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Today, we will talk exclusively about work and energy.

First, let's do a one-dimensional case.

The work that a force is doing, when that force is moving from point A to point B —

one-dimensional, here's point A and here is point B —

and the force is along that direction or...

either in this direction or in this direction but it's completely one-dimensional, that work is the integral in going from A to B of that force dx, if I call that the x-axis.

The unit of work, you can see, is newton-meters.

So work is newton-meters, for which we...

we call that "joule.

If there's more than one force in this direction, you have to add these forces in this direction vectorially, and then this is the work that the forces do together.

Work is a scalar, so this can be larger than zero, it can be zero, or it can be smaller than zero.

If the force and the direction in which it moves are in opposite directions, then it is smaller than zero.

If they're in the same direction, either this way or that way, then the work is larger than zero.

F = ma, so therefore, I can also write with this m dv/dt.

And I can write down for dx, I can write down v dt.

I substitute that in there, so the work in going from A to B is the integral from A to B times the force, which is m dv/dt, dx which is v dt.

And look what I can do.

I can eliminate time, and I can now go to a integral over velocity —

velocity A to the velocity B, and I get m times v times dv.

That's a very easy integral.

That is 1/2 m v squared, which I have to evaluate between vA and vB, and that is 1/2 m vB squared, minus 1/2 m vA squared.

1/2 m v squared is what we call in physics "kinetic energy." Sometimes we write just a K for that.

It's the energy of motion.

And so the work that is done when a force moves from A to B is the kinetic energy in point B —

you see that here —

minus the kinetic energy in point A, and this is called the work-energy theorem.

If the work is positive, then the kinetic energy increases when you go from A to B.

If the work is smaller than zero, then the kinetic energy decreases.

If the work is zero, then there is no change in kinetic energy.

Let's do a simple example.

Applying this work-energy theorem, I have an object that I want to move from A to B.

I let gravity do that.

I give it a velocity.

Here's the velocity v of A, and let the separation be h, and this could be my increasing y direction.

The object has a mass m, and so there is a force, gravitational force which is mg, and if I want to give it a vector notation, it's mg y roof, because this is my increasing value of Y.

When it reaches point B, it comes to a halt, and I'm going to ask you now what is the value of h.

We've done that in the past in a different way.

Now we will do it purely based on the energy considerations.

So I can write down that the work that gravity is doing in going from A to B, that work is clearly negative.

The force is in this direction and the motion is in this direction, so the work that gravity is doing in going from A to B equals minus mgh.

That must be the kinetic energy at that point B, so that this kinetic energy at point B minus the kinetic energy at point A, this is zero, because it comes to a halt here, and so you find that 1/2 m vA squared equals mgh.

m cancels, and so you'll find that the height that you reach equals vA squared divided by 2g.

And this is something we've seen before.

First, let's do a one-dimensional case.

The work that a force is doing, when that force is moving from point A to point B —

one-dimensional, here's point A and here is point B —

and the force is along that direction or...

either in this direction or in this direction but it's completely one-dimensional, that work is the integral in going from A to B of that force dx, if I call that the x-axis.

The unit of work, you can see, is newton-meters.

So work is newton-meters, for which we...

we call that "joule.

If there's more than one force in this direction, you have to add these forces in this direction vectorially, and then this is the work that the forces do together.

Work is a scalar, so this can be larger than zero, it can be zero, or it can be smaller than zero.

If the force and the direction in which it moves are in opposite directions, then it is smaller than zero.

If they're in the same direction, either this way or that way, then the work is larger than zero.

F = ma, so therefore, I can also write with this m dv/dt.

And I can write down for dx, I can write down v dt.

I substitute that in there, so the work in going from A to B is the integral from A to B times the force, which is m dv/dt, dx which is v dt.

And look what I can do.

I can eliminate time, and I can now go to a integral over velocity —

velocity A to the velocity B, and I get m times v times dv.

That's a very easy integral.

That is 1/2 m v squared, which I have to evaluate between vA and vB, and that is 1/2 m vB squared, minus 1/2 m vA squared.

1/2 m v squared is what we call in physics "kinetic energy." Sometimes we write just a K for that.

It's the energy of motion.

And so the work that is done when a force moves from A to B is the kinetic energy in point B —

you see that here —

minus the kinetic energy in point A, and this is called the work-energy theorem.

If the work is positive, then the kinetic energy increases when you go from A to B.

If the work is smaller than zero, then the kinetic energy decreases.

If the work is zero, then there is no change in kinetic energy.

Let's do a simple example.

Applying this work-energy theorem, I have an object that I want to move from A to B.

I let gravity do that.

I give it a velocity.

Here's the velocity v of A, and let the separation be h, and this could be my increasing y direction.

The object has a mass m, and so there is a force, gravitational force which is mg, and if I want to give it a vector notation, it's mg y roof, because this is my increasing value of Y.

When it reaches point B, it comes to a halt, and I'm going to ask you now what is the value of h.

We've done that in the past in a different way.

Now we will do it purely based on the energy considerations.

So I can write down that the work that gravity is doing in going from A to B, that work is clearly negative.

The force is in this direction and the motion is in this direction, so the work that gravity is doing in going from A to B equals minus mgh.

That must be the kinetic energy at that point B, so that this kinetic energy at point B minus the kinetic energy at point A, this is zero, because it comes to a halt here, and so you find that 1/2 m vA squared equals mgh.

m cancels, and so you'll find that the height that you reach equals vA squared divided by 2g.

And this is something we've seen before.

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