# Elastic Potential Energy and Combination of Springs

In this Physics tutorial, you will learn:

• What is elastic potential energy and how it is calculated?
• What does the area under the elastic force vs deformation graph represent?
• In which kinds of energy, the elastic potential energy can be converted?
• What does the Law of Energy Conversion say?
• How we combine the springs?
• How to find the total spring constant in a certain spring combination setup?
• How the spring constant changes when we combine two or more springs?

## Introduction

Have you ever seen a Middle Ages arc? Do you know what features must a good arc have to make the arrow go as far as possible?

What can you say about the energy of an arrow that goes farther compared to that of an arrow that falls nearer?

## The Meaning of Elastic Potential Energy

In the Physics tutorial "Work. Energy. Types of Energy", it was stated that energy is the ability to do work, i.e. to make an object move at a certain linear distance r by using a force F. When this linear distance lies according to a single basic direction, we can write Δx instead of r.

A compressed spring can also do work on an object when released, i.e. it can push the object by x units by exerting a force F on it. In the Physics tutorial "Types of Forces III (Elastic Force and Tension)", we have called this (restoring) force exerted by an extended or compressed spring as Elastic Force (Fe).

It was found that Fe = - k × x where k is the spring constant and x the deformation (extension or compression) of the spring. The sign minus means the elastic force is in the opposite direction to the direction of spring deformation. The above formula is known as the mathematical expression of the Hooke's Law, whose F vs x graph is shown below.

The Elastic force vs Spring Deformation (Fe vs x) graph for cases when the Hooke's Law is applied, looks like the one shown below:

Also, in the Physics tutorial "Dot (Scalar) Product of Two Vectors" it was stated that Work - as a dot product of force and linear distance - is geometrically represented by the area under the Force vs Position graph. This rule is also valid for the elastic force vs extension graph as well. Therefore, it is obvious that the area under the elastic force vs deformation graph represents the work done to extend or compress a spring but at the same time, it also represents the work done by the spring on objects in contact when it is released and tries to turn back to the unstretched position. This work contributes in changing the potential energy of the spring, which is otherwise known as Elastic Potential Energy, EPE. Look at the graph:

From the above graph, it is obvious that elastic force is not constant. It increases at the same rate from 0 to Fmax. During this process, the spring stretches (or compresses) from 0 to xmax.

From geometry, it is known that the area of a right triangle as the one represented by the graph, is

Area = Base × Height/2

Substituting the above values, we obtain for the elastic potential energy EPE of a spring:

EPE = W = Fe × x/2

Since Fe = k × x (we are dealing only with the value of elastic force, not with its direction; that's why the negative sign does not appear here), we obtain:

EPE = (k × x) × x/2

Or

EPE = k × x2/2

The above equation is the formula of the elastic potential energy stored in an elastic object when it is stretched or compressed. It is always positive because whatever sign we get for x, when it is raised at power two, it becomes positive.

### Example 1

A 2 kg object is hanged on an elastic spring and as a result, the spring stretches by 4 cm as shown in the figure.

What is the elastic potential energy stored in the spring during this process? For simplicity, take g = 10 N/kg.

### Solution 1

First, we need to calculate the spring constant k. Thus, since the object's weight is numerically balanced by the elastic force, we can write:

W = Fe
m × g = k × x

Thus,

k = m × g/x
= 2 kg × 10 N/kg/4 cm
= 2 kg × 10 N/kg/0.04 m
= 20 N/0.04 m
=500 N/m

Therefore, applying the equation of Elastic Potential Energy, we obtain

EPE = k × x2/2
= 500 N/m × (0.04 m)2/2
= 0.8 J/2
= 0.4 J

## Conversion of Elastic Potential Energy into Other Forms of Energy

As a stored energy, the elastic potential energy can be converted in other forms of energy as well. Some of the most widespread examples in this regard include:

### a. From elastic to kinetic energy and vice-versa

A horizontal compressed spring transmits its stored energy to an object attached to it when released, like the one shown in the figure

Initially, the spring is compressed by x units. As a result, it stores elastic potential energy. When released, it transmits its energy to the attached object. As a result, the object of mass m starts moving at velocity v. The chain of energy conversion in this case is

EPE of the spring → KE of the object

We can write:

k × x2/2 = m × v2/2

On the other hand, if a moving object compresses a spring like the one shown in the figure below,

the initial kinetic energy of the object converts entirely into elastic potential energy of the spring.

