Potential energy is an energy of position. The gravitational potential energy depends on the height of an object, and the elastic potential energy depends on a spring’s displacement. Other potential energies you will meet in the future will depend in some way on position. Functions of position are easy to represent as graphs. A graph showing a system’s potential energy and total energy as a function of position is called an energy diagram. Energy diagrams allow you to visualize motion based on energy considerations.

FIGURE 10.17 is the energy diagram of a particle in free fall. The gravitational potential energy $U_{\text{G}}=mgy$ is graphed as a line through the origin with slope mg. The potential-energy curve is labeled PE. The line labeled TE is the total energy line, $E=K+U_{\text{G}}$. It is horizontal because mechanical energy is conserved, meaning that the system’s mechanical energy $E$ has the same value at every position.

FIGURE 10.17 The energy diagram of a particle in free fall.

Suppose the particle is at position $y_1$. By definition, the distance from the axis to the potential-energy curve is the system’s potential energy $U_{\text{G}1}$ at that position. Because $K_1=E-U_{\text{G}1}$, the distance between the potential-energy curve and the total energy line is the particle’s kinetic energy.

The four-frame “movie” of FIGURE 10.18 illustrates how an energy diagram is used to visualize motion. The first frame shows a particle projected upward from $y_1=0$ with kinetic energy $K_1$. Initially the energy is entirely kinetic, with $U_{\text{G}1}=0$. A pictorial representation and an energy bar chart help to illustrate what the energy diagram is showing.

FIGURE 10.18 A four-frame movie of a particle in free fall.

In the second frame, the particle has gained height but lost speed. The potential energy $U_{\text{G}2}$ is larger, and the distance $K_2$ between the potential-energy curve and the total energy line is less. The particle continues rising and slowing until, in the third frame, it reaches the $y$-value where the total energy line crosses the potential-energy curve. This point, where $K=0$ and the energy is entirely potential, is a turning point where the particle reverses direction. Finally, we see the particle speeding up as it falls.

A system with this amount of total energy would need negative kinetic energy to be to the right of the point, at $y_3$, where the total energy line crosses the potential-energy curve. Negative $K$ is not physically possible, so the particle cannot be at positions with $\mathit{U}\mathbf{>}\mathit{E}$. Now, it’s certainly true that you could make the particle reach a larger value of $y$ simply by throwing it harder. But that would increase $E$ and move the total energy line higher.

FIGURE 10.19 shows the energy diagram of a mass on a horizontal spring, where $x$ has been measured from the wall where the spring is attached. The equilibrium length of the spring is $L_0$ and the displacement of the end of the spring is $\text{Δ}x=x-L_0$, so the elastic potential energy is $U_{\text{Sp}}=\frac{1}{2}k{\left(\text{Δ}x\right)}^2=\frac{1}{2}k{\left(x-L_0\right)}^2$. The potential-energy curve, a graph of $U_{\text{Sp}}$ versus $x$, is a parabola centered at the equilibrium position. You can’t change the PE curve—it’s determined by the spring constant—but you can set the TE to any height you wish simply by stretching the spring to the proper length. The figure shows one possible TE line.

FIGURE 10.19 The energy diagram of a mass on a horizontal spring.

If you pull the mass out to position $x_{\text{R}}$ and release it, the initial mechanical energy is entirely potential. As the restoring force of the spring pulls the mass to the left, the kinetic energy increases as the potential energy decreases. The mass has maximum speed at $x=L_0$, where $U_{\text{Sp}}=0$, and then it slows down as the spring starts to compress. You should be able to visualize that $x_{\text{L}}$, where the PE curve crosses the TE line, is a turning point. It’s the point of maximum compression where the mass instantaneously has $K=0$. The mass will reverse direction, speed up until $x=L_0$, then slow down until reaching $x_{\text{R}}$, where it started. This is another turning point, so it will reverse direction again and start the process over. In other words, the mass will oscillate back and forth between the left and right turning points at $x_{\text{L}}$ and $x_{\text{R}}$ where the TE line crosses the PE curve. We’ll study oscillations in Chapter 15, but we can already see from the energy diagram that a mass on a spring undergoes oscillatory motion.

FIGURE 10.20 applies these ideas to a more general energy diagram. We don’t know how this potential energy was created, but we can visualize the motion of a particle in a system that has this potential energy. Suppose the particle is released from rest at position $x_1$. How will it then move?

FIGURE 10.20 A more general energy diagram.

The initial conditions are $K=0$ at $x_1$, hence the TE line must cross the PE curve at this point. The particle cannot move to the left, because that would require $K<0$, so it begins to move toward the right. We see from the energy diagram that $U$ decreases from $x_1$ to $x_2$, so the particle is speeding up as potential energy is transformed into kinetic energy. The particle then slows down from $x_2$ to $x_3$ as it goes up the “potential-energy hill,” increasing $U$ at the expense of $K$. The particle doesn’t stop at $x_3$ because it still has kinetic energy. It speeds up from $x_3$ to $x_4$ (K increasing as $U$ decreases), reaching its maximum speed at $x_4$, then slows down between $x_4$ and $x_5$. Position $x_5$ is a turning point, a point where the TE line crosses the PE curve. The particle is instantaneously at rest, then reverses direction. The particle will oscillate back and forth between $x_1$ and $x_5$, following the pattern of slowing down and speeding up that we’ve outlined.