When you compress or extend a spring – or any elastic material – you’ll instinctively know what’s going to happen when you release the force you’re applying: The spring or material will return to its original length.

It’s as if there is a “restoring” force in the spring that ensures it returns to its natural, uncompressed and un-extended state after you release the stress you’re applying to the material. This intuitive understanding – that an elastic material returns to its equilibrium position after any applied force is removed – is quantified much more precisely by ​Hooke’s law​.

Hooke’s law is named after its creator, British physicist Robert Hooke, who stated in 1678 that “the extension is proportional to the force.” The law essentially describes a linear relationship between the extension of a spring and the restoring force it gives rise to in the spring; in other words, it takes twice as much force to stretch or compress a spring twice as much.

The law, while very useful in many elastic materials, called “linear elastic” or “Hookean” materials, doesn’t apply to ​every​ situation and is technically an approximation.

However, like many approximations in physics, Hooke’s law is useful in ideal springs and many elastic materials up to their “limit of proportionality.” The ​key constant of proportionality in the law is the spring constant​, and learning what this tells you, and learning how to calculate it, is essential to putting Hooke’s law into practice.

The spring constant is a key part of Hooke’s law, so to understand the constant, you first need to know what Hooke’s law is and what it says. The good news it’s a simple law, describing a linear relationship and having the form of a basic straight-line equation. The formula for Hooke’s law specifically relates the change in extension of the spring, ​x​, to the restoring force, ​F​, generated in it:

The extra term, ​k​, is the spring constant. The value of this constant depends on the qualities of the specific spring, and this can be directly derived from the properties of the spring if needed. However, in many cases – especially in introductory physics classes – you’ll simply be given a value for the spring constant so you can go ahead and solve the problem at hand. It’s also possible to directly calculate the spring constant using Hooke’s law, provided you know the extension and magnitude of the force.

The “size” of the relationship between the extension and the restoring force of the spring is encapsulated in the value the spring constant, ​k​. The spring constant shows how much force is needed to compress or extend a spring (or a piece of elastic material) by a given distance. If you think about what this means in terms of units, or inspect the Hooke’s law formula, you can see that the spring constant has units of force over distance, so in SI units, newtons/meter.

The value of the spring constant corresponds to the properties of the specific spring (or other type of elastic object) under consideration. A higher spring constant means a stiffer spring that’s harder to stretch (because for a given displacement, ​x​, the resulting force ​F​ will be higher), while a looser spring that’s easier to stretch will have a lower spring constant. In short, the spring constant characterizes the elastic properties of the spring in question.

Elastic potential energy is another important concept relating to Hooke’s law, and it characterizes the energy stored in the spring when it’s extended or compressed that allows it to impart a restoring force when you release the end. Compressing or extending the spring transforms the energy you impart into elastic potential, and when you release it, the energy is converted into kinetic energy as the spring returns to its equilibrium position.

You’ll have undoubtedly noticed the minus sign in Hooke’s law. As always, the choice of the “positive” direction is always ultimately arbitrary (you can set the axes to run in any direction you like, and the physics works in exactly the same way), but in this case, the negative sign is a reminder that the force is a restoring force. “Restoring force” means that the action of the force is to return the spring to its equilibrium position.

If you call the equilibrium position of the end of the spring (i.e., its “natural” position with no forces applied) ​x​ = 0, then extending the spring will lead to a positive ​x​, and the force will act in the negative direction (i.e., back towards ​x​ = 0). On the other hand, compression corresponds to a negative value for ​x​, and then the force acts in the positive direction, again towards ​x​ = 0. Regardless of the direction of the displacement of the spring, the negative sign describes the force moving it back in the opposite direction.

Of course, the spring doesn’t have to move in the ​x​ direction (you could equally well write Hooke’s law with ​y​ or ​z​ in its place), but in most cases, problems involving the law are in one dimension, and this is called ​x​ for convenience.

The concept of elastic potential energy, introduced alongside the spring constant earlier in the article, is very useful if you want to learn to calculate ​k​ using other data. The equation for elastic potential energy relates the displacement, ​x​, and the spring constant, ​k​, to the elastic potential ​PE​_el, and it takes the same basic form as the equation for kinetic energy:

As a form of energy, the units of elastic potential energy are joules (J).

