential energy. So: differentiate, and put a minus sign. That's it! With me? And going back the other way: U of x equals minus the integral of F d x, plus a constant C, okay? The tall S shaped sign is the integral sign. And why the constant? Because only differences matter; the constant is just your choice of zero. Clear?

2:50Trap: which way does the force point?

Now, a trap that catches many students. Suppose the potential energy increases as x increases, so the slope is positive. Which way does the force point: towards plus x, or towards minus x? Towards minus x! If you said plus x because the slope is positive, that is exactly the trap. The force always points towards decreasing potential energy, okay? See it in the formula. If d U over d x is greater than zero, then, and this double arrow means so, F is less than zero: the force is towards minus x. That minus sign in F equals minus d U over d x is the whole story. Never drop it! Got it?

3:33Three dimensions, and near the Earth

Now, three dimensions. Same rule, one direction at a time. F with an arrow on top is the force vector. The curly d is a partial derivative: differentiate U with respect to x, keeping y and z fixed. And i hat, j hat and k hat, the letters with little caps, are the unit vectors along x, y and z. So, F vector equals minus, in brackets, partial U over partial x times i hat, plus partial U over partial y times j hat, plus partial U over partial z times k hat. Okay? And near the Earth's surface, U equals m g h: the mass, times g, the acceleration due to gravity, times the height h above your reference level. Right? But what about a long rope hanging down? Its top and bottom are at different heights. So which h do we use? The centre of mass. Correct! For an extended body, h is the height of its centre of mass. Remember this for ropes, chains and poles. Clear?

4:36Example 1: book lifted to a shelf, change in U and the zero

Now, remember our book from the start? Let's settle it. So, a two kilogram book is lifted from the floor to a shelf one point five metres high. Find the change in its potential energy, and its potential energy on the floor if the shelf is taken as the zero. Take g equals ten. Pause the video and try it. Done? Okay, let's check. Delta U equals m g h: two times ten times one point five, that is plus thirty joules; the capital J stands for joules. Positive, because the book went up. Right? Now the second part. The shelf is our zero. The floor is below the shelf. So is h positive, or negative? Negative. Yes! The floor is one point five metres below the shelf, so h equals minus one point five metres. With me? So U on the floor, the word floor written small as a subscript, is two times ten times minus one point five: minus thirty joules. Okay? And look: the change is still U on the shelf minus U on the floor: zero minus, minus thirty, so plus thirty joules. So, our argument from the start: both students were right! One took the floor as zero, the other took the shelf. The value of the potential energy depends on your zero, but the change never does. Clear?

6:02Example 2: force at (1, 2) from U = 3x^2 y

Now a slightly harder one, in two dimensions. The potential energy is U equals three x squared y, in S I units. Find the force on a particle at the point one, two; that is, x equals one and y equals two. Pause the video and try it. Done? Okay, let's check. First, partial U over partial x. While we differentiate with respect to x, what do we do with y? Treat it as a constant. Correct! So partial U over partial x is six x y. And partial U over partial y: now x is held fixed, so we get three x squared. Got it? So the force is six x y i hat, plus three x squared j hat, right? No, no, wait! The minus sign! F vector equals minus, in brackets, six x y i hat plus three x squared j hat. This is exactly our trap: the force points towards decreasing potential energy. Okay? Now put x equals one and y equals two. Six times one times two is twelve, and three times one squared is three. So the force is minus twelve i hat minus three j hat newtons; the capital N stands for newtons. Both parts are negative: it pushes towards smaller x and smaller y. Right? And the two vertical bars around F mean its magnitude. Twelve squared is one hundred and forty four, three squared is nine, so we get the square root of one hundred and fifty three; the wavy equals sign means approximately: about twelve point four newtons. Nice and quick!

7:47Quick revision: change in U = -W, F = -dU/dx, U = mgh

So, quick revision. One: delta U equals minus W, only for conservative forces, and the zero is your choice. Two: F equals minus d U over d x, so the force points towards decreasing potential energy. Three: U equals m g h, with h the height of the centre of mass. Now go and try the practice problems on this page. You are ready for them now. All the best!