Elastic Potential Energy

What is Elastic Potential Energy?

It takes putting in energy to stretch an elastic string beyond its natural length
- This is the work done in stretching it
Once it is stretched and held in position, the energy put in is now stored in the string
- This is a type of potential energy, because the string has the potential to contract and release it
This stored energy is called the elastic potential energy (EPE)
- Note that the "work done to stretch it" and the "elastic energy stored in it" have the same value
  - Questions may use either phrase
It also takes energy to compress a spring from its natural length
- The compressed spring has energy stored in it, ready to "spring" open
- This is also elastic potential energy

How do I calculate Elastic Potential Energy?

Let an elastic string (or spring) of natural length $l$ metres and modulus of elasticity $λ$ N be stretched to a new length of $(l + x)$ metres, where $x$ metres is the extension
- The formula for elastic potential energy is $\frac{λ}{2 l} x^{2}$
  - Sometimes written $\frac{1}{2} (\frac{λ}{l}) x^{2}$ (in a form like kinetic energy)
  - The units are in Joules
  - It's also the same as "work done by stretching"
This formula works for the compression of a spring
- $x$ represents the length of compression from its natural length
The formula can be derived by knowing that, in general, work done is the area under a force-distance graph
- The area under the graph of $T = \frac{λ}{l} x$ can be found by integration to give $\frac{λ}{2 l} x^{2}$
  - Or by noting it's a triangle of base $x$ and height $\frac{λ}{l} x$ then using $\frac{1}{2}$ × base × height

Exam Tip

Be careful when selecting which formula to use, as they are very similar!
- $T = \frac{λ}{l} x$ is Hooke's Law, for finding a force
- $E P E = \frac{λ}{2 l} x^{2}$ is for finding the elastic potential energy

Worked example

An elastic string of natural length 1.5 metres and modulus of elasticity 60 N is stretched to a total length of 3.5 metres.

Find the energy stored in the string.

worked example showing how to find the elastic potential energy stored in a string

Work-Energy Principle with Elasticity

How do I include Elastic Potential Energy in the Work-Energy Principle?

The Work-Energy Principle is an energy balance
- Total final energy = total initial energy ± work done
  - - or, using subscripts for final and initial, $E_{f} = E_{i} \pm W D$
"Total energy" can now include elastic potential energy (EPE)
- e.g. the total initial energy is $E_{i} = G P E_{i} + K E_{i} + E P E_{i} = m g h_{i} + \frac{1}{2} m {v_{i}}^{2} + \frac{λ}{2 l} {x_{i}}^{2}$
The ± work done terms now refer to any non-gravitational and non-elastic forces
- e.g. friction (-), driving force (+), air resistance (-), etc
- But not weight ( $m g$ ) and no longer tensions that use Hooke's Law ( $\frac{λ}{l} x$ )
  - These are both already accounted for in the GPE and EPE parts

How do I know when to use the Work-Energy Principle?

You can use it when a particle moves from one position to a different position, for example:
- when it has been pulled down and released from rest
- when it has been projected at a certain speed
- when you need to find the distance it has moved
- This would not include a particle hanging at rest (in equilibrium)
  - Here, you could use Newton's 2nd Law + Hooke's Law
You can use it when a question involves finding speeds
- Speeds can be found from kinetic energy
  - Not from Newton's 2nd Law or Hooke's Law
You should not use it if asked to find the initial acceleration
- Acceleration is part of Newton's 2nd Law
It is not required if asked to find the equilibrium extension
- The particle is not moving, so Newton's 2nd Law + Hooke's Law works
SUVAT equations cannot be used for particles on elastic strings or springs as their acceleration is not constant
- They can only be used if particles detach from the elastic string (e.g. the string breaks, or becomes slack, etc)
  - The particle then becomes a projectile under gravity
  - Don't forget that the Work-Energy Principle can also be used on projectiles, and in some cases (e.g. finding speeds) it's much quicker than SUVAT

Exam Tip

The maximum speed for a particle on an elastic string occurs when it's acceleration is zero, which is when it passes through its equilibrium position
- You may need to find this position using Newton's 2nd Law + Hooke's Law in equilibrium

Worked example

A particle of mass $m$ kg is attached to one end of a light elastic spring of natural length $l$ metres and with modulus of elasticity $4 m g$ N. The other end of the spring is attached to the point at the top of a rough slope of length $3 l$ metres at 30° to the horizontal. The particle is held on the bottom of the slope and released from rest. The coefficient of friction between the particle and the ramp is $\frac{\sqrt{3}}{4}$ .

Find, in terms of $g$ and $l$ , the speed of the particle as it passes through the point halfway up the slope.

worked example showing how to use the work-energy principle in conjunction with elastic potential energy

part 2 of worked example showing how to use the work-energy principle in conjunction with elastic potential energy

Elastic Potential Energy (Edexcel A Level Further Maths: Further Mechanics 1)

Revision Note