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School Physics Notes: Forces Section 4.5 Elastic potential energy calculations

GCSE level physics exam revision notes on Forces 4

4.5 Elastic potential energy formula, energy stored, work done , calculations, problem solving

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[KEY POINTS and learning objectives for this page, after initial notes]

Index of physics notes on FORCES section 4 Elastic potential energy

This page contains online questions only.

Jot down your answers and check them against the worked out answers at the end of the page.

4.5a Elastic potential energy - energy stored and work done - calculations

How do you calculate the potential energy stored in a spring?

Elastic potential energy is the energy stored when some material is stretched or compressed and the energy released when the constriction is released eg the wound up spring of a clockwork clock, a pulled elastic rubber band, stretched coiled metal spring, the compressed spring in a an animal trap, stretched bow  before the arrow is released.

Since elastic potential energy is a form of stored energy, it does nothing until it is released and converted into another form of energy.

 

You can calculate the work done in stretching or compressing a spring = energy transferred = the energy added to 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 store, Ee, in joules, J

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

extension, e, in metres, m

 

This equation is only valid within the limit of proportionality, that is when the spring obeys Hooke's Law.

 

 

Using the experiment results from above we can calculate the stored elastic potential energy for the 0.06 m extension of the spring with a load of 3.0 N.

You can actually do this calculation in two ways.

(a) Using the equation Ee = 1/2 k e2 described above

 k = 50 N/m, e = 0.06 m

 Ee = 0.5 x 50 x 0.062 = 0.09 J

 

BUT, you can also do the stored energy calculation from the graph of results using the graph above and the principle explained by the graph below.

 

(b) Calculating the area under the graph of force versus extension.

This is the area under the shaded triangle of the graph of force versus extension.

Using the same values as above: x = 0.06 m, y = 3.0 N

 area under graph = xy / 2

  Ee = 0.06 x 3.0 / 2  = 0.09 J

Comments on the above two calculations:

(i) Work done = energy = force x distance = N x m = J, but don't forget the halving in calculation (b).

(ii) Method (b) does not directly involve the spring constant k.

(iii) Again, it should be pointed out that these calculations are only valid if the spring is not stretched beyond the elastic limit (limit of proportionality).

Remember, force must be in newtons and extension in metres (cm/100 = m, mm/1000 = m).

At a higher mathematical level what you are doing via the graph is called an integration, here of the work equation W = F x d.

BUT, you cannot use this simple equation to calculate the stored energy because for this equation to be valid the force must be constant, but for a spring the force must be steadily increased to increase the extension so please use Ee = 1/2 k e2 to calculate elastic potential energy unless given sufficient graphical data!

4.5b Questions on spring extension and elastic potential energy

Q1A student carried out an experiment by putting weights on the end of a spring.

After each weight was added, the length of the spring was carefully measured.

The 'idealised' results are summarised below.

Weight added to spring (N) 2 4 6 8 10
Length of spring (cm) 23 27 31 35 39

You can solve these problems with or without a graph.

(i) What is the spring extension per newton weight added?

(ii) What would be the spring length with an added weight of 7N?

(iii) What is the real length of the spring with no added weight on?

(iv) What would be the spring length be with an weight of 15N?

(v)  What assumption are you making in doing the (iii) and (iv) calculations?

Worked out ANSWERS

Q2 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?

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

Worked out ANSWERS

Q3 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.

Worked out ANSWERS

Q4 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?

Worked out ANSWERS

Q5 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?

Quite a tricky question!

Worked out ANSWERS

Q6 A spring stores an extra 20 J of elastic potential energy when stretched an extra 40 cm.

Calculate the spring constant.

Worked out ANSWERS

Q7 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?

Worked out ANSWERS

Q8 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?

Worked out ANSWERS

Q9 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.

A tricky question!

Worked out ANSWERS

 

Sub-index of physics notes: FORCES 4. Elastic potential energy

Key points on elastic potential energy - calculations

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 and CIE GCSE physics 9-1 level science examinations

A structured set of summary revision notes on Elastic Potential Energy, tailored to the GCSE/IGCSE Physics specifications across WJEC, CCEA, CIE, AQA, Edexcel, and OCR exam boards:

Core Concept for Elastic Potential Energy Calculations

  • Definition: The energy stored in an object when it is stretched or compressed, provided it remains within its elastic limit.
  • Energy Store: Part of the mechanical energy store in physics.
  • Examples: Springs, rubber bands, bungee cords, trampolines.

Key Equation for elastic potential energy calculations

Elastic potential energy is calculated using:

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

Ee = 1/2 k e2

Where:

  • Ee = elastic potential energy (Joules)
  • k = spring constant (N/m)
  • x = extension or compression from the natural length (m)

Only valid if the object is not stretched beyond its elastic limit.

Typical Exam Specifications on elastic potential energy calculations

Elastic Potential Energy Coverage
Formula use, energy transfers, spring constant, extension.
Application of formula, Hooke’s Law, energy transfers.
Elastic energy as part of energy stores, calculations, practicals.
Energy transfers in springs, bungee jump scenarios, worked examples.
Elastic energy in context of forces and energy, calculations.
Elastic potential energy in energy stores, formula application, practical examples.

Student Tips for elastic potential energy calculations

  • Memorise the formula and understand each variable.
  • Practice calculations using different spring constants and extensions.
  • Link to Hooke’s Law: ( F = kx ) — helps reinforce understanding of spring behavior.
  • Use Sankey diagrams or bar charts to show energy transfers (e.g. gravitational → kinetic → elastic).
  • Think conceptually: Elastic energy is stored due to deformation, not just movement.
  • Revise practical examples like trampolines, stress balls, and bungee jumps to contextualise energy changes.

Keywords, phrases and learning objectives for elastic potential energy

Be able to use the formula for elastic potential energy to solve problems including energy stored, and work done calculations.

Be able to solve a variety of problems involving elastic potential energy.

Be able to rearrange the potential energy equation.

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Sub-index of physics notes: FORCES 4. Elastic potential energy
ANSWERS to calculation questions on elastic potential energy

Q1A student carried out an experiment by putting weights on the end of a spring.

After each weight was added, the length of the spring was carefully measured.

The 'idealised' results are summarised below.

Weight added to spring (N) 2 4 6 8 10
Length of spring (cm) 23 27 31 35 39

You can solve these problems with or without a graph.

(i) What is the spring extension per newton weight added?

For every 2 N weight added, the spring extends by 4 cm.

Therefore the spring extends 2 cm per N.

(ii) What would be the spring length with an added weight of 7N?

Interpolating the graph/data 33 cm, mid-way between 31 and 35 cm.

(iii) What is the real length of the spring with no added weight on?

Extrapolating back from 23 cm for 2 N gives 23 - 4 = 19 cm for no added weight.

(iv) What would be the spring length be with an weight of 15N?

This weight is 5 N beyond the data limit, but you can use extrapolation with data or graph.

The spring extends 2 cm per N weight added.

Therefore the spring will extend another 5 x 2 = 10 cm,

so the predicted length of the spring is 30 + 10 = 49 cm

(v)  What assumption are you making in doing the (iii) and (iv) calculations?

You are assuming the string is perfectly elastic with no deformation from a linear relationship between spring extension and added weight.

Q2 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)

 

(d) 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).

Q3 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

Q4 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

Q5 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

Q6 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

Q7 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

Q8 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: e = √(2Eepe / k) = √(2 x 0.50 / 20) = 0.224 m = 22.4 cm

Q9 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)

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Sub-index of physics notes: FORCES 4. Elastic potential energy

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