**Pre Reqs:** [Work-Energy](http://scienceblogs.com/dotphysics/2008/10/basics-work-energy.php)
You need to be familiar with work and energy to understand this. If you are not familiar, look at the pre requisite link. Ok? Now, let’s begin.
Suppose a ball moves from point A (3 m, 3 m) to B (1 m, 1 m) at a constant speed as shown in the diagram below:
Suppose there is some other force (like my hand) also exerting a force on this ball to make it move along this path at a constant speed. What is the work done by gravity? [If you recall](http://scienceblogs.com/dotphysics/2008/10/basics-work-energy/) work is:
Where θ is the angle between the force and the displacement. In this example, θ is 45 degrees. This means that the work done by gravity is:
I will leave it like that, but note that this is a positive value since the gravitational force is mostly in the same direction as the displacement. Ok – now let me do the same thing, but take a different path:
What is the work from A to B done by the gravitational force along this path? I can break this into two parts (1 and 2). The work done by gravity along path 1 is zero. This is because the angle between the gravitational force and the displacement is 90 degrees. The cosine of 90 degrees is zero. For path 2, the angle is zero so the total work is:
I can relate L and L2 using the right triangle by combining the first and second path (L was the length of the first path)
This is the SAME as the work done by gravity along the first path. So, what does this mean?
This means that the work done by gravity does not depend on the path, but only the two points (the starting and ending point). I know what you are saying. But what if it is some wacked-out path? Why would someone do that? I don’t know, but they could.
Well, how do I calculate the work for this path? The angle between gravity and the displacement keeps changing. Also, I don’t know the total distance. To do this, I will do what you should always do when faced with a difficult problem: break it into a bunch of small steps. If I break the path into many small pieces, each piece is approximately a constant displacement – I can do that. Here is a sample small piece (blown up for viewability):
Here I have shown two small pieces. For the whole thing there would be many many pieces. Now let me look at just one piece. I can do two things to calculate the work. First, I could just find the angle between the small piece and gravity, or I could break that piece down even further like this:
If you remember your [vector addition](http://scienceblogs.com/dotphysics/2008/09/basics-vectors-and-vector-addition/), then you remember that a vector can be broken into an x-component and a y-component. Doing it this way, I can calculate the work done by gravity for the two components. The work done along the x-component is zero because the gravitational force is perpendicular to the displacement in the x-direction. The work done along the y-component is simply -mgΔy (because in this piece, the y displacement is up, but gravity is down). So if I break this path into a whole bunch of pieces, each piece has an x- and y-component. The work along the ALL of the x-components is zero. Now I just have to add up all the works done on the y-components. Since the force of gravity is constant, the y-pieces add up to the change in the y-position from A to B. Ok, you didn’t really need to know this – I just wanted to give further explanation as to why it does not matter about the path.
Great, what next? Well, if the work done by the force is path independent (technically called a conservative force), then I can write:
Here, Wstuff is work done by other forces on the object. You can see that all I did was to take the work done by gravity and move it to the other side (the energy side). Before, only the change in kinetic energy was there. If that work is on the energy side, I should call it some type of energy. I will call it gravitational potential energy:
So, if I move an object up – then Δy is positive and ΔUgrav would be positive also. This means that (if I want to think about gravitational potential energy) that as the object moves up, it increases in gravitational potential energy.
- You CAN NOT have work done by gravity AND gravitational potential energy. This would be like having your cake and eating it too. You can either HAVE your cake, or you can EAT your cake.
- Since the potential energy is due to work done by a force, and forces are interactions between TWO objects, the potential energy is for the system consisting of the two objects. In the example above, the system is the ball plus the Earth.
- You see that it is the CHANGE in potential energy that appears. This means that (for this potential) it doesn’t matter where U = 0 – so choose a position that makes you happy.
- For this calculation, I assumed the gravitational force was a constant mg. It isn’t. This form of gravitational potential energy is only valid close to the surface of the Earth (you know, normal everyday stuff close).
**Elastic Potential Energy**
The force exerted by a spring is:
This is a scalar equation and Δs is the stretch of the spring (or compression). The direction of the spring force is in the opposite direction as the way the spring is stretched. What if I want to calculate the work done by the spring in stretching the spring an amount Δs?
The work is negative because the force exerted by the spring is in the opposite direction as the displacement. There is a problem. The problem is that the force exerted by the spring is not a constant value. I can fix this though. The force by the spring when it is not stretched (at the beginning) is zero. The force at the final stretch is kΔs. Since the force changes linearly with displacement and the work depends on displacement, I can use the average force. This means that the work done by the spring is:
So, if I move this work done by the spring to the other side, I have the elastic potential energy is:
Notice that since the stretch is squared, there is energy stored in the spring system both with it is stretched and when it is compressed.
**Can’t the work done by all forces be made into a potential?**
No. Here is an example. Suppose a I push a block on a table from point A to B.
In order to determine the work done by friction, I need to know the frictional force. A pretty good model for friction is:
This says the magnitude of the frictional force is product of some coefficient times the force of the table pushing up on the block. The direction of this force is always in the opposite direction that the object moves. (this is kinetic friction). Well, then this looks pretty easy. If I call the distance from A to B L1 then:
I went ahead and put N as mg (remember N is not ALWAYS mg). Otherwise, everything looks great. Ok. Now I will do this again. I will start at A and end at B, but I will take a different path:
In this case, I push the block a distance L2 past point B and then back to B. The work done by friction while the block is moving to the right is:
But now to push it back to B, friction has to do more work. During this movement to the left, friction is now pushing to the right (because that is what friction does). So, the work while moving to the left is:
This gives a total work due to friction going from A to B as:
Clearly this is not the same work done by friction as the first path. So the work done by friction DOES depend on the path – it can not be made into a potential energy.
Notice what would happen if you did this with gravity (turn the table vertical). For the second case, the part of the path that goes past point B would be extra negative work. However, when you come back down to point B, gravity is still pointing down. This would be positive work that would cancel the extra negative work.
Ok. I am done with this.