Hooke's Law Formula, Explanation, and Solved Examples

Hooke's Law Formula: Hooke's Law is a fundamental concept in physics and material science that describes the behavior of elastic materials when subjected to mechanical stress. This law was formulated by Robert Hooke, an English scientist, in the 17th century. The central idea behind Hooke's Law is that, within the elastic limit of a material, the deformation (change in shape or size) of the material is directly proportional to the applied force. In other words, when you apply a force to an elastic material, it will stretch or compress in direct proportion to the amount of force applied, as long as you don't exceed the material's elastic limit.

Introduction

Hooke's Law Formula: Hooke's Law is a fundamental principle in the field of physics and material science, describing the behavior of elastic materials when subjected to mechanical stress. Named after the 17th-century scientist Robert Hooke, this law serves as a cornerstone in understanding how materials deform and return to their original shape when the applied force is removed. In this comprehensive article, we will delve into the details of Hooke's Law, provide an in-depth explanation, present the formula, and offer solved examples to help you grasp this concept effectively.

Hooke's Law Formula

Hooke's Law Formulan can be succinctly expressed through a simple formula:

F = -kx

Where:

F represents the force applied to the material.

-k is the spring constant, a measure of the material's stiffness.

x stands for the displacement or deformation of the material from its original position.

Let's break down this Hooke's Law Formula to understand its components better.

Force (F)

The force (F) in Hooke's Law is the external load or stress applied to the material. It can be tension (stretching) or compression (squishing). The direction of the force can be either positive or negative depending on whether the material is stretched or compressed.

Spring Constant (k)

The spring constant (k) is a measure of the material's stiffness. It's an inherent property of the material and is typically expressed in units of force per unit length (e.g., N/m). A higher spring constant indicates a stiffer material, while a lower spring constant signifies a more flexible one.

Displacement (x)

The displacement (x) represents how much the material is stretched or compressed from its original position due to the applied force. It is measured in meters (or other appropriate units of length) and can be either positive or negative based on the direction of deformation.

Hooke's Law Explanation

Hooke's Law describes the linear relationship between the applied force and the resulting deformation of an elastic material. In other words, it states that the deformation of an elastic material is directly proportional to the force applied, as long as the material remains within its elastic limit.

Elastic Limit: Every material has a limit beyond which it will undergo plastic deformation, losing its ability to return to its original shape. Hooke's Law applies strictly within the elastic limit.

Mathematically, this can be expressed as:

F = -kx

In this form, the formula underscores that the force is directly proportional to the displacement. This means that if you double the force applied, the displacement will also double, assuming the material remains within its elastic limit.

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Solved Examples of Hooke's Law Formula

Now, let's work through a couple of examples to illustrate how to use Hooke's Law in practical scenarios.

Example 1: Calculating Force

Suppose you have a spring with a spring constant (k) of 500 N/m, and you stretch it by 0.2 meters. What is the force required to produce this deformation?

Solution:

Using Hooke's Law, F = kx, we can calculate the force:

F = 500 N/m * 0.2 m = 100 N

So, you would need to apply a force of 100 Newtons to achieve a deformation of 0.2 meters in this spring.

Example 2: Finding Deformation

Imagine a force of 120 N is applied to a material with a spring constant (k) of 40 N/m. What will be the deformation (displacement) of the material? Use Hooke's Law Formula

Solution:

Using Hooke's Law, F = kx, we can solve for x:

x = F / k

x = 120 N / 40 N/m = 3 meters

So, the material will be stretched by 3 meters under a 120 N force.

Certainly, let's work through five additional solved examples of Hooke's Law to help you understand its practical applications better.

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Example 3: Determining the Spring Constant

Suppose you have a spring and want to find its spring constant (k). You apply a force of 200 N, and the spring stretches by 0.5 meters. What is the spring constant?Use Hooke's Law Formula

Solution:

Using Hooke's Law, F = kx, we can rearrange the formula to find k:

k = F / x

k = 200 N / 0.5 m = 400 N/m

So, the spring constant of this spring is 400 N/m.

Example 4: Calculating Deformation

You have a material with a known spring constant (k) of 300 N/m. If you apply a force of 150 N to this material, what will be the deformation (displacement)? Use Hooke's Law Formula

Solution:

Using Hooke's Law, F = kx, we can solve for x:

x = F / k

x = 150 N / 300 N/m = 0.5 m

The material will deform by 0.5 meters under a 150 N force.

Example 5: Spring Compression

Now, let's consider a compression scenario. You have a spring with a spring constant (k) of 600 N/m, and you compress it by 0.3 meters. What force is required to compress the spring? Use Hooke's Law Formula

Solution:

Using Hooke's Law, F = kx, we can calculate the force:

F = k * x

F = 600 N/m * 0.3 m = 180 N

You would need to apply a force of 180 N to compress the spring by 0.3 meters.

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Example 6: Tension Force and Deformation

Suppose you have a rope with a spring constant (k) of 80 N/m. You pull the rope with a force of 120 N. How much will the rope stretch?Use Hooke's Law Formula

Solution:

Using Hooke's Law, F = kx, we can solve for x:

x = F / k

x = 120 N / 80 N/m = 1.5 m

The rope will stretch by 1.5 meters under a 120 N tension force.

Example 7: Finding the Spring Constant

Imagine you have a spring and apply a force of 250 N, causing it to stretch by 0.4 meters. What is the spring constant (k) of this spring? Use Hooke's Law Formula

Solution:

Using Hooke's Law, F = kx, we can rearrange the formula to find k:

k = F / x

k = 250 N / 0.4 m = 625 N/m

The spring constant of this spring is 625 N/m.