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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:
-
(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 |
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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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