4.1.1.2b - Elastic Potential Energy in a Stretched Spring
A stretched spring stores energy in its elastic potential store. This lesson is about using the given elastic potential energy equation carefully: choosing the right values, using the right units, and checking that the spring has stayed within its limit of proportionality.
Elastic store in a stretched spring
When a spring is stretched, work is done on the spring and energy is stored in the spring's elastic potential store. If the spring is still behaving elastically, that stored energy can be released when the spring returns towards its original length.
Elastic Potential Energy
Elastic potential energy is energy stored in an elastic object, such as a stretched spring, because its shape has been changed.
For this lesson, focus on a stretched spring. A spring that has not been stretched has no extension, so this equation would give zero elastic potential energy from extension. As the extension increases, the elastic potential energy stored increases.
The extension is not the final length of the spring. It is the increase in length from the spring's unstretched length.
extension = stretched length - unstretched length
For example, a spring with unstretched length 0.12 m and stretched length 0.17 m has extension:
e = 0.17 - 0.12
e = 0.05 m
The equation and units
The Physics equation sheet gives the elastic potential energy equation. In an exam, the important skill here is to recognise when to use it, substitute correctly, and keep the units consistent.
Elastic Potential Energy in a Stretched Spring
This equation is given on the Physics equation sheet, so you apply it rather than recall it.
The symbols and units are:
| Symbol | Quantity | Unit |
|---|---|---|
E_e | elastic potential energy | joule, J |
k | spring constant | newton per metre, N/m |
e | extension | metre, m |
The spring constant tells you how stiff the spring is. For the same extension, a spring with a larger spring constant stores more elastic potential energy.
The extension is squared. This matters: doubling the extension does not double the stored energy. If the spring constant stays the same and the equation still applies, doubling the extension makes the elastic potential energy four times as large.
Worked calculations
A typical calculation gives you the spring constant and the extension. Check the extension is in metres before substituting.
Example: A spring has spring constant 80 N/m and is stretched by 0.25 m. Calculate the elastic potential energy stored.
E_e = 1/2 k e^2
E_e = 0.5 x 80 x (0.25)^2
E_e = 0.5 x 80 x 0.0625
E_e = 2.5 J
The answer is an energy, so the unit is joules.
Unit conversion is a common trap. If the extension is given in centimetres, divide by 100 before using the equation.
Example: A spring has spring constant 500 N/m and is stretched by 6.0 cm.
6.0 cm = 0.060 m
E_e = 1/2 k e^2
E_e = 0.5 x 500 x (0.060)^2
E_e = 0.5 x 500 x 0.0036
E_e = 0.90 J
The limit condition
The elastic potential energy equation has a condition attached:
Use E_e = 1/2 k e^2 only if the limit of proportionality has not been exceeded.
Below the limit of proportionality, the spring is in the proportional region: extension is directly proportional to the force applied, so the spring constant can be treated as constant. The equation is built for that region.
If the limit of proportionality has been exceeded, the force-extension relationship is no longer proportional. The same simple equation may no longer give the correct elastic potential energy, because the spring is no longer behaving in the linear way the equation assumes.
This is why a calculation question may say something like "the limit of proportionality has not been exceeded". That sentence is permission to use the given equation. If a question says the limit has been exceeded, be cautious: do not assume the equation applies unless the question gives a valid method.
After checking the condition, the rest is careful maths: substitute k in N/m, use extension in m, square the extension, and give the final energy in J.
Rearranging for unknown values
Sometimes the question gives the stored elastic potential energy and the spring constant, then asks for the extension. You are still applying the given equation, but you need to rearrange it.
Start with:
E_e = 1/2 k e^2
To make e the subject:
2E_e = k e^2
e^2 = 2E_e / k
e = sqrt(2E_e / k)
Example: A spring stores 3.2 J of elastic potential energy. The spring constant is 160 N/m. The limit of proportionality has not been exceeded. Calculate the extension.
e = sqrt(2E_e / k)
e = sqrt((2 x 3.2) / 160)
e = sqrt(6.4 / 160)
e = sqrt(0.040)
e = 0.20 m
The square root step is needed because the original equation contains e^2.
That final square root is easy to miss. Whenever you rearrange this equation for extension, find e^2 first and then take the square root to get e.
Elastic potential energy calculations depend on three checks: use the given equation, use extension in metres, and only apply it when the limit of proportionality has not been exceeded.