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A certain spring-loaded gun is cocked by compressing its spring by 5.0 cm. The gun fires a 4.0 g projectile with a speed of 8.0 m/s. What spring constant is required for the spring?

Solution

We will ignore friction and other non-conservative interactions, which means that the mechanical energy of the system will be constant. We will further make the usual assumption that the projectile stops interacting with the spring when the spring returns to its equilibrium position (the projectile has essentially been "fired" at that point). Thus, appropriate initial-state final-state diagram and energy bar graphs are:

We can now express the fact that the mechanical energy is constant through the Law of Change:

\begin{large}\[ E_{i} = U_{i} = \frac{1}{2}kx_{i}^{2} = E_{f} = K_{f} = \frac{1}{2}mv_{f}^{2}\]\end{large}

Solving for k gives:

\begin{large}\[ k = \frac{mv_{f}^{2}}{x_{i}^{2}} = \mbox{102 N/m}\]\end{large}