Hooke's law explains how spring force changes with displacement from equilibrium. For a spring that stays in its linear elastic range, the force magnitude is

F=kxF = kx

If you track direction along one axis, the restoring-force form is

Fx=kxF_x = -kx

Here, kk is the spring constant and xx is the extension or compression measured from equilibrium. The minus sign means the spring force points back toward equilibrium, opposite the displacement.

What the spring constant means

kk measures stiffness. A larger kk means you need more force to stretch or compress the spring by the same amount.

Its SI unit is N/m\mathrm{N/m}. For example, if two springs are both stretched by 0.02 m0.02\ \mathrm{m}, a spring with k=200 N/mk = 200\ \mathrm{N/m} needs four times the force of a spring with k=50 N/mk = 50\ \mathrm{N/m}.

When F=kxF = kx is valid

Hooke's law is not a rule for every spring in every situation. It works only while the material stays in its approximately linear elastic range.

In intro physics, that usually means the spring returns to its original shape when released and the force-displacement graph is still close to a straight line. If the spring is overstretched or permanently deformed, F=kxF = kx is no longer a reliable model.

Worked example: a spring stretched 3 cm

A spring has

k=200 N/mk = 200\ \mathrm{N/m}

and it is stretched by

x=0.03 mx = 0.03\ \mathrm{m}

Use the magnitude form first:

F=kx=200×0.03=6 NF = kx = 200 \times 0.03 = 6\ \mathrm{N}

So the spring force has magnitude 6 N6\ \mathrm{N}.

If you choose the stretch direction as positive, then the force component is

Fx=6 NF_x = -6\ \mathrm{N}

because the spring pulls back toward equilibrium.

This example shows the two parts students usually mix up: kk sets the stiffness, and the minus sign gives the restoring direction.

A quick intuition check

Hooke's law is linear. If the displacement doubles, the force doubles, as long as the spring is still in the linear elastic range.

That gives you a quick check on your answer. If 0.03 m0.03\ \mathrm{m} gives 6 N6\ \mathrm{N}, then 0.06 m0.06\ \mathrm{m} should give 12 N12\ \mathrm{N} under the same conditions.

Common Hooke's law mistakes

Using total length instead of displacement

xx is the change in length from equilibrium, not the full length of the spring.

Mixing up force magnitude and signed force

F=kxF = kx is commonly used for magnitude. If you are tracking direction in one dimension, write Fx=kxF_x = -kx.

Using centimeters with a spring constant in N/m

If kk is in N/m\mathrm{N/m}, convert the displacement to meters before calculating.

Applying the law after the spring is overstretched

Once the spring is outside its linear elastic range, the force may stop being proportional to displacement.

Where Hooke's law is used

Hooke's law appears in spring balances, force sensors, suspension systems, vibration models, and introductory simple harmonic motion. It also helps as a first model for elastic behavior in many small-deformation problems.

The main reason it matters is practical: once the force is proportional to displacement, many mechanical systems become much easier to analyze.

Try a similar spring problem

Keep the same spring, k=200 N/mk = 200\ \mathrm{N/m}, but change the stretch to 0.05 m0.05\ \mathrm{m}. Compute the new force magnitude, then decide its direction from the sign convention you chose.

If you want one more step, compare that result with a spring that has half the value of kk. That is the fastest way to see what the spring constant really controls.

Frequently Asked Questions

What does Hooke's law state?
Hooke's law says that for a spring in its linear elastic range, the force magnitude is F = kx, where k is the spring constant and x is the extension or compression measured from equilibrium. The restoring-force form includes a minus sign, meaning the spring force points back toward equilibrium, opposite the displacement.
What does the spring constant k measure?
The spring constant k measures stiffness, with SI unit newtons per meter. A larger k means you need more force to stretch or compress the spring by the same amount. For example, at the same 0.02 meter stretch, a spring with k of 200 N/m needs four times the force of one with k of 50 N/m.
When is Hooke's law valid?
Hooke's law works only while the spring stays in its approximately linear elastic range. In intro physics, that means the spring returns to its original shape when released and the force-displacement graph stays close to a straight line. If the spring is overstretched or permanently deformed, F = kx is no longer a reliable model.
How do you calculate spring force using Hooke's law?
Multiply the spring constant by the displacement from equilibrium. For a spring with k of 200 N/m stretched by 0.03 m, the force is 200 times 0.03, which equals 6 newtons. If you take the stretch direction as positive, the force component is negative 6 newtons because the spring pulls back toward equilibrium.
What happens to spring force if displacement doubles?
Hooke's law is linear, so if the displacement doubles, the force doubles, as long as the spring stays in its linear elastic range. This gives a quick check on your answer: if a 0.03 meter stretch gives 6 newtons, then a 0.06 meter stretch should give 12 newtons under the same conditions.

Need help with a problem?

Upload your question and get a verified, step-by-step solution in seconds.

Open GPAI Solver →