Elastic Potential Energy

When a spring or elastic string is stretched or compressed, it stores elastic potential energy (EPE) — energy held in the strained material that is recovered as the spring relaxes. Because EPE can be traded freely with kinetic and gravitational potential energy, it turns Hooke's-law systems into some of the richest energy-conservation problems in the whole Mechanics option: oscillating masses, bungee jumps, catapults and the "circular motion on a spring" hybrids of the final lesson all rest on the single formula derived here.

1 Where this sits in AQA 7367

Elastic strings and springs sit in the Mechanics option on Paper 3 (7367/3M), the work–energy strand, immediately after the work–energy principle. Paper 3 carries the problem-solving-heavy weighting AO1 40% / AO2 25% / AO3 35%, and EPE is a classic AO3 carrier: a typical question buries the Hooke's-law and EPE formulae inside a multi-stage energy balance (slack phase, stretch phase, gravity doing work) and asks for a speed or a maximum extension. The prerequisites are Hooke's law and the work done by a variable force $W = \int F\,\mathrm{d}x$ from the previous lesson, and conservation of mechanical energy from A-Level Maths Mechanics. We build EPE directly from that work integral, so the topic is genuinely derivable rather than a formula to memorise.

2 Core theory

Hooke's law

The tension in an elastic string or spring is proportional to its extension $x$ (the increase beyond the natural length $l$ ):

$T = \frac{\lambda}{l}\,x = kx,$

where the constants are

Symbol	Name	Note
$T$	tension	newtons; a pull along the string
$\lambda$	modulus of elasticity	newtons — the tension that would double the natural length $(x=l)$
$l$	natural length	the unstretched length
$x$	extension	$x>0$ stretched; for a spring $x<0$ is a compression
$k = \lambda/l$	stiffness (spring constant)	newtons per metre

The two forms $T = \dfrac{\lambda}{l}x$ and $T = kx$ are identical; AQA favours the modulus form, the A-Level Maths textbook often the stiffness form. Hooke's law holds only up to the elastic limit, beyond which the material deforms permanently — examiners assume you stay within it unless told otherwise.

Elastic potential energy

The energy stored when the string is stretched by an extension $x$ is

$\boxed{\;E = \frac{\lambda x^2}{2l} = \tfrac12 k x^2.\;}$

Note the square: doubling the extension quadruples the stored energy. This single fact catches more candidates than any other in the topic.

Derivation — EPE is the work done against tension

EPE is not a new axiom; it is the work done by the stretching force, recoverable on release. At an intermediate extension $s$ the tension is $T(s) = \dfrac{\lambda}{l}s$ , and stretching a further $\mathrm{d}s$ needs work $T(s)\,\mathrm{d}s$ . Integrating from $0$ to the final extension $x$ ,

\begin{aligned} E &= \int_0^x T(s)\,\mathrm{d}s = \int_0^x \frac{\lambda}{l}\,s\,\mathrm{d}s = \frac{\lambda}{l}\left[\frac{s^2}{2}\right]_0^x = \frac{\lambda x^2}{2l}. \end{aligned}

Equivalently, EPE is the area under the tension–extension graph. Because that graph is the straight line $T = \dfrac{\lambda}{l}x$ through the origin, the area is a triangle of base $x$ and height $T = \dfrac{\lambda}{l}x$ :

$E = \tfrac12 \,(\text{base})(\text{height}) = \tfrac12 \,x \cdot \frac{\lambda}{l}x = \frac{\lambda x^2}{2l}. \quad\checkmark$

The two routes agree — a reassuring link between the integral and the geometric picture.

Conservation of energy with elastic elements

In the absence of friction or air resistance, the total mechanical energy is constant:

$\underbrace{\tfrac12 mv^2}_{\text{KE}} + \underbrace{mgh}_{\text{GPE}} + \underbrace{\frac{\lambda x^2}{2l}}_{\text{EPE}} = \text{constant}.$

The whole skill in EPE problems is to list every form of energy at the start and at the end, set the totals equal, and solve. The EPE term simply joins KE and GPE as a third account in the same ledger.

3 Worked examples with mark scheme

Worked Example 1 — EPE of a stretched spring

A spring has natural length $0.5\ \mathrm{m}$ and modulus of elasticity $40\ \mathrm{N}$ . Find the EPE stored when it is stretched by $0.2\ \mathrm{m}$ .

E = \frac{\lambda x^2}{2l} = \frac{40 \times 0.2^2}{2 \times 0.5} = \frac{40 \times 0.04}{1} = 1.6\ \mathrm{J}.

Mark scheme: M1 quote $E = \dfrac{\lambda x^2}{2l}$ and substitute; A1 $1.6\ \mathrm{J}$ . (2 marks.) A frequent slip is to forget the square and write $40 \times 0.2 / 1 = 8$ — dimensionally wrong (that is a force, not an energy).

