Elastic (Spring) Potential Energy

The Problem:

An elastic (Hooke's Law) force is a conservative force - it stores energy. How much? In other words, what is the elastic potential energy of a spring that has been stretched a distance x?

The Solution:

Following the 2-step process for calculating potential energy in general:

  1. It would be convenient to pick the unstretched, at rest position of the spring to be the EPE = 0 point.
  2. The elastic potential energy (EPE) at a distance x, then, equals the negative of the work done by the Hooke's Law force as the spring is stretched from x = 0 to its final position. Since we know that we do work at least equal to (1/2)kx2 in stretching the spring a distance x, the spring must do -(1/2)kx2 work Therefore, the elastic potential energy of the spring at point x is given by:
    1. epe = (1/2)kx^2


Example 1: If the force to stretch a spring is given as F = (100 N/m)x, then what is the potential energy of the spring if it is stretched 2 meters from rest?

Solution: Here k = 100N/m and x = 2 m. Therefore:

EPE = (1/2)kx2 = (1/2)(100N/m)(2 m)2 = 200 Joules

Example 2: It takes a force of 20 Newtons to hold a spring stretched to a distance of 40 cm. What is the elastic potential energy of the spring at this position?

Solution: We know that F = 20 N when x = 40 cm = 0.40 m. Since F = kx, then k = F/x = (20 N)/(0.40 m) = 50 N/m. Now, EPE = (1/2)kx2 = (1/2)(50 N/m)(0.40 m)2 = 4 Joules.

last update December 26, 2005 by JL Stanbrough