Elastic Strings & Springs Revision Notes for Edexcel A-Level Further Mathematics

15.1.2 Elastic Potential Energy

Introduction

When a string or spring is stretched or compressed, energy is stored within it as elastic potential energy. This energy is due to the work done to deform the string or spring. The elastic potential energy is given by a formula derived from the principles of mechanics.

This note focuses on:

Understanding elastic potential energy.
Deriving the formula for elastic potential energy.
Applying the work-energy principle to solve problems involving kinetic energy, gravitational potential energy, and elastic potential energy.

Elastic Potential Energy

The elastic potential energy ( $E_e$ ) stored in a stretched or compressed string or spring is given by:

E_e = \frac{1}{2} \frac{\lambda x^2}{l}

where:

$\lambda$ is the modulus of elasticity ( $\text{N}$ )
$x$ is the extension or compression of the string/spring ( $\text{m}$ )
$l$ is the natural length of the string/spring ( $\text{m}$ )

Elastic potential energy is the work done to stretch or compress the spring from its natural length.

Deriving the Formula for Elastic Potential Energy

The work done to stretch a spring or string by an infinitesimal length $dx$ is:

dW = T \, dx

where $T = \frac{\lambda x}{l}$ is the tension in the string or spring at extension $x$ .

To find the total work done in stretching the spring from $x = 0$ (natural length) to $x$ , integrate $dW$ :

W = \int_0^x T \, dx = \int_0^x \frac{\lambda x}{l} \, dx

Perform the integration:

W = \frac{\lambda}{l} \int_0^x x \, dx = \frac{\lambda}{l} \left[ \frac{x^2}{2} \right]_0^x = \frac{\lambda}{l} \cdot \frac{x^2}{2}

Thus, the elastic potential energy stored is:

E_e = \frac{1}{2} \frac{\lambda x^2}{l}

Work-Energy Principle

The work-energy principle can be extended to include elastic potential energy. For a system involving motion and deformation of a spring or string:

\text{Work done by external forces} = \Delta KE + \Delta PE + \Delta E_e

where:

$\Delta KE$ : Change in kinetic energy ( $\frac{1}{2} mv^2$ )
$\Delta PE$ : Change in gravitational potential energy ( $mgh$ )
$\Delta E_e$ : Change in elastic potential energy ( $\frac{1}{2} \frac{\lambda x^2}{l}$ )

Worked Examples

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Example 1: Elastic Potential Energy Stored in a String

Problem

An elastic string has a natural length of 2 m and a modulus of elasticity λ = 50 N

The string is stretched to a total length of 2.5 m

Find the elastic potential energy stored in the string.

Step 1: Calculate the extension:

x = \text{stretched length} - \text{natural length} = 2.5 - 2 = :highlight[0.5 \, \text{m}]

Step 2: Use the formula for elastic potential energy:

E_e = \frac{1}{2} \frac{\lambda x^2}{l}

Step 3: Substitute the values ( $\lambda = 50 \, \text{N}, x = 0.5 \, \text{m}, l = 2 \, \text{m}$ ):

E_e = \frac{1}{2} \times \frac{50 \times (0.5)^2}{2}

Step 4: Simplify:

E_e = \frac{1}{2} \times \frac{50 \times 0.25}{2} = \frac{1}{2} \times \frac{12.5}{1} = :highlight[6.25 \, \text{J}]

Final Answer:

The elastic potential energy stored in the string is 6.25 J

infoNote

Example 2: Applying the Work-Energy Principle

Problem

A particle of mass 0.5 kg is attached to an elastic string with natural length 1 m and modulus of elasticity 40 N

The particle is released from rest with the string initially stretched to 1.5 m.

Ignoring air resistance, find the speed of the particle when the string returns to its natural length.

Step 1: Use the work-energy principle

The total energy at the start equals the total energy when the string returns to its natural length:

E_\text{initial} = E_\text{final}

At the start (stretched position):

E_\text{initial} = KE_\text{initial} + PE_\text{initial} + E_{e_\text{initial}}.

The particle starts from rest ( $KE_\text{initial} = 0$ ) and is at the natural level ( $PE_\text{initial} = 0$ ):

E_\text{initial} = E_{e_\text{initial}} = \frac{1}{2} \frac{\lambda x^2}{l}

Substitute $\lambda = 40 \, \text{N}, x = 0.5 \, \text{m}, l = 1 \, \text{m}$

E_\text{initial} = \frac{1}{2} \times \frac{40 \times (0.5)^2}{1} = \frac{1}{2} \times 40 \times 0.25 = :highlight[5 \, \text{J}]

At the natural length:

E_\text{final} = KE_\text{final} + PE_\text{final} + E_{e_\text{final}}

At the natural length, $E_{e_\text{final}} = 0$ (no extension), and $PE_\text{final} = 0$ (level ground):

E_\text{final} = KE_\text{final} = \frac{1}{2} mv^2

Set $E_\text{initial} = E_\text{final}$

5 = \frac{1}{2} (0.5)v^2

Step 2: Solve for $v$

5 = 0.25v^2 \quad \Rightarrow \quad v^2 = \frac{5}{0.25} = 20

v = \sqrt{20} \approx :highlight[4.47 \, \text{ms}^{-1}]

Final Answer:

The speed of the particle is 4.47 ms⁻¹

Note Summary

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Common Mistakes

Forgetting to calculate the extension ( $x$ ): Remember $x = \text{stretched length} - \text{natural length}$
Incorrectly applying the modulus of elasticity ( $\lambda$ ): Use the correct value for the given spring or string.
Mixing potential and elastic energy: Gravitational PE depends on height, while elastic PE depends on extension or compression.
Ignoring zero values: When motion is at the natural length, $E_e$ is zero.
Units confusion: Ensure energy is always in joules (J)

infoNote

Key Formulas

Elastic Potential Energy:

E_e = \frac{1}{2} \frac{\lambda x^2}{l}

Work-Energy Principle:

\text{Work done by forces} = \Delta KE + \Delta PE + \Delta E_e

Elastic Strings & Springs (Edexcel A-Level Further Mathematics): Revision Notes

15.1.2 Elastic Potential Energy

Introduction

Elastic Potential Energy

Deriving the Formula for Elastic Potential Energy

Work-Energy Principle

Worked Examples

Example 1: Elastic Potential Energy Stored in a String

Example 2: Applying the Work-Energy Principle

Note Summary

Common Mistakes

Key Formulas

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