#
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?

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The Solution:

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

- It would be convenient to pick the unstretched, at rest
position of the spring to be the EPE = 0 point.
- 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)kx
^{2} in
stretching the spring a distance x, the spring must do
-(1/2)kx^{2} work Therefore, the elastic potential energy
of the spring at point x is given by:

##

Examples:

**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)kx^{2} = (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)kx^{2} = (1/2)(50 N/m)(0.40 m)^{2} = **4
Joules**.

last update December 26, 2005 by JL
Stanbrough