# 9.3: Potential Energy Graphs

• • Christopher Duston, Merrimack College
• OpenStax
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##### Learning Objectives
• Create and interpret graphs of potential energy
• Explain the connection between stability and potential energy

Often, you can get a good deal of useful information about the dynamical behavior of a mechanical system just by interpreting a graph of its potential energy as a function of position, called a potential energy diagram. This is most easily accomplished for a one-dimensional system, whose potential energy can be plotted in one two-dimensional graph—for example, U(x) versus x—on a piece of paper or a computer program. For systems whose motion is in more than one dimension, the motion needs to be studied in three-dimensional space. We will simplify our procedure for one-dimensional motion only.

First, let’s look at an object, freely falling vertically, near the surface of Earth, in the absence of air resistance. The mechanical energy of the object is conserved, E = K + U, and the potential energy, with respect to zero at ground level, is U(y) = mgy, which is a straight line through the origin with slope mg . In the graph shown in Figure $$\PageIndex{1}$$, the x-axis is the height above the ground y and the y-axis is the object’s energy. Figure $$\PageIndex{1}$$: The potential energy graph for an object in vertical free fall, with various quantities indicated.

The line at energy E represents the constant mechanical energy of the object, whereas the kinetic and potential energies, KA and UA, are indicated at a particular height yA. You can see how the total energy is divided between kinetic and potential energy as the object’s height changes. Since kinetic energy can never be negative, there is a maximum potential energy and a maximum height, which an object with the given total energy cannot exceed:

$K = E - U \geq 0,$

$U \leq E \ldotp$

If we use the gravitational potential energy reference point of zero at y0, we can rewrite the gravitational potential energy U as mgy. Solving for y results in

$y \leq \frac{E}{mg} = y_{max} \ldotp$

We note in this expression that the quantity of the total energy divided by the weight (mg) is located at the maximum height of the particle, or ymax. At the maximum height, the kinetic energy and the speed are zero, so if the object were initially traveling upward, its velocity would go through zero there, and ymaxwould be a turning point in the motion. At ground level, y0 = 0, the potential energy is zero, and the kinetic energy and the speed are maximum:

$U_{0} = 0 = E - K_{0},$

$E = K_{0} = \frac{1}{2} mv_{0}^{2},$

$v_{0} = \pm \sqrt{\frac{2E}{m}} \ldotp$

The maximum speed ±v0 gives the initial velocity necessary to reach ymax, the maximum height, and −v0 represents the final velocity, after falling from ymax. You can read all this information, and more, from the potential energy diagram we have shown.

Consider a mass-spring system on a frictionless, stationary, horizontal surface, so that gravity and the normal contact force do no work and can be ignored (Figure $$\PageIndex{2}$$). This is like a one-dimensional system, whose mechanical energy E is a constant and whose potential energy, with respect to zero energy at zero displacement from the spring’s unstretched length, x = 0, is U(x) = $$\frac{1}{2}$$kx2. Figure $$\PageIndex{2}$$: (a) A glider between springs on an air track is an example of a horizontal mass-spring system. (b) The potential energy diagram for this system, with various quantities indicated.

You can read off the same type of information from the potential energy diagram in this case, as in the case for the body in vertical free fall, but since the spring potential energy describes a variable force, you can learn more from this graph. As for the object in vertical free fall, you can deduce the physically allowable range of motion and the maximum values of distance and speed, from the limits on the kinetic energy, 0 ≤ K ≤ E. Therefore, K = 0 and U = E at a turning point, of which there are two for the elastic spring potential energy,

$x_{max} = \pm \sqrt{\frac{2E}{k}} \ldotp$

The glider’s motion is confined to the region between the turning points, −xmax ≤ x ≤ xmax. This is true for any (positive) value of E because the potential energy is unbounded with respect to x. For this reason, as well as the shape of the potential energy curve, U(x) is called an infinite potential well. At the bottom of the potential well, x = 0, U = 0 and the kinetic energy is a maximum, K = E, so vmax = ± $$\sqrt{\frac{2E}{m}}$$.

However, from the slope of this potential energy curve, you can also deduce information about the force on the glider and its acceleration. We saw earlier that the negative of the slope of the potential energy is the spring force, which in this case is also the net force, and thus is proportional to the acceleration. When x = 0, the slope, the force, and the acceleration are all zero, so this is an equilibrium point. The negative of the slope, on either side of the equilibrium point, gives a force pointing back to the equilibrium point, F = ±kx, so the equilibrium is termed stable and the force is called a restoring force. This implies that U(x) has a relative minimum there. If the force on either side of an equilibrium point has a direction opposite from that direction of position change, the equilibrium is termed unstable, and this implies that U(x) has a relative maximum there.

This page titled 9.3: Potential Energy Graphs is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Christopher Duston, Merrimack College (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.