Work Done by a Variable Force

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Imagine a force pushing a box along a line, but the force keeps changing: zero newtons at the start, sixteen newtons after two metres. Work is force times displacement, right? So which force do you take: zero, sixteen, or something in between? Not so easy, is it? Today we find out. This idea comes up often in JEE, so let's begin.

0:29One dimension: add up tiny bits of work

So, look at the graph. Capital F, the force, is on the vertical axis, and small x, the position, is along the horizontal one. The force changes as x changes. But force times displacement needs a constant force. So what can we do? Here is the idea: cut the motion into tiny steps! Over one tiny step, d x, the force is almost constant. So the tiny bit of work, d W, equals F of x times d x. Simple, right? Now add up all the tiny bits; that is exactly what an integral does. So W equals the integral from x one to x two of F of x, d x. The tall S is the integral sign; x one and x two, with small subscripts, are the start and the end. Clear? On the graph, F times d x is the area of one thin strip. Add all the strips: the work is the signed area under the F x graph, the graph of force against position. On the left, the shaded area from zero up to x two, W one, is greater than zero: positive work. With me? But now look at the right side. After x two, the graph goes below the axis, and the label there says W two is less than zero. Why should this area count as negative? Think about the direction. Below the axis, the force points backwards while the particle moves forwards, so this work is negative. That is signed area: above the axis positive, below the axis negative. So the total work from zero to x four is W one plus W two, with W two negative. Do not just add the two areas as sizes; that is a common slip. Got it?

2:11Along any path: use components

Now, what if the particle moves along a curved path? Same idea, but each tiny step is now a vector, d r, with an arrow on top. So W is the integral of F vector dot d r vector, from r i to r f, the initial and final positions. The dot is the dot product: only the part of the force along the step does work. Okay? In components: F x d x, plus F y d y, plus F z d z, where F x, F y and F z are the parts of the force along the three axes. And notice: along the actual path, the one the particle really takes. Right? But there is a problem here. In F x d x plus F y d y, both x and y are changing together. So how do you integrate that? Use the path! On the curve y equals f of x, d y equals f prime of x, d x; the little dash, called prime, means the derivative. Put this in, write y as f of x, and everything is in x alone: an ordinary integral. Clear? One more form. If the force is given in time, t, and you know the velocity, each tiny step, d r, is v times d t. So W is the integral of F vector dot v vector, d t, where v with an arrow is the velocity. With me?

3:36Topper's trick: when the path does not matter

Now a topper's trick that saves time. Look at the force. Suppose F x depends only on x, F y only on y, and F z only on z. No mixing of coordinates at all. Right? Then, as the double arrow says, the force is conservative: the path does not matter! Okay? So forget the curve. Integrate each component from its starting value to its final value, and add. That's it! But check the condition first: if F x has a y in it, the shortcut does not apply. Got it?

4:09Example 1: work by F = 3x^2 + 2x from x = 0 to 2 m

Now, let's finish our puzzle from the start. Example one: the force F equals three x squared plus two x newtons, along the x axis. Find the work done from x equals zero to x equals two metres. Pause the video, and try it first. Done? Okay, let's check. At x equals two, the force is three times four, plus two times two: sixteen newtons. So the work is sixteen times two, thirty two joules, right? No, no, wait! Sixteen newtons is the force only at the end; at x equals zero, it is zero. The force is not constant, so force times displacement fails. That is the trap. Clear? So we integrate. W equals the integral from zero to two of three x squared plus two x, d x. Okay? The integral of three x squared is x cubed, and of two x, x squared. The square brackets, with zero below and two above, mean: put in two, then subtract the value at zero. Eight plus four: twelve joules, not thirty two! Puzzle solved: we must add up the tiny bits of work. Beautiful, isn't it?

5:24Example 2: work along y = 2x and along y = 2x^2

Now a harder one, JEE level. The force, F with an arrow, equals two x y i hat plus x squared j hat, in newtons; i hat and j hat, with little caps, are the unit vectors along the x and y axes. Find the work done from zero comma zero to one comma two, in metres: first along y equals two x, then along y equals two x squared. Pause and try. Hint: write everything in x. Done? Okay, let's check. First, can we use our topper's trick and forget the path? No! F x is two x y: it has a y in it. The condition fails, so no shortcut. We must go along each path. Right? First path: the line y equals two x. So d y equals two d x. Okay? Now look at F x d x: two x y d x. Can we keep y fixed at its final value, two, and integrate in x? Oh no, careful! On this path, y changes with x, so replace y by two x. Then F x d x is four x squared d x, and F y d y is x squared times two d x. Six x squared, from zero to one, gives two joules. Clear? Second path: the curve y equals two x squared. So d y equals four x d x. With me? Now F x d x is two x times two x squared: four x cubed d x. And F y d y is x squared times four x d x: again four x cubed d x. Eight x cubed, from zero to one: two joules again! Two different paths, the same work. Coincidence? No, it is not! Differentiate x squared y with respect to x: two x y. With respect to y: x squared. Exactly our force! So F equals nabla of x squared y; nabla is the upside down triangle. The big square bracket means the value at the end, one comma two, minus the value at the start. One squared times two, minus zero: two joules, whatever the path. So even when the shortcut fails, the path may not matter. Interesting, right?

7:44Quick revision: integral, F-x graph area, path rules

So, quick revision. In one dimension, work is the integral of F d x, the signed area under the F x graph. Along a path, integrate F x d x plus F y d y, with d y equals f prime of x d x; for a force given in time, integrate F dot v d t. And if each component depends only on its own coordinate, the path does not matter. Now go and try the practice problems on this page. All the best!