## Introduction

We know that the energy in certain physical equipment such as springs, the energy is stored in the form of potential energy. The elastic potential energy is the energy stored due to the deformation when an external force is applied. The energy is stored until the force is removed and the object bounces back to its original shape (or size), doing work in the process. The deformation of the object may incorporate processes such as compressing, stretching or twisting the object. In this article, we will learn what is elastic potential energy and elastic potential energy formula along with solved problems that will boost our understanding. Let us start by understanding elastic potential energy in the coming section.

### Elastic Potential Energy Formula

Before we start with the elastic potential energy formula, let us have a look at the definition and the meaning of elastic potential energy.

The elastic potential energy is the total energy obtained as a result of deformation in the shape of an object under consideration.

As the name itself suggests, it is the elastic potential energy, which implies any object which is capable of regaining its original shape and size after deformation will possess the elastic potential energy.

The deformation of the objects is only possible as a result of the acquired potential energy which is known as the elastic potential energy. Thus, it can be said that elastic potential energy is the stored energy of a compressible or stretchable object or in other words stored energy of the perfectly elastic materials.

All the elastic materials such as the twisted rubber, spring, a bouncy ball, get compressed at the moment it strikes the brick wall. All these materials will store the energy in the form of elastic potential energy.

Any object which is particularly designed to store elastic potential energy must have a high elastic limit, however, all the elastic materials or the objects have a certain limit to the applying the load such that they can sustain. When the elastic materials get deformed beyond the elastic limit, the object can not return to its original shape and size, it will get deformed permanently.

### What is the Elastic Potential Energy Formula?

Elastic Potential Energy Formula Derivation: Let us understand what the potential energy formula is?

So, the elastic potential energy formula can be derived from the fundamental equation of elastic force. We know that the external force applied to an elastic body is directly proportional to the total displacement in the position. Mathematically, we write:

F= k x…….(1)

Where,

k - The spring constant

x - The displacement in the body as a result of applied external force

It is computed as the work done to stretch the spring which depends on the spring constant k and the displacement stretched. The spring force is a conservative force and conservative forces have potential energies associated with them. Therefore, the total work done in stretching the elastic material will be stored in the form of potential energy and it is known as the elastic potential energy formula. The elastic potential energy is denoted by the letter U.

Now, the elastic potential energy formula is given by:

U = \[\frac{1}{2}\] (Force \[\times\] Displacement)

U = \[\frac{1}{2}\] (kx) (x)

U = \[\frac{1}{2}\] k x^{2} joules………..(2)

Where,

k - The spring constant

x - The displacement in the body as a result of applied external force

Equation (2) is known as the elastic potential energy formula. If we apply the force in a direction opposite to the motion we will end up with negative potential energy, but we should always remember sign convention is just to identify the direction of applied force does mean the energy stored is negative.

### Elastic Potential Energy Examples:

Let us have a look at the solved numerical problems using the elastic potential energy formula.

1. Two balls are connected by a 0.60 m spring that hits a brick wall, compressing the spring to 0.30 m. The value of spring constant k is 200 N/m. Then, calculate the elastic potential energy stored due to the compression of the spring.

Sol: Given,

The total length of the spring = l = 0.60 m

The total compression in the length of spring = l' = 0.30 m

The total displacement as a result of compression = x = 0.60 - 0.30 = 0.30 m

The value of the spring constant = k = 200 N/m

Now our aim is to calculate the elastic potential energy stored due to the compression of the spring. We know that the elastic potential energy is given by the formula:

U = \[\frac{1}{2}\] k x^{2} joules

Where,

k - The spring constant

x - The displacement in the body as a result of applied external force

Substituting all the given data in the above equation and simplify it. We get:

U = \[\frac{1}{2} k x^{2}\]

U = \[\frac{1}{2} \frac{(200 N/m)}{0.30 m^{2}}\]

U = 9 Nm = 9 joules

Therefore, the elastic potential energy stored due to the compression of the spring is 9 joules.

2. The vertical spring is attached to a load of mass 8kg which is compressed by12 m. Then, calculate the force constant of the spring.

