The primary goal of this section is to calculate the work done by a spring. FIGURE 9.20 shows a spring acting on an object as it moves from $s_{\text{i}}$ to $s_{\text{f}}$. The spring force on the object varies as the object moves, but we can calculate the spring’s work by using Equation 9.21 for a variable force. Hooke’s law for the spring is $(F_{\text{Sp}}){}_s=-k\text{Δ}s=-k(s-s_{\text{eq}})$. Thus

FIGURE 9.20 The spring does work on the object.

This is an integration best carried out with a change of variables. Define $u=s-s_{\text{eq}}$, in which case $ds=du$. This changes the integrand from $(s-s_{\text{eq}})ds$ to $udu$. When we change variables, we also have to change the integration limits. At the lower limit, where $s=s_{\text{i}}$, the new variable u is $s_{\text{i}}-s_{\text{eq}}=\text{Δ}{s}_{\text{i}}$. The lower limit becomes the initial displacement. Similarly, $s=s_{\text{f}}$ makes $u=s_{\text{f}}-s_{\text{eq}}=\text{Δ}{s}_{\text{f}}$ at the upper limit. With these changes, the integral is

With a small rearrangement of the right side, we see that the work done by a spring is

Because the displacements are squared, it makes no difference whether the initial and final displacements are stretches or compressions.

The work done by a spring is energy transferred to the object by the force of the spring. We can use this—and the energy principle—to solve problems that we were unable to solve with a direct application of Newton’s laws.

EXAMPLE 9.9 Using the energy principle for a spring

The “pincube machine” was an ill-fated predecessor of the pinball machine. A 100 g cube is launched by pulling a spring back 12 cm and releasing it. The spring’s spring constant is 65 N/m. What is the cube’s launch speed as it leaves the spring? Assume that the surface is frictionless.

MODEL Let the system consist of only the cube, which we model as a particle. Two vertical forces, the normal force and gravity, are perpendicular to the cube’s displacement, and we’ve seen that perpendicular forces do no work. Only the spring force does work.

VISUALIZE FIGURE 9.21 is a before-and-after pictorial representation in which, for horizontal motion, we’ve replaced the generic s-axis with an x-axis.

FIGURE 9.21 Pictorial representation of the pincube machine.

FIGURE 9.21 Full Alternative Text

SOLVE We can solve this problem with the energy principle, $\text{Δ}K=K_1-K_0=W$, where W is the work done by the spring. The initial displacement from equilibrium is $\text{Δ}{x}_0=-0.12\text{m}$. The cube will separate from the spring when the spring has expanded back to its equilibrium length, so the final displacement is $\text{Δ}{x}_1=0\text{m}$. From Equation 9.25, the spring does work

$$\begin{array}{ll}W\hfill & =-\left(\frac{1}{2}k{\left(\text{Δ}{x}_1\right)}^2-\frac{1}{2}k{\left(\text{Δ}{x}_0\right)}^2\right)=\frac{1}{2}(65\text{ N}\text{/}\text{m}){(-0.12\text{ m)}}^2-0\hfill \\ \hfill & =0.468\text{ J}\hfill \end{array}$$

The initial kinetic energy is zero, so the final kinetic energy is simply the energy transferred to the system by the work of the spring: $K_1=W=0.468\text{J}$. Then, from the definition of kinetic energy,

$$v_1=\sqrt{\frac{2K_1}{m}}=\sqrt{\frac{2(0.468\text{ J})}{0.10\text{ kg}}}=3.1\text{ m}\text{/}\text{s}$$

REVIEW $\text{3.1 m/s}\approx 6\text{ mph}$ seems a reasonable final speed for a small, spring-launched cube.