Section 1 of 5Storing and Releasing Energy
Any object that is stretched, squashed or bent can hold energy in its elastic potential store.
Work done on a spring
- Stretching or compressing a spring moves its ends through a distance, so work is done and energy is transferred into the store.
Energy release
- When the spring is let go, the stored energy is transferred mechanically to the kinetic store of whatever the spring is attached to.
Limit of proportionality
- The energy equation only holds while extension stays proportional to force,
- so the spring must not be stretched beyond this point.
Elastic limit
- Beyond the elastic limit a spring is permanently deformed and will not return to its original length once the force is removed.
Section 2 of 5Calculating the Stored Energy
Work through these steps in any stretched spring question.
1
Find the extension by subtracting the natural length from the stretched length.
2
Convert the extension from centimetres to metres by dividing by 100 before using it.
3
Substitute the spring constant in N/m and the extension in metres into E = ½ × k × e².
4
Round the answer in joules to the number of significant figures given in the question.
Section 3 of 5Key Relationships
Key Relationships
Spring constant k
- Measured in newtons per metre; a stiffer spring has a larger k and stores more energy for the same extension.
Extension e
- The difference between stretched and natural length, always entered in metres inside the equation.
Squared term
- Energy depends on extension squared, so doubling the extension multiplies the stored energy by four.
Stored energy
- Equals the work done on the spring when it is stretched within its limit of proportionality, with no energy dissipated.
Section 4 of 5Worked Example
Question
- A spring with spring constant 40 N/m has a natural length of 12.0 cm.
- A force stretches it to a length of 16.0 cm.
- Calculate the elastic potential energy stored in the spring.
- Extension e = 16.0 − 12.0 = 4.0 cm
- Convert to metrese = 4.0 ÷ 100 = 0.040 m
- Write the equationE = ½ × k × e²
- SubstituteE = ½ × 40 × (0.040)² = 20 × 0.0016
Answer E = 0.032 J (to 2 significant figures).
Section 5 of 5Exam Checklist & Tips
2.1.2 – Key Points for the Exam
- Elastic potential store The store holding energy in an object that has been stretched, compressed or bent by work done on it.
- Spring constant The force needed per unit extension of a spring, measured in newtons per metre.
- Extension The increase in length of a spring compared with its natural length.
- Limit of proportionality The point beyond which force and extension are no longer directly proportional.
- Elastic limit The point beyond which a material stays permanently deformed when the load is removed.
Common Mistakes
- Using centimetres directly in E = ½ × k × e², which makes the squared term 10 000 times too large.
- Using the stretched length instead of the extension, which is the difference from the natural length.
- Forgetting to square the extension, which gives an answer that is far too large.
- Applying E = ½ × k × e² to a spring stretched beyond its limit of proportionality.