The chain of energy conversion in this case is

KE of the object → EPE of the spring

We can write:

m × v2/2 = k × x2/2

### b. From gravitational potential energy to elastic potential energy and vice-versa

If an object is released from a height and it compresses a spring when falling on the ground, this is an example of GPE → EPE conversion. Look at the figure:

In fact, GPE of the object is first converted into KE as an intermediate stage and then, KE converts into EPE of the spring. Therefore, the chain of energy conversion is

GPE of the object → KE of the object → EPE of the spring

In symbols, we have the following equation

m × g × h = m × v2/2 = k × x2/2

Neglecting the intermediate stage, we obtain

m × g × h = k × x2/2

On the other hand, a compressed spring in the vertical position can raise the object, which has caused this compression, up to a certain height h as shown in the figure.

Obviously, this process contains an intermediate stage as well, i.e. the stage in which the initial elastic potential energy stored in the compressed spring convers into kinetic energy of the object and then, this kinetic energy converts entirely into gravitational potential energy when the object reaches its highest position h.

The chain of energy conversion in this case is

EPE of the spring → KE of the object → GPE of the object

In symbols,

k × x2/2 = m × v2/2 = m × g × h

Again, if we are not interested on the intermediate stage, but only on what happens at the beginning and at the end of the process, we obtain

k × x2/2 = m × g × h

### c. From elastic into thermal energy

First of all, it would be worth pointing out that friction is a process during which mechanical energy of a moving object converts into thermal energy of both surfaces in contact. Therefore, if a compressed spring (which is a source of stored elastic potential energy) pushes an object attached to it when released, it happens that the object after moving for a certain distance eventually stops because of friction with the ground.

At the end of process, the object has no energy left. This is because the kinetic energy of the object produced by the pushing effect of the compressed spring gradually converts into thermal energy because of friction. We call it the "work done by the frictional force" (in short, Wf), which takes out all the kinetic energy of the object. The chain of energy conversion in this case is

EPE of spring → ME of object → W of friction → 0 (nothing)

The above scheme does not contradict the law of mechanical energy conservation rather, it is a generalization of this law. This generalized law is known as "The Law of Conservation of Energy". It states that:

"Energy can neither be created, nor destroyed; it only can be converted from one form of energy to another."

The above law is one of fundamental laws in physics.

The chain of energy conversions written above, can be mathematically expressed as:

EPE = ME = Wf

Or

EPE = KE + GPE = Wf
k × x2/2 = m × v2/2 + m × g × h = μ × N × ∆x

where x is the extension or compression of the spring and Δx is the displacement of the object. They express very different things.

Let's see a comprehensive example in which all the abovementioned energies are considered.

#### Example 2

A 800 g ball starts rolling at 25 m/s along a horizontal surface by friction coefficient 0.2 as shown in the figure. After moving for 4 m, the ball starts raising up along a smooth slope, and when it reaches the height of 3 m, it continues moving horizontally until it hits a spring. As a result, the ball compresses the spring by 2 cm. Calculate the spring constant. For simplicity, take g = 10 N/kg.

#### Solution 2

First, we must analyse the energy conversions taking place in this process. Thus, the ball initially has only KE because it is at ground level. In the first 4 m of motion, some of this energy converts into thermal energy due to the friction with the ground (we write it as Wf). The remaining KE after this distance, splits into KE and GPE when the object raises at h = 3m. At the end, only the last KE converts to EPE of the spring, as the height doesn't change.

If we put the index 1 at the starting position, 2 after 4 m of displacement, 3 when the ball is just raised at h = 3m and 4 at the end of process, when the ball has compressed the spring as shown in the figure below,

we obtain the following scheme for the energy conversions taking place in the process.

