Elastic potential energy Last week, we saw that the work done by a perfectly elastic spring, in one dimension, depends only on the spring constant and the squares of the displacements from the unstretched position, as given by Hooke's law. Therefore, we can define the difference of elastic potential energy for a spring force as the negative of the work done by the spring force in this equation, before we consider systems that embody this type of force. Thus, ?U =- WAB=??(x)––2), where the object travels from point A to point B. The potential energy function corresponding to this difference is U(z) - 2 kez2+const. If the spring force is the only force acting, it is simplest to take the zero of potential energy at z = 0, when the spring is at its unstretched length. Then, the constant in the equation above is zero. (Other choices may be more convenient if other forces are acting.)
Spring Potential Energy A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm. (a) How much elastic potential energy does the spring contribute when its length is 23 cm? (b) How much more potential energy does it contribute if its length increases to 26 cm? Strategy When the spring is at its unstretched length, it contributes nothing to the potential energy of the system, so we can use the equation above with the constant equal to zero. The value of x is the length minus the unstretched length. When the spring is expanded, the spring's displacement or difference between its relaxed length and stretched length should be used for the x-value in calculating the potential energy of the spring. Solution 1. The displacement of the spring is x = 23 cm - 20 cm = 3 cm, so the contributed potential energy is U = } kz2 = }(4N/cm) (3 cm)2 = 0.18 J. 2. When the spring's displacement is @ = 26 cm - 20 cm = 6 cm, the potential energy is U =4kz2 = {4N/cm) (6 cm)2 = 0.72 J, which is a 0.54-J increase over the amount in part (a). Significance Calculating the elastic potential energy and potential energy differences involves solving for the potential energies based on the given lengths of the spring. Since U depends on x2, the potential energy for a compression (negative x) is the same as for an extension of equal magnitude. Gravitational and elastic potential energy A simple system embodying both gravitational and elastic types of potential energy is a one-dimensional, vertical mass-spring system. This consists of a massive particle (or block), hung from one end of a perfectly elastic, massless spring, the other end of which is fixed, as illustrated below. 1 y= 0 ---- m A NV C m A vertical mass-spring system, with the y-axis pointing downward. The mass is initially at an unstretched spring length, point A. Then it is released, expanding past point B to point C, where