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School Physics notes: (b) Elastic potential energy stores and calculations

GCSE level Physics exam revision notes: Types of energy store

Elastic potential energy defined and examples explained

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ENERGY INDEX: Types of energy & energy stores, energy transfers & selected energy calculations


(b) Elastic potential energy stores and calculations

This page will help you with ... What is a elastic potential energy store?  How solve numerical problems involving elastic potential energy?  How do you explain examples of elastic potential energy  ?  How to use the formula and do calculations involving elastic potential energy There are elastic potential energy questions with worked out answers for exam revision practice

Index of types of energy stores/transfers


This page contains online questions only. Jot down your answers and check them against the worked out answers at the end of the page


Elastic potential energy stores and elastic potential energy transfers

  • This is energy stored when some material is stretched or compressed and the energy released when the constriction is released - usually as kinetic energy.

    • eg the wound up compressed spring of a clockwork clock - provides the energy store to drive the mechanical motor,

    • a pulled elastic rubber band - a stretched catapult stores elastic potential energy, on release the projectile is fired forward as its kinetic energy store is increased,

    • stretched coiled metal spring,

    • the compressed spring in a an animal trap

    • the springs on a car suspension absorb the impact energy of the wheels on the road to give a smoother ride,

    • stretched bow before the arrow is released - the 'twang' increases the kinetic energy store of the arrow..

  • Since elastic potential energy is a form of stored energy, it does nothing until it is released and converted into another form of energy - often converting to kinetic energy.

  • When the forces causing the stretching is removed, the spring or elastic returns to its original length (shape)

  • The more an elastic material is stretched, the greater the elastic potential energy store.

  • The amount of elastic potential energy stored in a stretched spring can be calculated using the equation:

    • elastic potential energy = 0.5 × spring constant × (extension)2,   

    • Ee = 1/2 k e2

    • (assuming the limit of proportionality has not been exceeded)

    • elastic potential energy, Ee, in joules, J

    • spring constant, k, in newtons per metre, N/m

    • extension or compression, e, in metres, m

For more details on elastic potential energy see

Elasticity and energy stored in a spring - experiments & calculations


How to solve elastic potential energy store problems

elastic potential energy = 0.5 × spring constant × (extension)2

Ee = 1/2 k e2

You may also need the spring relationship: force = a spring constant x extension

F = ke

where F = the applied force in newtons (N), e = the spring extension in metres (m)

and k is the spring (elastic) constant in N/m.

You need to be able to rearrange the elastic potential energy formula.


Questions on elastic potential energy

In all questions, unless otherwise indicated, you assume the limit of proportionality is not exceeded, unless this is indicated by extra information.

Q1 A spring with spring constant of 5.00 N/m is stretched for an extra 10.0 cm.

ANSWERS

 

Q2 A spring has a spring constant of 2000 N/m.

(a) If the elastic potential energy store of the spring is 50.0 J, how far is the spring compressed?

(b) What force is needed to compress the spring by 15.0 cm in length?

ANSWERS

 

Q3 It takes 5.0 J of work to stretch a spring 20 cm.

How much extra work must be done to stretch it another 20 cm?

(i) You need to work out the spring constant.

(ii) Then work out the total work to stretch the spring a total of 40 cm.

(iii) You then subtract (i) from (ii) to get the extra work done.

ANSWERS

 

Q4 A spring stores an extra 20 J of elastic potential energy when stretched an extra 40 cm.

Calculate the spring constant.

ANSWERS

 

Q5 A stretched string has a total length of 60 cm and a spring constant of 240 N/m.

If the stretched spring is storing 20 J of energy, what is the length of the unstretched spring to the nearest cm?

ANSWERS

 

Q6 A spring is fixed firmly in a vertical position. When a mass of 120.0 g is attached to the spring it extends in length by 3.2 cm.

(a) Assuming the gravitational field strength is 9.8 N/kg, calculate the spring constant k.

(b) Calculate the extra elastic potential energy stored in the spring as a result of the added weight.

(c) If an extra 200 g mass is placed on the spring, how much longer will it get?

(d) What force is needed to extend the spring by 30 cm?

(e) What important assumption have you made concerning the calculations in (a) to (c)?

ANSWERS

 

Q7 A spring has a spring constant of 20.0 N/m.

What would the extension be in cm, if 0.50 J of work was done on stretching the spring?

ANSWERS

 

Q8 For this question you need to know the formula for gravitational potential energy (GPE) and gravitational field constant g = 9.8 m/s2 or 9.8 N/kg). Watch out for units, remember in the end to work in J, m and kg.

A very difficult question!

A fun toy consists of a spring and a funny head on the end.

When the toy spring is compressed 5.0 cm, and released, it leaps vertically up into the air.

The toy spring has a mass of 20 g.

When released it leaps up to a maximum height of 75 cm.

If we assume all the elastic potential energy (EPE) is transferred to the GPE store of the toy spring, deduce the spring constant of the spring.