Mathematically, we can write:

KE1 - Wf = KE2 + GPE2

Substituting KE2 with EPE, we obtain

KE1 - Wf = EPE + GPE2

Therefore, we have:

m × v2/2 - μ × m × g × ∆x = k × x2/2 + m × g × h

Substituting the values (we must pay attention in writing m = 0.8 kg instead of 800 g and x = 0.02 m instead of 2 cm), we obtain:

0.8 × 252/2 - 0.2 × 0.8 × 10 × 4 = k × 0.022/2 + 0.8 × 10 × 3
250 - 6.4 = k × 0.0002 + 24
250 - 6.4-24 = k × 0.0002
219.6 = k × 0.0002
k = 219.6/0.0002
= 1 098 000 N/m

## Combination of Springs

The large value of spring constant obtained in the previous exercise, means the spring is very stiff and therefore, it can store a lot of elastic energy when stretched or compressed in maximum.

To change the level of stiffness in springs, we can combine two or more spring in two basic ways:

### a. In series

In this combination, the springs are placed one after another. As a result, when we hang an object on such a system, it pulls each spring if there was only one spring available. Therefore, the same weight causes a double extension in the system composed by two springs in series than if it was one spring only. Look at the figure below. For simplicity, the springs are taken as identical, i.e. they have a constant k each.

Since there is the same force F = m × g that acts in both cases, we have for the single spring system:

k1 = F/x

and for the double spring system:

k2 = F/x + x
= F/2x
= k1/2

This allows us to find a formula for the spring constant ks in series combination of springs. It is

1/ks = 1/k1 + 1/k2

where k1 and k2 are the constants of each spring respectively.

This rule can also extend for a system composed by more than two springs or when the springs are not identical, i.e. when their individual constants are different.

#### Example 3

Two springs are combined in series as shown in the figure. The maximum extension of each spring without losing the elastic properties is 20 cm and 40 cm respectively.

1. What is the spring constant of this combination if the constant of the first spring is 4000 N/m and that of the second spring is 5000 N/m?
2. What is the maximum weight they can hold in such a combination?

#### Solution 3

1. Using the formula for the series combination of springs, we obtain
1/ks = 1/k1 + 1/k2
= 1/4000 + 1/5000
= 5/20000 + 4/20000
= 9/20000

Therefore,

ks = 20000/9 = 2222 N/m
1. We must find the maximum weight each spring can hold and then, considering only the smallest value from the two. We have:
F1 max = k1 × x1 max
= 4000 N/m × 0.2 m
= 800 N

and

F1 max = k1 × x1 max
= 5000 N/m × 0.25 m
= 1250 N

Therefore, the system can hold 800 N at maximum so that both springs preserve their elastic properties.

Remark! Series combination of springs is usually applied when the springs available are stiffer than needed because combining two springs in series reduces the stiffness of the system, making it more elastic.

### b. In Parallel

When two or more springs are connected in parallel, they distribute the load amongst them and as a result, they will extend less than if there was only one spring available. Such a combination creates a system of springs with a higher stiffness. The equation for the spring constant of the system of two parallel springs is

kp = k1 + k2

where kp is the spring constant of the parallel setup, while k1 and k2 are the constants of the individual springs.

In other words, if a single spring with constant k stretches by x units when an object m is hanged on it, then when we use two springs of this kind connected in parallel to hang the same object, they will extend by x/2 because the stiffness of the system increases by a factor of two. As a result, it can store double of energy it could store when a single spring was used. Look at the figure.

#### Example 4

Two identical springs by constants 2000 N/m each, are combined in parallel.

1. What is their extension when a 50 kg object is hanged on them?
2. What is the amount of energy stored in the springs?

For simplicity, take g = 10 N/kg.

