10.6 Force and Potential Energy

As you’ve seen, we can find the energy of an interaction—potential energy—by calculating the work the interaction force does inside the system. Can we reverse this procedure? That is, if we know a system’s potential energy, can we find the interaction force?

We defined the change in a system’s potential energy to be $Δ U = - W_{int}$ . Suppose that only one object in the system moves while other objects—such as the earth or a rigid wall—are essentially at rest. The moving object has the only displacement, so $W_{int}$ is the work done on the moving object. We can allow this object to undergo a very small displacement $Δ s$ , so small that the interaction force $\vec{F}$ is essentially constant. The work done by a constant force is $W = F_{s} Δ s$ , where $F_{s}$ is the force component parallel to the displacement. During this small displacement, the system’s potential energy changes by

Δ U = - W_{int} = - F_{s} Δ s

(10.23)

which we can rewrite as

F_{s} = - \frac{Δ U}{Δ s}

(10.24)

In the limit $Δ s \to 0$ , the force on the object is

F_{s} = lim_{Δ s \to 0} (- \frac{Δ U}{Δ s}) = - \frac{d U}{d s}

(10.25)

That is, the interaction force on an object is the negative of the derivative of the potential energy with respect to position.

Graphically, as FIGURE 10.24 shows, force is the negative of the slope, at position $s$ , of the potential-energy curve in an energy diagram:

F_{s} = - \frac{d U}{d s} = the negative of the slope of the PE curve at s

(10.26)

The figure illustrates a Potential energy (U) versus position (s) graph.

FIGURE 10.24 Full Alternative Text

In practice, of course, we’ll usually use either $F_{x} = - d U / d x$ or $F_{y} = - d U / d y$ . Thus

A positive slope corresponds to a negative force: to the left or downward.
A negative slope corresponds to a positive force: to the right or upward.
The steeper the slope, the larger the force.

As an example, consider the elastic potential energy $U_{Sp} = \frac{1}{2} k x^{2}$ for a horizontal spring with $x_{eq} = 0$ so that $Δ x = x$ . FIGURE 10.25a shows that the potential-energy curve is a parabola, with changing slope. If an object attached to the spring is at position $x$ , the force on the object is

F_{x} = - \frac{d U_{Sp}}{d x} = - \frac{d}{d x} (\frac{1}{2} k x^{2}) = - k x

This is just Hooke’s law for an ideal spring, with the minus sign indicating that Hooke’s law is a restoring force. FIGURE 10.25b is a graph of force versus $x$ . At each position $x$ , the value of the force is equal to the negative of the slope of the PE curve.

The figure consists of two parts labeled ("a") and (b), respectively, that illustrate potential energy (U) versus position (x) graph of elastic potential energy and a force (F subscript x) versus position (x) graph.

FIGURE 10.25 Full Alternative Text

We already knew Hooke’s law, of course, so the point of this particular exercise was to illustrate the meaning of Equation 10.26. But if we had not known the force, we see that it’s possible to find the force from the PE curve. For example, you’ll learn in Part VIII that FIGURE 10.26a is a possible potential-energy function for a charged particle, one that we could create with suitably shaped electrodes. What force does the particle experience in this region of space? We find out by measuring the slope of the PE curve. The result is shown in Figure 10.26b. On the left side of this region of space $(x < 2 m)$ , a negative slope, and thus a positive value of $F_{x}$ , means that the force pushes the particle to the right. A negative force on the right side $(x > 3 m)$ tells us that $F_{x}$ pushes the particle to the left. And there’s no force at all in the center. This is a restoring force because a particle trying to leave this region is pushed back toward the center, but it’s not a linear restoring force like that of a spring.

The figure consists of two parts labeled ("a") and (b), respectively, that illustrate potential energy (U) versus position (x) graph and a force (F subscript x) versus position (x) graph.

FIGURE 10.26 Full Alternative Text

EXAMPLE 10.9 Finding equilibrium positions

A system’s potential energy is given by $U (x) = (2 x^{3} - 3 x^{2}) J$ , where $x$ is a particle’s position in m. Where are the equilibrium positions for this system, and are they stable or unstable equilibria?

SOLVE You learned in Chapter 6 that a particle in equilibrium has $\vec{F}_{net} = \vec{0}$ . Then, in the previous section, you learned that the maxima and minima of the PE curve are points of equilibrium. These may seem to be two different criteria for equilibrium, but actually they are identical. The interaction force on the particle is $F_{x} = - d U / d x$ . The force is zero—equilibrium—at positions where the derivative is zero. But you’ve learned in calculus that positions where the derivative of a function is zero are the maxima and minima of the function. At either a maximum or minimum of the PE curve, the slope is zero and hence the force is zero.

For this potential-energy function,

F_{x} = - \frac{d U}{d x} = (- 6 x^{2} + 6 x) N

when $x$ is in m. The force is zero—minima or maxima of $U$ —when $6 x_{eq}^{2} = 6 x_{eq}$ . This has two solutions:

x_{eq} = 0 m and x_{eq} = 1 m

These are positions of equilibrium, where a particle at rest will remain at rest. But how do we know if these are positions of stable equilibrium or unstable equilibrium?

A minimum in the PE curve is a stable equilibrium. Recall, from calculus, that a minimum of a function has a first derivative equal to zero and a second derivative that’s positive. Similarly, a maximum of a function has a first derivative equal to zero and a second derivative that’s negative. The second derivative of $U$ is

\frac{d^{2} U}{d x^{2}} = \frac{d}{d x} (6 x^{2} - 6 x) = (12 x - 6) N / m

The second derivative evaluated at $x = 0 m$ is $- 6 N / m < 0$ , so $x = 0 m$ is a maximum of the PE curve. At $x = 1 m,$ the second derivative is $+ 6 N / m > 0$ , hence a minimum in the PE curve. Thus this system has an unstable equilibrium if the particle is at $x = 0 m$ and a stable equilibrium if it is at $x = 1 m$ . There’s a force on the particle at all other positions.

10.6 Force and Potential Energy

FIGURE 10.24 Relating force to the PE curve.

FIGURE 10.25 Elastic potential energy and force graphs.

FIGURE 10.26 A potential-energy curve and the associated force curve.

EXAMPLE 10.9 Finding equilibrium positions