Worked Example 2 — using the stiffness form

A spring with stiffness $k = 200\ \mathrm{N\,m^{-1}}$ is compressed by $0.1\ \mathrm{m}$ . Find the EPE stored.

E = \tfrac12 k x^2 = \tfrac12 (200)(0.1)^2 = \tfrac12 (200)(0.01) = 1\ \mathrm{J}.

Mark scheme: M1 $E = \tfrac12 kx^2$ ; A1 $1\ \mathrm{J}$ . The energy depends on $x^2$ , so a compression stores exactly as much energy as an equal extension — the sign of $x$ is irrelevant to EPE (though it matters for the direction of the force).

Worked Example 3 — horizontal energy conversion

A $0.5\ \mathrm{kg}$ mass on a smooth horizontal table is attached to a spring $(k = 50\ \mathrm{N\,m^{-1}}$ , natural length $0.3\ \mathrm{m})$ . It is pulled $0.2\ \mathrm{m}$ beyond the natural length and released from rest. Find the speed as it passes through the natural-length position.

There is no change in height, so GPE plays no part — EPE converts entirely to KE.

\begin{aligned} \text{At release:}\quad & E = \tfrac12 (50)(0.2)^2 = 1\ \mathrm{J},\quad \text{KE} = 0,\\ \text{At natural length:}\quad & E = 0,\quad \text{KE} = \tfrac12 (0.5)v^2,\\ \text{Conservation:}\quad & \tfrac12(0.5)v^2 = 1 \;\Rightarrow\; v^2 = 4 \;\Rightarrow\; v = 2\ \mathrm{m\,s^{-1}}. \end{aligned}

Mark scheme: M1 EPE at release; A1 $1\ \mathrm{J}$ ; M1 equate EPE to KE at natural length; A1 $v = 2\ \mathrm{m\,s^{-1}}$ . At the natural length the extension is zero, so all the elastic energy has become kinetic — the natural-length position is where the mass is fastest.

Worked Example 4 — vertical spring (EPE, GPE and KE together)

A $2\ \mathrm{kg}$ mass hangs from a spring $(\lambda = 100\ \mathrm{N},\ l = 0.5\ \mathrm{m})$ . It is pulled down $0.1\ \mathrm{m}$ below the equilibrium position and released. Find its speed as it passes back through equilibrium. (Take $g = 9.8\ \mathrm{m\,s^{-2}}$ .)

Step 1 — equilibrium extension. At equilibrium the tension balances the weight:

$\frac{\lambda}{l}\,e = mg \;\Rightarrow\; \frac{100}{0.5}\,e = 2(9.8) \;\Rightarrow\; 200e = 19.6 \;\Rightarrow\; e = 0.098\ \mathrm{m}.$

Step 2 — the two positions. At equilibrium the extension is $e = 0.098\ \mathrm{m}$ ; pulled down a further $0.1\ \mathrm{m}$ the extension is $0.198\ \mathrm{m}$ . Measure GPE from the lowest (pulled-down) point, where KE $= 0$ :

\begin{aligned} \underbrace{\frac{100(0.198)^2}{2(0.5)}}_{\text{EPE, low}} + 0 + 0 &= \underbrace{\frac{100(0.098)^2}{2(0.5)}}_{\text{EPE, equil}} + \tfrac12(2)v^2 + \underbrace{2(9.8)(0.1)}_{\text{GPE gain}}\\ 3.9204 &= 0.9604 + v^2 + 1.96\\ v^2 &= 3.9204 - 0.9604 - 1.96 = 1.0\\ v &= 1\ \mathrm{m\,s^{-1}}. \end{aligned}

Mark scheme: M1 equilibrium condition $\tfrac{\lambda}{l}e = mg$ ; A1 $e = 0.098\ \mathrm{m}$ ; M1 energy equation with EPE at both positions and the GPE change; A1 correct numerical EPEs; A1 $v = 1\ \mathrm{m\,s^{-1}}$ . (5 marks.) The classic error is to forget the EPE at the equilibrium position (it is not zero there — the spring is still stretched by $e$ ).

Worked Example 5 — elastic string with a slack phase

A particle of mass $0.5\ \mathrm{kg}$ is attached to one end of an elastic string of natural length $1\ \mathrm{m}$ and modulus $20\ \mathrm{N}$ ; the other end is fixed to a point on a smooth horizontal table. The particle is projected directly away from the fixed point at $4\ \mathrm{m\,s^{-1}}$ . How far does it travel before first coming to instantaneous rest?