#### Solution 4

We can take the figure shown above in theory as a reference. First, we must calculate the constant of the parallel system of springs. We have

kp = k1 + k2
= k + k
= 2k
= 2 × 2000 N/m
= 4000 N/m

a. The force caused in this system of spring is equal to the weight of the hanged object. Thus,

F = W = m × g
= 50 kg × 10 N/kg
= 500 N

Therefore, the extension of the parallel system of springs is

x = F/kp
= 500 N/4000 N/m
= 0.125 m
= 12.5 cm

b. The energy stored in this system of springs is calculated through the formula of elastic potential energy. We have:

EPE = kp × x2/2
= 4000 N/m × (0.125 m)2/2
= 31.25 J

Remark! The energy stored in this system of springs is different (is smaller) than if only one spring was used. Indeed, for the same force, if we had only one spring the extension would be

x = F/k = 500 N/2000 N/m = 0.25 m

As a result, the energy stored in this spring would be

EPE = k × x2/2
= 2000 × 0.252/2
= 62.5 J

As you see, the energy stored is double of the one stored in the parallel setup, confirming the theory explained earlier.

### c. Mixed combination of springs

In this case, we start by calculating the constant of the parallel part and then, we think as having a single spring in that part. It is combined in series with the rest of the springs to get the total spring constant of the entire system. Look at the figure:

For simplicity, we will consider only identical springs. If the mass of the springs and the support are not considered, there is only one force acting on the system. It is the object's weight. Therefore, the constants will be 2k for the parallel part and k for the lower spring, which is in series with the other two. As a result, the total spring constant of the system is

1/ktot = 1/kp + 1/k
= 1/2k + 1/k
= 1/2k + 2/2k
= 3/2k

Therefore,

ktot = 2k/3

As for the total extension of the system, we have

xtot = x/2 + x
= x/2 + 2x/2
= 3x/2

## Summary

The area under the elastic force vs deformation graph represents the work done to extend or compress a spring but at the same time, it also represents the work done by the spring on objects in contact when it is released and tries to turn back to the unstretched position. This work contributes in changing the potential energy of the spring, which is otherwise known as Elastic Potential Energy, EPE.

Giving that the area of a right triangle as the one represented by the graph, is

Area = Base × Height/2

Substituting the relevant quantities, we obtain for the elastic potential energy EPE of a spring:

EPE = k × x2/2

As a stored energy, the elastic potential energy can be converted into other forms of energy as well. Some of the most widespread examples in this regard include:

1. From elastic to kinetic energy and vice-versa
2. From gravitational potential energy to elastic potential energy and vice-versa
3. From elastic into thermal energy
4. etc.

The Law of Conservation of Energy (as one of fundamental laws in physics) states that:

"Energy can neither be created, nor destroyed; it only can be converted from one form of energy to another."

To change the level of stiffness in springs, we can combine two or more spring in two basic ways:

### a. In series

In this combination, the springs are placed one after another. The formula for the spring constant of such a system is

1/ks = 1/k1 + 1/k2

where k1 and k2 are the constants of each spring respectively.

### b. In Parallel

When two or more springs are connected in parallel, they distribute the load amongst them and as a result, they will extend less than if there was only one spring available. Such a combination creates a system of springs with a higher stiffness. The equation for the spring constant of the system of two parallel springs is

kp = k1 + k2

where kp is the spring constant of the parallel setup, while k1 and k2 are the constants of the individual springs.

We can combine the above two basic setups of springs to obtain another setup.

### c. Mixed combination of springs

In this case, we start by calculating the constant of the parallel part and then, we think as having a single spring in that part. It is combined in series with the rest of the springs to get the total spring constant of the entire system.

## Elastic Potential Energy and Combination of Springs Revision Questions

1. A 2 kg object falls from 8 m above the ground on a spring placed vertically and as a result, it compresses the spring by 4 cm. What is the value of spring constant? For simplicity, take g = 10 N/kg.

1. 200 000 N/m
2. 400 000 N/m
3. 20 N/m
4. 40 N/m

2. A 2 kg object starts sliding from rest down a frictionless slope of maximum height 4 m and when it reaches the bottom, it continues sliding for other 5 m along a horizontal plane by friction coefficient 0.3. Then, the object compresses a spring with constant 2000 N/m. Calculate the spring compression in cm. For simplicity, take g = 10 N/kg.

1. 5 cm
2. 22 cm
3. 62.5 cm
4. 0.22 cm

3. Two springs by constants 60 N/m and 300 N/m are first combined in parallel, then in series as shown in the figure.

What is the ratio kp / ks for the two total spring constants obtained in these combinations?

1. 1
2. 0.14
3. 6
4. 7.2