00:01
Here i'll say a few words about finding the work done by a force in displacing an object.
00:08
And in particular, we're going to look at a situation in which the force is constant versus which the force is a variable force.
00:17
And then we'll take a look at a specific example.
00:21
So work is defined as a dot product, a vector dot product, between the force and the displacement.
00:27
So that means that the force actually has to push the object in the direction of motion in order for there to be work done.
00:38
That's the idea of the dot product.
00:43
So here i've written down a formula if the force is constant.
00:48
We simply take the dot product between the force and the displacement factor.
00:55
If it is a non -constant force, it is a very similar expression, but what we have to do is find the area under the force.
01:07
Displacement.
01:10
So we're moving from point one to point two.
01:14
We'll call that r1.
01:16
Actually, point one to point two.
01:18
And we'll say something about a special type of force.
01:26
Okay, so this is a variable force.
01:29
We have to actually integrate under the force dotted into the displacement all along the path.
01:38
So this is something called a line integral, and typically depends on the path that you take, unless the force is what is called conservative.
01:49
So this is something called a line integral.
01:57
Depends on path.
02:03
Unless the force is what is called conservative.
02:11
And there's a whole mythology, if you will, built up on how to determine if a force is conservative and classes of forces that are conservative.
02:30
I won't get into that, but i will note that one class of conservative force is what is called a central force.
02:45
In other words, the force just depends on a distance out from a center.
02:56
And we typically write that as a function times r -hat.
03:02
Very common examples are gravity and electricity, gravitation and electrical force from a point charge.
03:23
So without further ado, and this isn't necessary really to understand.
03:28
That to solve the following example, but it does kind of lead to why the example is brought up.
03:38
So the idea is we have an object on the surface of the earth, and we're going to calculate the work done to move it from one radius, we'll call that r1, out to r2.
03:58
Just like it takes work to lift an object on the surface of the earth, it is going to take positive work for somebody applying a force to move that object.
04:15
And the force of gravity points radially inwards.
04:23
We're going to assume that it has a particular value in si units, 4 times 10 to the 15.
04:32
Newton's over r squared where r is in meters.
04:38
So we're using si units...