An ideal spring follows Hooke’s law. The spring force is a restoring force, so it always points toward the relaxed
position \(x=0\). If the spring is stretched or compressed, it stores elastic potential energy.
Hooke’s law
The force exerted by an ideal spring is:
\[
F_s=-kx
\]
Here \(k\) is the spring constant in \(\mathrm{N/m}\), and \(x\) is the displacement from the relaxed length.
The minus sign means the force points opposite the displacement.
Elastic potential energy
The elastic potential energy stored in the spring is:
\[
U_s=\frac12kx^2
\]
This energy depends on \(x^2\), so both stretching and compression store positive energy.
Work done by the spring from \(x_1\) to \(x_2\)
Work done by the spring force is the integral of the spring force over displacement:
\[
W_s=\int_{x_1}^{x_2}F_s\,dx
\]
Using \(F_s=-kx\):
\[
\begin{aligned}
W_s
&=
\int_{x_1}^{x_2}(-kx)\,dx\\
&=
-\frac12k\left(x_2^2-x_1^2\right)\\
&=
\frac12k\left(x_1^2-x_2^2\right)
\end{aligned}
\]
Connection with potential energy
The change in elastic potential energy is:
\[
\Delta U_s=U_2-U_1=\frac12k(x_2^2-x_1^2)
\]
Therefore:
\[
W_s=-\Delta U_s
\]
This means the work done by the spring is positive when the spring releases stored energy, and negative when energy
is being stored in the spring.
Sign convention
| Motion |
Energy change |
Work by spring |
Meaning |
| Farther from equilibrium |
\(\Delta U_s>0\) |
\(W_s<0\) |
The spring absorbs energy. |
| Toward equilibrium |
\(\Delta U_s<0\) |
\(W_s>0\) |
The spring gives energy to the block. |
| Same \(|x|\) at start and end |
\(\Delta U_s=0\) |
\(W_s=0\) |
No net spring-energy change. |
Work done on the spring
If an external agent slowly moves the spring from \(x_1\) to \(x_2\), the external work done on the spring is:
\[
W_{\mathrm{on\ spring}}=\Delta U_s
\]
Therefore:
\[
W_{\mathrm{on\ spring}}=-W_s
\]
This is why a spring stretched farther has negative work by the spring but positive work done on the spring.
Worked example
Suppose \(k=200\ \mathrm{N/m}\), and the spring is stretched from \(x_1=0.02\ \mathrm{m}\) to
\(x_2=0.08\ \mathrm{m}\).
Step 1: compute the initial and final elastic potential energies.
\[
U_1=\frac12(200)(0.02)^2=0.04\ \mathrm{J}
\]
\[
U_2=\frac12(200)(0.08)^2=0.64\ \mathrm{J}
\]
Step 2: compute the change in elastic potential energy.
\[
\Delta U_s=0.64-0.04=0.60\ \mathrm{J}
\]
Step 3: use \(W_s=-\Delta U_s\).
\[
W_s=-0.60\ \mathrm{J}
\]
The work done by the spring is negative because the spring is stretched farther from equilibrium. Energy is being
stored in the spring.
Formula summary
| Concept |
Formula |
Meaning |
| Hooke’s law |
\(F_s=-kx\) |
Spring restoring force. |
| Elastic potential energy |
\(U_s=\frac12kx^2\) |
Energy stored in spring deformation. |
| Work by spring from \(0\) to \(x\) |
\(W_s=-\frac12kx^2\) |
Spring work when moved away from relaxed length. |
| Work by spring from \(x_1\) to \(x_2\) |
\(W_s=\frac12k(x_1^2-x_2^2)\) |
Signed work by the spring force. |
| Potential-energy relation |
\(W_s=-\Delta U_s\) |
Spring work equals negative change in elastic PE. |
| Work on spring |
\(W_{\mathrm{on}}=\Delta U_s=-W_s\) |
External work stored in or removed from the spring. |
The key idea is that the spring force is restoring. Moving farther from equilibrium stores energy; moving toward equilibrium releases it.
