Elastic potential energy
Materials - OCR A-Level Physics
Key Definition
Elastic potential energy
The energy stored in a spring or material when it is stretched or compressed within its elastic limit.
The energy stored in a spring or material when it is stretched or compressed within its elastic limit.
$$E = \frac{1}{2}kx^2$$
$$\begin{aligned}
E &= \frac{1}{2}Fx \\
&= \frac{1}{2}kx^{2}
\end{aligned}$$
- E = $\frac{1}{2}$$kx^{2}$ is the area under the linear region of a force-extension graph (a triangle)
- E = $\frac{1}{2}$Fx only applies when the force-extension relationship is linear (within the Hooke's lawThe extension of a spring is directly proportional to the applied force, provided the limit of proportionality is not exceeded. region)
- For a non-linear graph, the energyThe capacity to do work. Measured in joules (J). stored is the total area under the curve (use counting squares or trapezium rule)
- When a spring is released within the elastic limit, all stored elastic PE is converted back to kinetic energyThe capacity to do work. Measured in joules (J).The energyThe capacity to do work. Measured in joules (J). an object possesses due to its motion.
Worked Example [2 marks]
A spring with spring constantThe force per unit extension of a spring. A measure of the stiffness of the spring. Measured in N m⁻¹. 40 $N m^{-1}$ is stretched by 0.15 m. Calculate the elastic potential energyThe energy stored in a stretched or compressed spring (or other elastic object). stored.
Show Solution
1
$E = \frac{1}{2}kx^{2} = \frac{1}{2} \times 40 \times (0.15)^{2}$
[1]2
$E = \frac{1}{2} \times 40 \times 0.0225 = 0.45 \text{ J}$
[1]Answer
0.45 J
Examiner Tips and Tricks
- If the question gives F and x, use $E = 1/2 Fx$.
- If it gives k and x, use $E = 1/2 kx^2$.
- If it gives F and k, find x first from $F = kx$, then use either formula.