Hooke’s law – Defintion | formula | Examples

Hooke’s law is a general law of physics which is also known as the law of elasticity. After the distortion of objects or materials, the elasticity can restore them into original shapes. The property of the elasticity can be described by Hooke’s law as a great example. Hooke’s law explains the restoring force that has the ability to return the original shape to the object. In this article, we will learn about the basics of Hooke’s law along with a lot of examples.

Defintion of Hooke’s law

The force needed to compress or extend a spring by some distance (x) scales linearly with respect to that distance. While the spring has the ability to return its original shape when the force is applied and released on it.

There are different forms of springs, metal coil spring is the well-known form of the spring. Spring is almost applied either in the small objects like ball-point or large objects like racing cars engines.

When the force is applied on the spring or other elastic objects, Hooke’s law is used to respond to the objects as a first-order linear approximation. We know that, when a force is applied to an object, the object or materiel must compress or stretch in response to the force.

Hooke’s law is applied for comparatively small bends of an object, the distance or size of the bend is directly proportional to the bending load or force. And can be written as:

F – ∆x

The formula of Hooke’s law

The formula for Hooke’s law is given below.

F = – k∆x

  • F is the force of the spring that is applied.
  • K is the constant of spring.
  • ∆x is the change in distance from initial to final.

The equation of Hooke’s law can also be written as

F = – k (x – x0)

Where x – x0 is a change in distance.   

The formula for Hooke’s law can also be defined for distance and spring constant.

  • For spring constant

K = – F / (x – x0)

  • For distance x

x = -F/k + x

  • For distance x0

x0 = F/k + x

You can use Hooke’s law calculator to get the results according to the above formulas. Following are some steps to use this calculator.

Step 1: Select the option that you want to calculate i.e., force, constant, or distance.

calculation of Hooke's law

Step 2: Select the distance values option.

calculation of Hooke's law

Step 3: Enter the input values.

calculation of Hooke's law

Step 4: Press the calculate button to get the result.

calculation of Hooke's law

Step 5: The step-by-step solution will show below the calculate button.

You can press the show more button to get the solution with steps.

How to calculate Hooke’s law?

By using the formula of Hooke’s law, you can calculate the problems of Hooke’s law. Following are a few examples solved by using the formula of this law.

Example 1: For Hooke’s law

Calculate the force of spring by using Hooke’s law, if the spring force constant is 30 N/m, the distance from equilibrium (x) is 50 m, and the spring equilibrium position (x0) is 35 m.

Solution

Step 1: Write the given data.

Spring constant = k = 30 N/m

Distance form equilibrium = x = 50 m

Spring equilibrium position = x0 = 35 m

Force of spring = F = ?

Step 2: Now take the general equation of Hooke’s law.

 F = – k (x – x0)

Step 3: Put the given values in the above equation.

F = – k (x – x0)

F = -30 N/m (50 m – 35 m)

F = -30 N/m (15 m)

F = -450 N

Hence the spring force of the given problem is -450 N.

Example 2

Calculate the force of spring by using Hooke’s law, if the spring force constant is 70 N/m, the change in distance (∆x) is 40 m.

Solution

Step 1: Write the given data.

Spring constant = k = 70 N/m

Change in the distance = ∆x = 40 m

Force of spring = F = ?

Step 2: Now take the general equation of Hooke’s law.

 F = – k∆x

Step 3: Put the given values in the above equation.

F = – k∆x

F = -70 N/m * 40 m

F = -(70 * 40) N/m * m

F = -280 N

Hence the spring force of the given problem is -280 N.

Example 3: For the spring force constant

Calculate the spring force constant by using Hooke’s law, if the force of the spring is -70 N, the distance from equilibrium (x) is 20 m, and the spring equilibrium position (x0) is 50 m.

Solution

Step 1: Write the given data.

Spring constant = k = ?

Distance form equilibrium = x = 20 m

Spring equilibrium position = x0 = 50 m

Force of spring = F = -70 N

Step 2: Now take the general equation to determine the spring force constant.

K = – F / (x – x0)

Step 3: Put the given values in the above equation.

K = – F / (x – x0)

K = -(-70) N / (20 m – 50 m)

K = 70 N / (-30 m)

K = -7/3 (N/m)

K = -2.33 N/m

Hence the spring force constant of the given problem is -2.33 N/m.

Example 4: For the distance from equilibrium (x)

Calculate the distance from equilibrium (x) by using Hooke’s law, if the force of the spring is 80 N, the spring force constant is 30 N/m, and the spring equilibrium position (x0) is 40 m.

Solution

Step 1: Write the given data.

Spring constant = k = 30 N/m

Distance form equilibrium = x = ?

Spring equilibrium position = x0 = 40 m

Force of spring = F = 80 N

Step 2: Now take the general equation to determine the distance from equilibrium (x).

x = -F/k + x

Step 3: Put the given values in the above equation.

x = -F/k + x

x = -80 N/30 (N/m) + 40 m­

x = -8 N/3 (N/m) + 40 m­

x = -8/3 m + 40 m­

x = -8/3 m + 40 m­

x = -2.667 m + 40 m­

x = 37.3333 m­

Hence the distance from equilibrium (x) of the given problem is 37.3333 m­.

Summary

Hooke’s law is very applicable for small compresses of objects or materials. We have discussed all the basics of Hooke’s law in this post. Now you can solve any problem of Hooke’s law by following this post.

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