**Pre Reqs:** [What is a Force](http://blog.dotphys.net/2008/09/basics-what-is-a-force/)

[Previously, I talked about the momentum principle](http://blog.dotphys.net/2008/10/basics-forces-and-the-momentum-principle/). Very useful and very fundamental idea. The other big (and useful) idea in introductory physics is the work-energy theorem. Really, with work-energy and momentum principle, you will be like a Jedi with a lightsaber and The Force – extremely powerful.

Well, what is work? What is energy? How are they related? In [another post, I talked about energy.](http://blog.dotphys.net/2008/10/what-is-energy/) I think it is interesting to look at how most textbooks define energy:

*Energy is the ability to do work*

This is really a stupid definition. Kind of circular logic, if you ask me. In the post I mentioned earlier, I claim there are two kinds of energy, particle energy and field energy. At low speeds (not near the speed of light), particle energy can be written as:

![Screenshot 53](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-53.jpg)

Where *m* is the mass of the particle, *c* is the speed of light. So, if you just look at a particle, that is it for the energy. Now, what about the “work” portion? Work is defined as:

![Screenshot 54](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-54.jpg)

Where *F* is the net force on the particle, ?r is the vector displacement of the particle. The “dot” in between F and ?r represents the “dot product” operation between vectors (also known as the scalar product). In a [previous post](http://blog.dotphys.net/2008/09/basics-vectors-and-vector-addition/) I showed that you could multiply a scalar quantity by a vector quantity. Here I need to do “something” with two vectors. You can’t multiply two vectors in the same sense that you multiply scalars. A general definition of the dot product for two vectors:

![Screenshot 55](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-55.jpg)

That looks a little more messy than I wanted, but it can not be helped. Really, it is not that complicated. The dot product is simply the projection of one vector on the other. Let me explain in terms of work.

Suppose I pull a block 2 meters with a force of 10 Newtons as shown in this diagram:

![Screenshot 56](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-56.jpg)

Since both force and displacement are in the same direction, this would give a work of 20 Joules. However, I will do this the long way. Assume the x-axis is horizontal, then:

![Screenshot 57](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-57.jpg)

Note that I am calculating the work done by THAT force. There could be other forces acting on that block, but even if they aren’t it doesn’t change the work done. Now let me look at another example that is similar. Suppose I again push with a force of 10 Newtons, and I again move the block 2 meters:

![Screenshot 58](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-58.jpg)

In this case, I push perpendicular to the direction the block moves (to do this, there would need to be other forces acting on the block). How much work would be done?

![Screenshot 59](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-59.jpg)

So, forces perpendicular to the displacement do no work, only forces in the same (or opposite) direction do. So, what if there is a force not perpendicular, but also not in the same direction?

![Screenshot 60](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-60.jpg)

I could calculate the work done by the force in the normal fashion (dot product) or I could say I just need the component of force in the direction of motion:

![Screenshot 61](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-61.jpg)

Or:

![Screenshot 62](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-62.jpg)

Either way, same thing. This is why sometimes you will hear people explain the dot product as a projection of one vector onto the other. It is only the components of the two vectors that are in the same direction that matter.

So, that is work. Now for the good part. The work energy theorem says:

*The work done on a particle is equal to it’s change in energy*

![Screenshot 63](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-63.jpg)

Notice that it is the CHANGE in energy, not the energy. At low speeds, the mass-energy (mc^{2}) doesn’t really change, so I will typically just relate the work to the change in kinetic energy.

But wait!! What about potential energy? Yes, I know. But for a PARTICLE, there is no potential energy. You can have potential energy for a system. Instead of talking about potential energy, I will give a short example of work-energy. I will save potential for another day.

Here is an example I used [previously](http://blog.dotphys.net/2008/10/basics-numerical-calculations/). Suppose I throw a ball straight up with a speed of 10 m/s (and the ball has a mass of 0.5 kg, but that doesn’t really matter). How high will it go? I did this problem both analytically and numerically using the momentum principle. The momentum principle deals with force and TIME – but in this case, I don’t want the time. Work-energy deals with force and displacement. This will be perfect for this problem. So, while the ball is going up, here is the free body diagram:

![Screenshot 01](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-011.jpg)

I can now write an expression for the work (the total work) done on this ball as it rises:

![Screenshot 64](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-64.jpg)

In this case, the ball is moving up, so ?y is a positive number. The gravitational force is down but in the same direction (opposite direction – so that there is no cosine term). I can set the initial position of the ball to * y = 0 m*. As for the change in energy, the mass energy does not change so I just have change in kinetic energy.

![Screenshot 65](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-65.jpg)

Since I know the initial velocity, I can get a value for the final y:

![Screenshot 66](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-66.jpg)

If you look back at the analytical solution using the momentum principle, this probably looks easier. It should, because it is. Remember:

- Work-Energy deals with force and change in position. This problem specifically was looking for the position – a perfect match.
- Work-Energy does not give a vector answer. The kinetic energy has v
^{2}which is technically the square of the magnitude of velocity - The stuff I have done here deals with a PARTICLE. If you have something that can not be represented as a particle (like a car with internal combustion engine) then you will need to do something else.