Much of what you’ve just learned about gravitational potential energy carries over to the elastic potential energy of a spring. FIGURE 10.11 shows a spring exerting a force on a block while the block moves on a frictionless, horizontal surface. In Chapter 9, we analyzed this problem by defining the system to consist of only the block, and we calculated the work of the spring on the block. Now let’s define the system to be block $\text{+}$ spring $\text{+}$ wall. That is, the system is the spring and the objects that are interacting with the spring. The surface and the earth exert forces on the block—the normal force and gravity—but those forces are always perpendicular to the displacement and do not transfer any energy to the system.

FIGURE 10.11 The block $\text{+}$ spring $\text{+}$ wall system has an elastic potential energy.

Let’s model the spring as being massless, which is reasonable as long as the spring’s mass is much less than the mass of the block. With this model, the spring has no kinetic energy but instead provides an interaction between the block and the wall. Because the interaction is inside the system, it has an interaction energy, the elastic potential energy, given by

where $W_{\text{B}}$ is the work the spring does on the block and $W_{\text{W}}$ is the work done on the wall. But the wall is rigid and has no displacement, so $W_{\text{W}}=0$ and thus $\text{Δ}{U}_{\text{Sp}}=-W_{\text{B}}$.

We calculated the work done by an ideal spring—one that produces a linear restoring force for all displacement—in 《 SECTION 9.4. If the block moves from an initial position $s_{\text{i}}$, where the spring’s displacement is $\text{Δ}{s}_{\text{i}}=s_{\text{i}}-s_{\text{eq}}$, to a final position $s_{\text{f}}$ with displacement $\text{Δ}{s}_{\text{f}}=s_{\text{f}}-s_{\text{eq}}$, the spring does work

With the minus sign of Equation 10.15, we have

Thus the elastic potential energy is

where $\text{Δ}s$ is the displacement of the spring from its equilibrium length. Elastic potential energy, like gravitational potential energy, is an energy of position. It depends on how much the spring is stretched or compressed, not on how fast the block is moving. Although we derived Equation 10.18 for a spring, it applies to any linear restoring force if $k$ is the appropriate “spring constant” for that force.

The energy principle for a system with elastic potential energy and no external interactions is either $\text{Δ}{E}_{\text{sys}}=\text{Δ}K+\text{Δ}{U}_{\text{Sp}}=0$ or, recognizing that mechanical energy is again conserved,

EXAMPLE 10.5 An air-track glider compresses a spring

In a laboratory experiment, your instructor challenges you to figure out how fast a 500 g air-track glider is traveling when it collides with a horizontal spring attached to the end of the track. He pushes the glider, and you notice that the spring compresses 2.7 cm before the glider rebounds. After discussing the situation with your lab partners, you decide to hang the spring on a hook and suspend the glider from the bottom end of the spring. This stretches the spring by 3.5 cm. Based on your measurements, how fast was the glider moving?

MODEL Let the system consist of the track, the spring, and the glider. The spring is inside the system, so the elastic interaction will be treated as a potential energy. Gravity and the normal force of the track on the glider are perpendicular to the glider’s displacement, so they do no work and do not enter into an energy analysis. An air track is essentially frictionless, and there are no other external forces.

VISUALIZE FIGURE 10.12 shows a before-and-after pictorial representation of the collision, an energy bar chart, and a free-body diagram of the suspended glider.

FIGURE 10.12 Pictorial representation of the experiment.

FIGURE 10.12 Full Alternative Text

SOLVE The glider’s kinetic energy is gradually transformed into elastic potential energy as it compresses the spring. The point of maximum compression—After in Figure 10.12—is a turning point in the motion. The velocity is instantaneously zero, the glider’s kinetic energy is zero, and thus—as the bar chart shows—all the energy has been transformed into potential energy. The spring will expand and cause the glider to rebound, but that’s not part of this problem. The energy principle with elastic potential energy, in conservation form, is

$$K_{\text{i}}+U_{\text{Sp i}}=\frac{1}{2}m{v}_0{}^2+0=K_{\text{f}}+U_{\text{Sp f}}=0+\frac{1}{2}k{\left(\text{Δ}{x}_1\right)}^2$$

where we utilized our knowledge that the initial elastic potential energy and the final kinetic energy, at the turning point, are zero. Solving for the glider’s initial speed, we find

$$v_0=\sqrt{{\displaystyle \frac{k}{m}}}\text{Δ}{x}_1$$

It was at this point that you and your lab partners realized you needed to determine the spring constant $k$. One way to do so is to measure the stretch caused by a suspended mass. The hanging glider is in equilibrium with no net force, and the free-body diagram shows that the upward spring force exactly balances the downward gravitational force. From Hooke’s law, the magnitude of the spring force is $F_{\text{Sp}}=k\left|\text{Δ}y\right|$. Thus Newton’s first law for the suspended glider is

$$F_{\text{Sp}}=k\left|\text{Δ}y\right|=F_{\text{G}}=mg$$

from which the spring’s spring constant is

$$k={\displaystyle \frac{mg}{\left|\text{Δ}y\right|}}={\displaystyle \frac{(0.50\text{kg)(}9.80{\text{m/s}}^2)}{0.035\text{m}}}=140\text{N}\text{/}\text{m}$$

Knowing $k$, you can now find that the glider’s speed was

$$v_0=\sqrt{{\displaystyle \frac{k}{m}}}\text{Δ}{x}_1=\sqrt{{\displaystyle \frac{140\text{N/m}}{0.50\text{kg}}}}(0.027\text{m})=0.45\text{m}\text{/}\text{s}$$

REVIEW A speed of $\approx$0.5 m/s is typical for gliders on an air track.