Edexcel IAL - Mechanics 3- 2.2 Elastic Potential Energy- Study notes - New syllabus
Edexcel IAL – Mechanics 3- 2.2 Elastic Potential Energy -Study notes- New syllabus
Edexcel IAL – Mechanics 3- 2.2 Elastic Potential Energy -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 2.2 Elastic Potential Energy
Energy Stored in an Elastic String or Spring
When an elastic string or a spring is stretched, work is done against the elastic force. This work is stored as elastic potential energy.
Problems in this topic commonly combine kinetic energy, gravitational potential energy, and elastic potential energy using the work–energy principle.
Elastic Potential Energy of a Spring
For a light spring obeying Hooke’s law \( F = ke \), the elastic potential energy stored when the extension is \( e \) is
\( \dfrac{1}{2}ke^2 \)
This result follows from the work done in stretching the spring from its natural length.
Elastic Potential Energy of an Elastic String
An elastic string has natural length \( l \) and modulus of elasticity \( \lambda \).
If the string is stretched to length \( x \), where \( x > l \), the extension is \( x – l \) and the elastic potential energy stored is
\( \dfrac{\lambda}{2l}(x – l)^2 \)
Using the Work–Energy Principle
In problems involving elastic systems, the work–energy principle is often written as
$\text{Initial kinetic energy + initial potential energies = final kinetic energy + final potential energies}$
Potential energies may include:
- Gravitational potential energy \( mgh \)
- Elastic potential energy of a spring or string
This method avoids the direct use of equations of motion.
Example :
A spring with spring constant \( 500 \,\text{N m}^{-1} \) is stretched by \( 0.08 \) m. Find the elastic potential energy stored in the spring.
▶️ Answer/Explanation
Elastic potential energy of a spring is
\( \dfrac{1}{2}ke^2 \)
\( = \dfrac{1}{2} \times 500 \times 0.08^2 \)
\( = 1.6 \)
Conclusion: The elastic potential energy stored is \( 1.6 \) J.
Example :
A particle of mass \( 2 \) kg is attached to a vertical spring of spring constant \( 400 \,\text{N m}^{-1} \). It is released from rest when the spring is unstretched. Find the maximum extension of the spring.
▶️ Answer/Explanation
At maximum extension, the particle is momentarily at rest.
Loss of gravitational potential energy:
\( mgx \)
Elastic potential energy stored in the spring:
\( \dfrac{1}{2}kx^2 \)
Using energy conservation:
\( mgx = \dfrac{1}{2}kx^2 \)
\( x = \dfrac{2mg}{k} \)
\( x = \dfrac{2 \times 2g}{400} = \dfrac{g}{100} \)
Conclusion: The maximum extension of the spring is \( \dfrac{g}{100} \) m.
Example :
A particle of mass \( 1.5 \) kg is attached to an elastic string of natural length \( 1.2 \) m and modulus of elasticity \( 30 \) N. The string is stretched to a length of \( 1.5 \) m. Find the elastic potential energy stored in the string.
▶️ Answer/Explanation
Extension of the string:
\( x – l = 1.5 – 1.2 = 0.3 \)
Elastic potential energy of an elastic string:
\( \dfrac{\lambda}{2l}(x – l)^2 \)
\( = \dfrac{30}{2 \times 1.2}(0.3)^2 \)
\( = \dfrac{30}{2.4} \times 0.09 = 1.125 \)
Conclusion: The elastic potential energy stored is \( 1.125 \) J.