# Derivation of Energy Function

1. Jun 17, 2013

### GarrettB

A potential energy function for a two-dimensional force is of the form U = 3.21x3 y - 5.79x. Calculate the force that acts at the point (1.47m,1.42m). Enter the x-component first and then the y-component. -2.38×101 N -1.02×101 N

I know in order to find the force components of this energy function you need to derive relative to x and y. Can anyone provide insight to why that is though, perhaps with a real example? I find physics sticks better if I have some sort of example to relate to, rather than mindlessly using equations. Thanks in advance

2. Jun 18, 2013

### CompuChip

Actually the idea of potential energy is particularly useful for so-called conservative forces. Crudely put, these are the forces whose work done only depends on your position, not how you got there. For example, gravity is a conservative force: whether you move straight up and down, or in a circle, or in a spiral path to Mars and back - when you end up at your initial position you will have exactly the same energy. In contract, a force like drag or friction is not conservative: if you move along a longer path you will have lost more energy when you finally get back to your starting point.

Now if you look at the work done by a conservative force between to points a and b, you can basically take any path C from a to b and integrate along it:
$$W = \int_C \vec F \cdot \mathrm d\vec x$$
Note that since the result is independent of the choice of C, as long as it starts at a and ends at b, you can write the result as $W = U(b) - U(a)$.

A little algebra will show you that in fact $\vec F_x = \frac{\partial U}{\partial x}$ and similarly for the y- and z-components. However, I would like to make these two more physical points:

(1) W = U(b) - U(a) is the work done in moving from a to b. If you would move back to a (along any path) you would do work U(a) - U(b) = -W. This is precisely this idea of a conservative force. In a sense, you can say that we "stored" an energy W in the object when we moved it to b, which we can release when we move it back. Classical example is storing "gravitational energy" in an object by lifting it; when you let it go this energy will be converted to kinetic energy (hint: don't stand underneath it). Hence the term "potential energy": it is energy which is "stored" in the object, which does not "do" anything at the moment but has the potential of doing work. Usually, the "creation" of potential energy is external - e.g. you lift an object, compress a spring. As soon as you let go of this artificial constraint, the potential energy will be released - e.g. the object will fall, the spring decompress.

(2) Note that U really only makes sense when you talk about differences like U(b) - U(a). You cannot claim that U is the potential energy function, because then I will claim that U' = U + c works equally well for any constant c. The fact is, you cannot observe U itself, you can only observe U(b) - U(a) which satisfies U(b) - U(a) = U'(b) - U'(a). We often use this in calculations to choose a "convenient" reference point, e.g. the potential energy of a spring is usually set to 0 when the spring is relaxed and gravitational potential energy may be zero on the earth's surface, at the initial height of the object, or at some other point. What you are really doing of course, is choosing an arbitrary point $\vec x_0$ with which you define $$U(\vec x) = \int_{\vec x_0}^{\vec x} \vec F \cdot \mathrm d\vec x'$$. If I choose another point $\vec x_0'$ that will just give me
$$U'(\vec x) = \int_{\vec x_0'}^{\vec x} \vec F \cdot \mathrm d\vec x' = = \underbrace{\int_{\vec x_0}^{\vec x} \vec F \cdot \mathrm d\vec x'}_{U(x)} + \underbrace{\int_{\vec x_0}^{\vec x_0'} \vec F \cdot \mathrm d\vec x'}_{c}$$

Hopefully I haven't gone off on too many (mathematical) tangents and this answers your question.

3. Jun 18, 2013