ANSWERS


Key points for Physics - elastic potential energy store conversions - energy transfers involving elastic potential energy

Information sources for Doc Brown's key points: IGCSE-GCSE physics are based on textbooks & syllabus-specifications for students taking the UK AQA, Edexcel, OCR 21st Century Science, OCR Gateway science suite, WJEC, CCEA & CIE GCSE physics 9-1 level science examinations.

A syllabus-aligned summary of Elastic Potential Energy Stores, tailored for GCSE/IGCSE Physics students across WJEC, CCEA, CIE, AQA, Edexcel, and OCR exam boards:


Elastic Potential Energy Stores – GCSE Physics Revision Notes

What is Elastic Potential Energy?

  • Definition: The energy stored in an object when it is stretched or compressed.
  • Found in elastic materials like springs, rubber bands, trampolines, and bungee cords.
  • It is a type of potential energy because it depends on the object’s position or shape.

Elastic Potential Energy Formula

Ee = 1/2 k x2

  • Eelastic = elastic potential energy (Joules)
  • k = spring constant (N/m)
  • x = extension or compression from the equilibrium position (m)

Only valid if the object obeys Hooke’s Law and is within its elastic limit.

Elasticity and energy stored in a spring - experiments & calculations


Real-World Uses of Elastic Potential Energy

Application Description
Trampolines Springs stretch and store energy when someone lands, then release it to bounce them up.
Bungee Jumping The cord stretches, storing energy, then pulls the jumper back up.
Catapults/Slingshots Elastic bands store energy when pulled back, then release it to launch objects.
Car Suspension Springs absorb shocks by storing and releasing energy.
Toys (e.g. wind-up) Springs store energy when wound, then release it to power movement.

Energy Transfers Involving Elastic Potential Energy

Scenario Energy Transfer Pathway
Stretching a spring Kinetic → Elastic Potential
Releasing a stretched spring Elastic Potential → Kinetic
Bungee jump Gravitational Potential → Kinetic → Elastic Potential → Kinetic
Trampoline bounce Kinetic → Elastic Potential → Kinetic

Exam Tips for All Boards

Use correct terminology:

  • Say “energy is transferred to the elastic potential store” not “energy is stored in the spring.”

Know when to apply the formula:

  • Only use Ee = 1/2 k x2 if the object is not stretched beyond its elastic limit.

Draw energy transfer diagrams:

  • Use arrows to show energy flow between stores (e.g., Sankey diagrams or bar charts).

Understand Hooke’s Law:

  • F = kx if the force is proportional to extension x up to the elastic limit.

Practice with graphs:

  • The area under a force-extension graph = elastic potential energy.

Extra examples involving elastic potential energy

Let’s stretch our imagination a bit further - here are more real-world examples of elastic potential energy in action:

Elasticity and energy stored in a spring - experiments & calculations


Everyday Objects and Activities

  • Wind-up toys: Coiled springs store energy when wound, then release it to power movement.
  • Archery bows: Drawing the string stores energy, which is transferred to the arrow when released.
  • Diving boards: Flex and store energy just before launching a diver into the air.
  • Rubber bands: Classic example - stretching stores energy, which is released when snapped or used to launch something.

Engineering and Design

  • Shock absorbers in vehicles: Springs compress and extend to absorb bumps, storing and releasing energy to smooth the ride.
  • Mechanical watches: Springs store energy when wound and release it gradually to keep time.
  • Door closers: Compress a spring when opened, then release energy to close the door gently.

Medical and Safety Equipment

  • Orthodontic braces: Use elastic bands to apply continuous force, storing and releasing energy to move teeth.
  • Prosthetic limbs: Some designs use elastic materials to store energy during movement and release it for a more natural gait.
  • Crash barriers: Flexible materials deform on impact, storing energy to reduce force on vehicles.

Sports and Recreation

  • Tennis rackets: Strings stretch slightly on impact, storing energy and releasing it to return the ball.
  • Pole vaulting: The pole bends and stores energy, then straightens to launch the athlete upward.
  • Spring-loaded gym equipment: Resistance bands and springs store energy during use and release it to assist or resist motion.

Keywords, phrases and learning objectives on elastic potential energy

Be able to do calculations and solve problems using the formulae for elastic potential energy

Ee = 1/2 k e2    and  F = ke

Be able to describe and explain examples of elastic potential energy stores.


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 INDEX ENERGY: Types, stores, transfers, energy calculations

ANSWERS to the Questions on elastic potential energy

In all questions, unless otherwise indicated, you assume the limit of proportionality is not exceeded, unless this is indicated by extra information.

Q1 A spring with spring constant of 5.00 N/m is stretched for an extra 10.0 cm.

How much extra energy is stored in the elastic potential energy store of the spring by this extension.

Eepe = 1/2 k e2,  10.0 cm ≡ 10.0 / 100 = 0.10 m

Eepe = 0.5 x 5.0 x 0.102

Eepe = 0.025 J

 

Q2 A spring has a spring constant of 2000 N/m.

(a) If the elastic potential energy store of the spring is 50.0 J, how far is the spring compressed?

