Elastic force is the restoring force produced by a deformed spring or elastic object. For an ideal spring, the force is
proportional to the displacement from the natural length. This linear relation is called Hooke’s law.
Hooke’s law
Let \(x\) measure the displacement from the spring’s natural length. In this calculator, \(x>0\) means the block is displaced
to the right, while \(x<0\) means the block is displaced to the left or the spring is compressed.
Elastic restoring force.
\[
\begin{aligned}
F_{\mathrm{el}} &= -kx
\end{aligned}
\]
The minus sign is the important physical idea. It means the elastic force points toward the equilibrium position \(x=0\).
If the spring is stretched to the right, the spring pulls left. If the spring is compressed to the left, the spring pushes right.
Solving for different quantities
Hooke’s law can be rearranged depending on which quantity is unknown. To solve for the spring constant, divide by the
signed displacement:
Spring constant from force and displacement.
\[
\begin{aligned}
F_{\mathrm{el}} &= -kx \\
k &= -\frac{F_{\mathrm{el}}}{x}
\end{aligned}
\]
To solve for the extension or compression, divide by the spring constant:
Displacement from force and spring constant.
\[
\begin{aligned}
x &= -\frac{F_{\mathrm{el}}}{k}
\end{aligned}
\]
A physical ideal spring has \(k>0\). If you enter force and displacement with the same sign, the computed \(k\) would be
negative, which means the signs are inconsistent with a restoring spring force.
Magnitude and energy
Sometimes only the size of the spring force is needed. The magnitude removes the sign:
Force magnitude.
\[
\begin{aligned}
\left|F_{\mathrm{el}}\right| &= k\left|x\right|
\end{aligned}
\]
A deformed spring also stores elastic potential energy. The energy is always nonnegative because it depends on \(x^2\):
Elastic potential energy.
\[
\begin{aligned}
U &= \frac{1}{2}kx^2
\end{aligned}
\]
The energy is zero at the natural length and grows quadratically as the spring is stretched or compressed farther from
equilibrium.
Worked example
Suppose a spring has \(k=250\ \mathrm{N\,m^{-1}}\) and is stretched by \(x=0.040\ \mathrm{m}\).
Compute the elastic force.
\[
\begin{aligned}
F_{\mathrm{el}} &= -kx \\
&= -(250)(0.040) \\
&= -10.0\ \mathrm{N}
\end{aligned}
\]
The negative sign means the restoring force points left, back toward the wall and the natural length.
Compute the stored energy.
\[
\begin{aligned}
U &= \frac{1}{2}kx^2 \\
&= \frac{1}{2}(250)(0.040)^2 \\
&= 0.200\ \mathrm{J}
\end{aligned}
\]
Formula summary
| Goal |
Formula |
Meaning |
| Elastic force |
\(F_{\mathrm{el}}=-kx\) |
Signed restoring force |
| Spring constant |
\(k=-F_{\mathrm{el}}/x\) |
Stiffness, valid for \(x\ne0\) |
| Displacement |
\(x=-F_{\mathrm{el}}/k\) |
Signed extension or compression |
| Force magnitude |
\(|F_{\mathrm{el}}|=k|x|\) |
Size of the restoring force |
| Elastic potential energy |
\(U=\tfrac12kx^2\) |
Energy stored in the deformed spring |
This model assumes an ideal spring, small enough deformation for the force–displacement relation to remain linear, and motion constrained along one axis.
