To calculate potential energy for a conservative force or vector field, we need to calculate the work done by that force... but only once! The result will then be used to define the potential energy. This section will be a review of the last chapter, but I want to be sure that you understand a few things.[br][br]Calculation of the work may be done with the simplest possible choice of path since the path chosen doesn't change the result. The simplest possible choice is just a path moving vertically parallel to the force of gravity. We do need to choose a coordinate system. I will use the +y axis to point upward. It is important to remember this since the convention must be assumed when using the potential energy expression! [br][br]The work done by gravity as a mass falls is just:[br][center][math][br]W_g = \int \vec{F}_g\cdot\vec{ds} \\[br]W_g = \int (-mg\hat{j})\cdot(dx\hat{i}+dy\hat{j} \\[br]W_g = \int -mg\; dy \\[br]W_g = -mg\Delta y.[br]\text{We now change the sign of this expression and call it gravitational potential energy: } \\[br]\Delta U_g = mg\Delta y. \\[br]\text{Since it's crucial to remember that +y is upward, and yet it's easy to forget, the 'y' is } \\[br]\text{sometimes replace with 'h' for height above the ground.} \\[br]\Delta U_g = mg\Delta h.[br][/math][/center]It is worth noting that I have stopped at defining a[i] change [/i]in potential energy. The reality is that we always calculate using changes. That should be clear from our work/energy relationship from this chapter: [math]\Sigma W_{NC} = \Delta K+\Sigma \Delta U.[/math] [b]Any time a potential energy is written without a [math]\Delta[/math] it is assumed to be measured relative to some reference, and therefore still represent a change with respect to that reference location.[/b] We will see that in the next section on universal gravitation.[br][br]The potential energy is simply a function of height of an object above the ground in the earth/ball system. This potential energy is seen as being stored energy in a system that depends on the relative positions of the parts of the system. [b]Potential energy always depends on relative locations of the parts of a system, whether that be the current case, or the relative stretch of a spring or the separation of atoms in a molecular bond.[/b][br][br]In this case the height of the ball describes the relative separation distance between earth and the ball. The farther the earth is separated from the ball, the larger the potential energy - similar to stretching an invisible spring. In the following sections we will consider other conservative vector fields that allow us to define other potential energy terms. There are not very many in nature, so the list is fairly short.

[color=#1e84cc]EXAMPLE: Consider a person of mass 80kg who starts at the top of a mountain road from rest on a longboard (a long skateboard used for such activities due to better stability at high speeds), and descends down the hill subject to all the realistic forces of friction and air drag. Suppose the vertical descent is 100m and the speed at the bottom of the mountain is 24m/s. How much energy is lost to the non-conservative forces of friction and air drag?[br][br]SOLUTION: [math]\Sigma W_{NC}=\Delta K + \Delta U_g.[/math] We are looking for the left side. While those work integrals are not possible to calculate by hand, we can still find the value of the integrals if we evaluate the right side of the expression. This requires only that we know the vertical descent and the the speeds at start and finish, which are known. The initial kinetic energy is zero. Therefore the change of kinetic energy is just [math]\Delta K = K_f-K_i = K_f-0 = K_f = \tfrac{1}{2}mv_f^2 = \tfrac{1}{2}(80kg)(24m/s)^2 = 23kJ.[/math] The change in potential energy is [math]\Delta U_g = mg\Delta h = (80kg)(10m/s^2)(-100m)=-80kJ.[/math] Therefore [math]\Sigma W_{NC} = 23kJ-80kJ = -57kJ.[/math] The fact that the result is negative means a loss of energy by the system. [/color]