A string exerts no force until it is taut, so for the first $1\ \mathrm{m}$ (while the particle is within the natural length) the table is smooth and the speed stays $4\ \mathrm{m\,s^{-1}}$ . Only once the string is taut does it begin to stretch and store energy.

\begin{aligned} \text{At natural length:}\quad & \text{KE} = \tfrac12(0.5)(4^2) = 4\ \mathrm{J},\quad E = 0,\\ \text{At max extension } x:\quad & \text{KE} = 0,\quad E = \frac{20\,x^2}{2(1)} = 10x^2,\\ \text{Conservation:}\quad & 4 = 10x^2 \;\Rightarrow\; x^2 = 0.4 \;\Rightarrow\; x = \sqrt{0.4} \approx 0.632\ \mathrm{m}. \end{aligned}

Total distance $= 1 + 0.632 = 1.632\ \mathrm{m}$ . Mark scheme: B1 state the speed is unchanged during the slack $1\ \mathrm{m}$ ; M1 KE at natural length $=$ EPE at maximum extension; A1 $x = \sqrt{0.4}$ ; A1 total $1.63\ \mathrm{m}$ . The two-phase structure — free motion then stretch — is the signature of every elastic-string problem.

Worked Example 6 — mass dropped onto a spring

A particle of mass $0.5\ \mathrm{kg}$ is released from rest $0.4\ \mathrm{m}$ directly above the top of a vertical spring of stiffness $k = 200\ \mathrm{N\,m^{-1}}$ standing on the ground. Find the maximum compression of the spring. (Take $g = 9.8$ .)

The particle falls freely $0.4\ \mathrm{m}$ onto the spring, then compresses it by a further $x$ ; at maximum compression it is instantaneously at rest. All the gravitational PE lost (over the total drop $0.4 + x$ ) converts to elastic PE:

\begin{aligned} mg(0.4 + x) &= \tfrac12 k x^2\\ 0.5(9.8)(0.4 + x) &= \tfrac12 (200)x^2\\ 1.96 + 4.9x &= 100x^2\\ 100x^2 - 4.9x - 1.96 &= 0. \end{aligned}

By the quadratic formula,

$x = \frac{4.9 \pm \sqrt{4.9^2 + 4(100)(1.96)}}{200} = \frac{4.9 \pm \sqrt{808.01}}{200} = \frac{4.9 \pm 28.43}{200},$

giving $x \approx 0.167\ \mathrm{m}$ (the negative root is rejected — a compression cannot be negative). Mark scheme: M1 energy equation including the drop and the compression in the GPE term; A1 correct quadratic; M1 solve; A1 $x = 0.167\ \mathrm{m}$ , rejecting the negative root. The classic error is to take the GPE drop as just $0.4\ \mathrm{m}$ and forget the extra $x$ the particle descends while compressing the spring.

4 Specimen-style exam question

(Specimen-style.) A light elastic string of natural length $0.8\ \mathrm{m}$ and modulus $49\ \mathrm{N}$ hangs from a fixed point $A$ . A particle of mass $2\ \mathrm{kg}$ is attached to the lower end and released from rest at $A$ (i.e. from the level of the fixed point, with the string slack). Taking $g = 9.8\ \mathrm{m\,s^{-2}}$ , find the greatest distance the particle falls below $A$ .

Let the particle fall a total distance $d$ below $A$ . For the first $0.8\ \mathrm{m}$ the string is slack; thereafter the extension is $x = d - 0.8$ . At the lowest point the particle is momentarily at rest, so KE $= 0$ at both ends. Taking the release point $A$ as the GPE zero:

\begin{aligned} \text{loss in GPE} &= \text{gain in EPE}\\ mgd &= \frac{\lambda (d-0.8)^2}{2l}\\ 2(9.8)\,d &= \frac{49\,(d-0.8)^2}{2(0.8)}\\ 19.6\,d &= 30.625\,(d-0.8)^2. \end{aligned}

Expanding and forming a quadratic:

\begin{aligned} 30.625\,(d^2 - 1.6d + 0.64) &= 19.6\,d\\ 30.625\,d^2 - 49\,d + 19.6 &= 19.6\,d\\ 30.625\,d^2 - 68.6\,d + 19.6 &= 0. \end{aligned}

Solving with the quadratic formula,

$d = \frac{68.6 \pm \sqrt{68.6^2 - 4(30.625)(19.6)}}{2(30.625)} = \frac{68.6 \pm \sqrt{2304.96}}{61.25} = \frac{68.6 \pm 48.01}{61.25}.$

This gives $d \approx 1.904\ \mathrm{m}$ or $d \approx 0.336\ \mathrm{m}$ . The second root is less than the natural length $0.8\ \mathrm{m}$ , so the string would not even be taut there — it is spurious (it solves the algebra but violates $x = d - 0.8 > 0$ ). The greatest fall is therefore

$d \approx 1.90\ \mathrm{m}.$

Elastic Potential Energy

Elastic Potential Energy

1 Where this sits in AQA 7367

2 Core theory

Hooke's law

Elastic potential energy

Derivation — EPE is the work done against tension

Conservation of energy with elastic elements

3 Worked examples with mark scheme

Worked Example 1 — EPE of a stretched spring

Worked Example 2 — using the stiffness form

Worked Example 3 — horizontal energy conversion

Worked Example 4 — vertical spring (EPE, GPE and KE together)

Worked Example 5 — elastic string with a slack phase

Worked Example 6 — mass dropped onto a spring

4 Specimen-style exam question

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