Sol: Given,

The total mass attached to the vertical spring = m = 8 kg

The total displacement of the spring = x = 12 m

Now our aim is to calculate the spring constant of the vertical spring. We know that force applied can be calculated using the equation:

F = m a

Where,

m - The total mass

a - The total acceleration, here since we are using a vertical spring the only acceleration will be, acceleration due to gravity.

Therefore, we write:

F = mg = 89.8 = 78.4 N

Now, the spring force is given by:

F = k x

k = Fx = \[ \frac{78.4}{12} \] = 6.53 N/m

Therefore, the value of the spring constant is 6.53 N/m.

3. The vertical spring is connected with a load of mass 5 kg that is compressed by 10m. Determine the Force constant of the spring.

Solution:Given: Mass m = 5 kg

Distance x = 10 cm

Force formula is given by F = ma = 5 kg × 9.8 m/s²

F = 49 N

Also, Force in the stretched spring is

F = k x

Force Constant k is given by

k = \[\frac{F}{x} = \frac{49}{10}\]

k = 4.9 N/m

Therefore spring constant is 4.9 N/m.

5. A compressed spring has a potential energy of 50 Joule and its spring constant is 200 N/m. Determine the displacement of the spring as a result of this potential energy.

Solution: Given parameters in the problem:Potential energy P.E = 50 J,

Spring Constant, k = 200 N/m,

The Potential energy formula is given by:

P.E = \[\frac{1}{2} k x^{2}\] Rearranging the equation

We get \[x = \sqrt{\frac{2 \times P.E}{K}} \]

Now, Substituting the values, \[x = \sqrt{\frac{2 \times 50}{200}} = \sqrt{0.5} \]

\[x = 0.707\] meter

Therefore displacement will be 0.707 meters.

## FAQs on Elastic Potential Energy Formula

**1. What is The Formula For Elastic Potential Energy or Elastic Energy Formula in Physics?**

The elastic potential energy formula is given by,

U = \[\frac{1}{2}\] k x^{2} joules

**2. Define Elastic Potential Energy?**

This is the energy possessed by the deformation of the shape of the object. Any object that can be deformed and then restored to its original shape can have elastic potential energy. Rubber bands, sponges, and bungee cords are examples of such objects. When we deform these objects, they will return to their original shapes on their own. It is only possible because the accumulated potential energy is elastic potential energy. As a result, elastic potential energy is the energy stored in a compressible or stretchable object.

**3. What is the potential energy?**

The energy possessed by a body as a result of its location with respect to a reference point is known as potential energy. An object can store energy as a function of its position. The heavy ball of a demolition machine saves energy when it is kept in an elevated position. Potential energy is the name given to this stored positional energy. A drawn bow, similarly, may store energy as a function of its stance.

When the bow is in its usual posture, no energy is stored in it (i.e., not pulled). The bow, on the other hand, may store energy when its position is changed from its normal equilibrium condition. Potential energy is the name given to this stored positional energy. The stored energy of an object's location is known as potential energy.

**4. You have an elastic spring that has a spring constant of 1.5 x 10 ^{-2} Newtons per meter, and the spring is compressed by 15.0 cm. What is the PE of the spring?**

The spring constant k = 1.5 x 10^{-2} Newtons/m and the Δs = 15.0 cm = 0.15 m.

PE = \[\frac{1}{2}\] ks^{2}

PE = \[\frac{1}{2}\] x (1.5 x 10^{-2}) Newtons/m (0.15 m)^{2}

PE = 1.69 x 10^{-4} Newtons/m = J

**5. What is Hooke’s Law?**

Hooke's Law asserts that within the elastic limit of a material, the strain of the material is proportional to the applied stress.

F = –k.x

where

F is the force

x is the extension length

k is spring constant in N/m

**6. You attach a Hooke's law spring to a board, and use 3 J to stretch the spring 99 cm. What is the value of the spring constant?**

We know that W = 3 J and s = 99 cm = 0.99 m. From this determine k, the spring constant.

W = \[\frac{1}{2}\] k s2

\[\frac{1}{2}\] k = \[\frac{W}{s^2} \]

k = \[\frac{W}{(\frac{1}{2} s^2)}\]

k = \[(\frac{3J}{[0.5 \times 0.99m^{2}]})\]

k = \[\frac{3J}{0.49m^{2}}\]

K = \[\frac{3 \frac{N}{m}}{0.49m^{2}}\]

k = 6.122 N/m

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