The **Hooke's Law** is used in physics to calculate the elongation of an elastic material (a spring, for example) from the force we apply and the constant of elasticity of the material on which we apply the force.

With the calculator below, you will be able to calculate any of the three variables involved in the mathematical formula for **solving Hooke's Law**. Just select the one you want to calculate, fill in the corresponding data and click on the calculate button to get the result.

Article sections

## Definition of Hooke's Law

As we have said before, Hooke's Law establishes that **the stretch of a spring is proportional to the modulus of force** we apply on it.

Taking into account the definition of this law, we can deduce that in its formula we will have to deal with three unknowns that relate the force, the elongation and, finally, the constant of elasticity that marks the properties of the material.

It is important to know that the springs return to their original position when we stop applying force on them. If this does not happen, it means that we have exceeded their** elasticity limit** and, therefore, has been permanently deformed.

## Hooke's Law Formula

We are going to specify a little more at the theoretical level by looking at the **formula for Hooke's Law** in more detail:

F = k - (x - x_{0}) = k - Δx

The following is explained **what each term corresponds to** of the equation:

- F is the modulus of the force that we apply on the spring and because of this, we should never introduce in the formula a negative value of this variable in Newtons.
- k is the constant of elasticity of the material from which the spring is made. This constant gives us the properties of the object, relating the stretch to the force we apply on it. It is not a fixed value, so each spring can have its own elastic constant.
- x is the length of the spring once it has been stretched
- x
_{0}is the size of the spring before it has been stretched

The term (x - x_{0}) can also be summarized as Δx (increment of x), i.e., we know the length that the spring has been stretched and we do not need to know its initial position and its final position to calculate it.

## Example and solved exercise

We are given a spring on which we apply a force of 20 Newtons and which, due to this force, stretches 20 centimeters in total.

### Calculate the spring elasticity constant:

Here the first thing we have to do is to clear from the equation of Hooke's Law the variable that corresponds to the elastic constant of the spring, i.e. k:

k = F / Δx = 20 Newtons / 20 cm = 1 N/cm

### Calculate the elongation of the spring when a force of 100 N is applied.

Now that we have the elasticity constant, it is easy to solve this section. Again, we clear Δx from the initial equation and we have that:

Δx = F / k = 100 N / 1 N/cm = 100 centimeters = 1 meter

### Find the force that must be applied for the spring to stretch 70 centimeters.

In this case, we apply the equation of Hooke's Law as we have it at the beginning of the explanation. Remember that the elastic constant was calculated in the first section of the exercise.

F = k - Δx = 1 N/cm - 70 cm = 70 N

## Applications of Hooke's Law

The **applications of Hooke's Law** are many and varied. Since it is used to measure the elasticity of a material when a force is applied to it, engineers use Hooke's Law to design mechanical elements.

An example of the application of Hooke's Law is the **dynamometer, an instrument that is used to measure force** applied on a spring or to know how much an object weighs.

In short,** this Law can have multiple uses** will always be present whenever forces intervene on an elastic material that deforms.

EXCELLENT. CONGRATULATIONS

Thank you very much José, we are glad you liked our page focused on Hooke's Law.

Greetings!

thank you very much for your help

hola que tal amig@ quisiera saber si me pueden ayudar a resolver el sig problema como es que debo empezar e leido detalladamente la formula pero no puedo resolver el problema.

a scale has a spring whose constant of elasticity is 1000 N /m . a 10 kg weight is placed on it, how much is its spring compressed ?

consider that ; g = 10 m/ s2

greetings.

Hello José Manuel,

The formula for Hooke's Law is:

F = k - (x - x0) = k - Δx

And from what you comment, in this case the Force is that of gravity, so:

F = m x a = m x g

Therefore, we are left with the following:

mg = k - Δx

We deduct the increase in the length of the spring and we have:

Δx = mg / k = 10Kg x 10m/s

^{2}/ 1000 N / m = 0.1 metersI need help to solve this question.

A spring elongates 2cm when a force of 24N is exerted on it.

What is the value of the elastic constant (k) of the spring?

The new elongation of the spring when a force of 60N is exerted on it and its elastic constant is 12.00, what is its new elongation or elongation?

Hello José,

In the first case, the elastic constant of the spring will be 12.

In the second question, the stretch will be 5 cm.

Greetings!

Hello, I need to solve this exercise please:

We have a 2.5 m pendulum.

Length of 30cm calculate :

A) the velocity of the pendulum in the lower part of the pendulum

B)the acceleration at the ends of the trajectory .

Thank you.