Elastic Force and Hooke's Law

Hooke's Law states that, within certain limits, the force exerted by an ideal spring is directly proportional to its extension or compression from its equilibrium position. $$ F = -kx $$ where $ F $ represents the force applied to the spring, $ k $ is the spring's stiffness constant, and $ x $ is the displacement of the spring.

Picture holding a spring in your hand: as you stretch it, you feel the resistance increase the farther you pull.

This "elastic" force $ F $ acts in the opposite direction of the displacement $ x $, working to return the spring to its original length $ L $.

example of elastic force

This force doesn’t just occur when you stretch the spring—it also comes into play when you compress it.

In both cases, the elastic force $ F $ opposes the displacement $ x $, striving to restore the spring to its natural length $ L $.

example of spring compression

In the 17th century, English physicist and naturalist Robert Hooke introduced a simple yet revolutionary law to describe this elastic behavior in materials.

Hooke's Law states that the force exerted by an ideal spring is directly proportional to its deformation (stretching or compression) relative to its natural length.

Mathematically, Hooke's Law of a "ideal spring" is expressed as:

$$ F = -kx $$

Where:

$ F $ is the elastic force (directed toward the equilibrium position)
$ x $ is the displacement of the spring from its natural length (positive for stretching, negative for compression)
$ k $ is the spring's elastic constant (also called Hooke's constant), measured in newtons per meter (N/m). A higher $ k $ value indicates a stiffer spring. For instance, a rigid steel spring has a much higher $ k $ than a soft plastic spring.

The negative sign in the formula $ F = -kx $ signifies that the elastic force $ F $ always acts in the direction opposite to the displacement $ x $, a key characteristic that allows the spring to "snap back" to its original shape.

What is an "ideal spring"? An ideal spring is a theoretical construct in physics designed to simplify the analysis of elastic systems. It is assumed to have no mass or friction and behaves perfectly in accordance with Hooke's Law under all conditions, provided the elastic limit is not exceeded. This model enables a focus on the relationship between the applied force and displacement, without factoring in the spring's mass or the real-world properties of the materials it is made from.

The direction of the elastic force changes depending on whether the spring is stretched or compressed. For this reason, Hooke's Law is often written in vector form:

$$ \vec{F} = -k \vec{x} $$

In this equation, $ \vec{F} $ is the elastic force vector, and $ \vec{x} $ is the displacement vector.

Today, Hooke's Law remains a cornerstone of classical physics. Even after more than three centuries, it continues to provide an accurate description of how springs and other elastic systems respond to applied forces.

Limits of Hooke's Law. While powerful, Hooke's Law has its boundaries. It applies only when the deformation is elastic—meaning the spring returns to its original shape after being stretched or compressed. If the displacement exceeds a certain threshold, the material may undergo permanent deformation, breaking the proportionality and making the law invalid. Every spring has a well-defined elastic limit, beyond which the material may deform plastically (resulting in permanent deformation) or even break.

A Practical Example

Imagine you have a spring with an elastic constant $ k = 200 \, \text{N/m} $.

You want to calculate the elastic force the spring exerts when stretched by $ x = 0.1 \, \text{m} $ (10 cm).

Hooke's Law in scalar form is written as:

$$ F = k \cdot x $$

Substituting the values:

$$ F = 200 \, \text{N/m} \cdot 0.1 \, \text{m} $$

$$ F = 20 \, \text{N} $$

Thus, the elastic force exerted by the spring is 20 newtons.

This means that if you apply a force of 20 N to stretch the spring by 10 cm, the spring exerts an opposing elastic force of 20 N, directed back toward the equilibrium position.

In this case, the constant $ k = 200 \, \text{N/m} $ indicates a relatively stiff spring: stretching it by one meter would require a force of 200 newtons.

Example 2

To further illustrate Hooke's Law, consider a spring dynamometer—a device that measures applied force by translating spring elongation into a numerical reading.

dynamometer at rest

When you attach a weight to the end of the spring, it stretches in direct proportion to the weight and the gravitational force acting on the object.

Now, let’s use a softer spring with $ k = 50 \, \text{N/m} $ on the dynamometer.

We hang a weight of $ m = 2 \, \text{kg} $.

a 2 kg mass

How much does the spring stretch?

First, calculate the force $ F $ exerted by gravity on the weight:

$$ F = m \cdot g = 2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} $$

Next, using Hooke's Law ($ F = k \cdot x $), calculate the displacement $ x $:

$$ x = \frac{F}{k} = \frac{19.6 \, \text{N}}{50 \, \text{N/m}} = 0.392 \, \text{m} $$

In this case, the spring stretches by 39.2 cm.

The dynamometer is just one of the many practical applications of Hooke's Law. The principles it describes extend far beyond simple spring mechanics, influencing fields such as vehicle suspension design, earthquake-resistant construction, medical devices like pacemakers, and even molecular vibration studies in chemistry and physics.

In conclusion, these examples highlight how, over 350 years since its discovery, Hooke's Law continues to be an essential tool for solving real-world problems and understanding the behavior of elastic systems.