The elastic potential energy is equal to the work done (ignoring losses to heat or other wastage), and you can easily calculate it based on the distance the spring has been stretched if you know the spring constant for the spring. Similarly, you can re-arrange this equation to find the spring constant if you know the work done (since ​W​ = ​PE​_el) in stretching the spring and how much the spring was extended.

There are two simple approaches you can use to calculate the spring constant, using either Hooke’s law, alongside some data about the strength of the restoring (or applied) force and the displacement of the spring from its equilibrium position, or using the elastic potential energy equation alongside figures for the work done in extending the spring and the displacement of the spring.

Using Hooke’s law is the simplest approach to finding the value of the spring constant, and you can even obtain the data yourself through a simple setup where you hang a known mass (with the force of its weight given by ​F​ = ​mg​) from a spring and record the extension of the spring. Ignoring the minus sign in Hooke’s law (since the direction doesn’t matter for calculating the value of the spring constant) and dividing by the displacement, ​x​, gives:

Using the elastic potential energy formula is a similarly straightforward process, but it doesn’t lend itself as well to a simple experiment. However, if you know the elastic potential energy and the displacement, you can calculate it using:

In any case you’ll end up with a value with units of N/m.

A spring with a 6 N weight added to it stretches by 30 cm relative to its equilibrium position. What is the spring constant ​k​ for the spring?

Tackling this problem is easy provided you think about the information you’ve been given and convert the displacement into meters before calculating. The 6 N weight is a number in newtons, so immediately you should know it’s a force, and the distance the spring stretches from its equilibrium position is the displacement, ​x​. So the question tells you that ​F​ = 6 N and ​x​ = 0.3 m, meaning you can calculate the spring constant as follows:

For another example, imagine you know that 50 J of elastic potential energy is held in a spring that has been compressed 0.5 m from its equilibrium position. What is the spring constant in this case? Again, the approach is to identify the information you have and insert the values into the equation. Here, you can see that ​PE​_el = 50 J and ​x​ = 0.5 m. So the re-arranged elastic potential energy equation gives:

A 1800-kg car has a suspension system that cannot be allowed to exceed 0.1 m of compression. What spring constant does the suspension need to have?

This problem might appear different to the previous examples, but ultimately the process of calculating the spring constant, ​k​, is exactly the same. The only additional step is translating the mass of the car into a ​weight​ (i.e., the force due to gravity acting on the mass) on each wheel. You know that the force due to the weight of the car is given by ​F​ = ​mg​, where ​g​ = 9.81 m/s², the acceleration due to gravity on Earth, so you can adjust the Hooke’s law formula as follows:

However, only one quarter of the total mass of the car is resting on any wheel, so the mass per spring is 1800 kg / 4 = 450 kg.

Now you simply have to input the known values and solve to find the strength of the springs needed, noting that the maximum compression, 0.1 m is the value for ​x​ you’ll need to use:

This could also be expressed as 44.145 kN/m, where kN means “kilonewton” or “thousands of newtons.”

It’s important to stress again that Hooke’s law doesn’t apply to ​every​ situation, and to use it effectively you’ll need to remember the limitations of the law. The spring constant, ​k​, is the gradient of the straight-line ​portion​ of the graph of ​F​ vs. ​x​; in other words, force applied vs. displacement from the equilibrium position.

However, after the “limit of proportionality” for the material in question, the relationship is no longer a straight-line one, and Hooke’s law ceases to apply. Similarly, when a material reaches its “elastic limit,” it won’t respond like a spring and will instead be permanently deformed.

Finally, Hooke’s law assumes an “ideal spring.” Part of this definition is that the response of the spring is linear, but it’s also assumed to be massless and frictionless.

These last two limitations are completely unrealistic, but they help you avoid complications resulting from the force of gravity acting on the spring itself and energy loss to friction. This means Hooke’s law will always be approximate rather than exact – even within the limit of proportionality – but the deviations usually don’t cause a problem unless you need very precise answers.