Eepe = 1/2 k e2,  rearranging gives e2 = 2Eepe / k,  e = √(2Eepe / k)  and  e = the compression

e = √(2Eepe / k) = √(2 x 50 / 2000) = √0.063 = 0.224 m  (22.4 cm, 3 s.f.)

(b) What force is needed to compress the spring by 15.0 cm in length?

F = ke = 2000 x 15/100 = 300 N

 

Q3 It takes 5.0 J of work to stretch a spring 20 cm.

How much extra work must be done to stretch it another 20 cm?

(i) You need to work out the spring constant.  20 cm ≡ 20 / 100 = 0.20 m

Eepe = 1/2 k e2,  rearranging gives k = 2Eepe / e2

k = (2 x 5.0) / (0.20 x 0.20) = 250 N/m

(ii) Then work out the total work to stretch the spring a total of 40 cm.

The total work done on the spring equals its elastic potential energy store when fully stretched 40 cm (which is 0.40 m). Since you now know the spring constant, you use the same equation again, but solving for the total elastic potential energy.

Eepe = 1/2 k e2 = 0.5 x 250 x 0.402 = 20 J

(iii) You then subtract (i) from (ii) to get the extra work done.

Therefore the extra work done = 20 - 5 = 15 J

 

Q4 A spring stores an extra 20 J of elastic potential energy when stretched an extra 40 cm.

Calculate the spring constant.

40 cm = 0.40 m.

Eepe = 1/2 k e2, rearranging:

k = Eepe x 2 / e2 = 20 x 2 / 0.402 = 250 N/m

 

Q5 A stretched string has a total length of 60 cm and a spring constant of 240 N/m.

If the stretched spring is storing 20 J of energy, what is the length of the unstretched spring to the nearest cm?

Eepe = 1/2 k e2, rearranging and 60 cm = 0.60 m

e = √(2Eepe / k) = √(2 x 20 / 240) = √(1/6) = 0.408 m

0.408 m = 40.8 cm, ~41 cm = extra length added to the stretched string

Therefore original length of spring = 60 - 41 = ~19 cm

 

Q6 A spring is fixed firmly in a vertical position. When a mass of 120.0 g is attached to the spring it extends in length by 3.2 cm.

(a) Assuming the gravitational field strength is 9.8 N/kg, calculate the spring constant k.

100 g is equivalent to a weight (force) of 9.8 x 120/1000 = 1.176 N

The extension e = 3.2/100 = 0.032 m. Force F = 1.176 N

(The equation F = ke has already been dealt with in detail further up this page)

F = ke, so the spring constant k = F ÷ e = 1.176 ÷ 0.032 = 36.8 N/m  (3 sf)

 

(b) Calculate the extra elastic potential energy stored in the spring as a result of the added weight.

Using the equation for elastic stored energy: Ee = 1/2 k e2   (dealt with in detail already on this page)

Ee = 1/2 k e2  = 1/2 x 36.75 x 0.032 x 0.032 = 0.019 J

 

(c) If an extra 200 g mass is placed on the spring, how much longer will it get?

200 g equates to a weight of 9.8 x 200/1000 = 1.96 N, k = 36.75 N/m

F = ke, so the spring extension e = F ÷ k = 1.96/36.75 = 0.053 m (5/3 cm)

 

(d) What force is needed to extend the spring by 30 cm?

force required = F = ke = 36.75 x 30/100 = 11.0 N (1 dp, 3 sf)

 

(e) What important assumption have you made concerning the calculations in (a) to (c)?

You have assumed the spring behaves truly elastic i.e. within the elastic limit (limit of proportionality).

 

Q7 A spring has a spring constant of 20.0 N/m.

What would the extension be in cm, if 0.50 J of work was done on stretching the spring?

Elastic potential energy formula: Eepe = 1/2 k e2

rearranging: extension e = √(2Eepe / k) = √(2 x 0.50 / 20) = 0.224 m = 22.4 cm

 

Q8 For this question you need to know the formula for gravitational potential energy (GPE) and gravitational field constant g = 9.8 m/s2 or 9.8 N/kg). Watch out for units, remember in the end to work in J, m and kg.

A fun toy consists of a spring and a funny head on the end.

When the toy spring is compressed 5.0 cm, and released, it leaps vertically up into the air.

The toy spring has a mass of 20 g.

When released it leaps up to a maximum height of 75 cm.

If we assume all the elastic potential energy (EPE) is transferred to the GPE store of the toy spring, deduce the spring constant of the spring.

GPE = mgh = (20 / 1000) x 9.8 x (75 / 100) = 0.147 J

Neglecting air resistance we can say GPE = EPE for the energy store transfer

Elastic potential energy formula: Eepe = 1/2 k e2

rearranging gives: k = Eepe x 2 / e2 = 0.147 x 2 / (5.0 /100)2 = 118 N/m (3 sf)

 INDEX ENERGY: Types, stores, transfers, energy calculations

Elasticity and energy stored in a spring - experiments & calculations

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