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The elastic potential energy formula is Ee = ½ke², where k is the spring constant in N/m and e is the extension in metres. Free GCSE Physics notes.
Elastic Potential Energy Calculations is part of Energy changes in a system, and the ways energy is stored before and after such changes in GCSE AQA Higher Physics.
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The elastic potential energy formula is Ee = ½ke², where k is the spring constant in N/m and e is the extension in metres. Free GCSE Physics notes.
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Use this lesson to stay inside Energy changes in a system, and the ways energy is stored before and after such changes while connecting the detail back to the wider Energy topic.
Read the core explanations from Elastic Potential Energy Calculations before testing yourself from memory.
When you pull back a catapult or compress a spring in a pen, you're doing work against the elastic forces. That energy doesn't disappear — it gets stored in the stretched material, ready to be released.
Think about a trampoline. As you land on it, your weight stretches the springs underneath. The energy from your fall gets stored in those stretched springs. When they snap back, that stored energy launches you upwards again.
This stored energy is called elastic potential energy, and it's everywhere in engineering — from car suspension systems to archery bows. Understanding how to calculate it is crucial for predicting how these systems behave.
To calculate elastic potential energy, you need to understand three key quantities. Each one affects how much energy gets stored, but not in the way you might expect.
A stiff spring (high k value) needs more force to stretch it the same distance as a weak spring (low k value). Car springs have k values around 20,000 N/m, while a slinky toy might be only 1 N/m.
Extension is not the total length — it's how much longer the spring becomes. If a spring's natural length is 10 cm and it stretches to 15 cm, the extension is 5 cm (0.05 m).
Here's the crucial insight: doubling the extension doesn't double the stored energy — it quadruples it. This is because energy depends on extension squared, making small increases in stretch create large increases in stored energy.
The formula for elastic potential energy comes from the work done stretching a spring. As you pull harder, the spring fights back with increasing force, so the work done isn't force times distance.
Notice the factor of ½ in the formula. This appears because the force increases linearly with extension (Hooke's Law), so the average force during stretching is half the final force. The energy stored is this average force multiplied by the distance.
The e² term is what makes this formula powerful but dangerous in calculations. Small measurement errors in extension get magnified when you square them. Always double-check your extension values.
This formula works while the spring behaves elastically, meaning it returns to its original length when the force is removed. A spring can go beyond the limit of proportionality and still be elastic, but once it goes beyond the elastic limit it becomes permanently deformed. At that point the simple GCSE modelling assumptions no longer apply cleanly.
Let's tackle some typical exam questions. The key is methodical substitution and careful attention to units — many students lose marks by mixing centimetres and metres.
Worked example: Basic Elastic Potential Energy Calculation
A spring with spring constant 200 N/m is stretched by 8.0 cm. Calculate the elastic potential energy stored. [6 marks]
Answer: 0.64 J
Notice how the extension was squared: (0.080)² = 0.0064. This is where many students make errors — they forget to square the extension or they square it incorrectly.
Worked example: Rearranging for Spring Constant
A spring stores 2.5 J of elastic potential energy when extended by 5.0 cm. Calculate the spring constant. [6 marks]
Answer: 2000 N/m
When rearranging, remember that the factor of ½ becomes a factor of 2 when you multiply both sides. This is a common source of errors in exam questions.
Elastic potential energy isn't just a physics curiosity — it's the principle behind countless technologies you use every day. Understanding the calculations helps explain why these systems work the way they do.
Car suspension systems use springs to absorb energy from bumps in the road. The springs compress when you hit a pothole, storing the impact energy as elastic potential energy. This energy is then released gradually, giving you a smooth ride instead of a jarring impact.
Archery provides a perfect example of energy conversion. Drawing a bow stores elastic potential energy in the bent bow limbs. When released, this energy converts to kinetic energy of the arrow. A bow with higher 'draw weight' (spring constant) stores more energy for the same draw length.
Even your mattress uses elastic potential energy. The springs inside compress under your weight, storing energy that helps support you throughout the night. The spring constant determines whether the mattress feels firm or soft.
Elastic potential energy questions are mark-rich in GCSE physics, but they're also where students commonly lose points through careless errors. The calculations aren't complex, but the details matter enormously.
Unit conversion is the biggest trap. Extensions are often given in centimetres or millimetres, but the formula requires metres. Always convert first, then substitute. Write your conversion clearly — examiners give method marks for showing this step.
The squaring operation catches many students. When extension doubles, energy quadruples, not doubles. Check your arithmetic carefully, especially when dealing with decimal numbers. (0.05)² = 0.0025, not 0.25.
Formula rearrangement questions are common. Practice isolating k and e from the standard formula. Remember that when you move the ½ to the other side, it becomes 2. Show each algebraic step clearly for method marks.
Watch out for questions that ask about the validity of the formula. The elastic potential energy equation only works when Hooke's law applies, so force is proportional to extension. Beyond the limit of proportionality, the graph stops being a straight line. The spring may still return to its original shape until it reaches the elastic limit, which is the point beyond which permanent deformation begins.
In exam questions, it is helpful to separate three ideas: extension, limit of proportionality and elastic limit. Extension tells you how much the spring has stretched. The limit of proportionality tells you when Hooke's law stops giving a straight-line relationship. The elastic limit tells you when the spring stops returning to its original shape. Keeping these ideas separate prevents common mark-scheme mistakes.
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Move to the related lessons or back to the topic page when you need the wider sequence before exam-style practice.
The elastic potential energy formula is Ee = ½ke², where k is the spring constant in N/m, e is the extension in metres and energy is in joules (J). It only applies while the spring obeys Hooke's Law, before the limit of proportionality.
Elastic potential energy is energy stored in a stretched or compressed elastic object that can be released when the object returns to its original shape. It is measured in joules (J) and calculated with Ee = ½ke².
The spring constant is measured in newtons per metre (N/m), because it is the force needed per unit extension. Car springs have k values around 20,000 N/m, while a slinky toy might be only 1 N/m.
The spring constant (k) is a measure of how stiff a spring is, defined as the force needed per unit extension and measured in newtons per metre (N/m). A stiff spring has a high k value; car springs have k values around 20,000 N/m, while a slinky toy might be only 1 N/m.
To calculate elastic potential energy, convert the extension to metres, then use Ee = ½ke². A spring with spring constant 200 N/m stretched by 8.0 cm (0.080 m) stores ½ × 200 × 0.0064 = 0.64 J.
Examples of elastic potential energy include a pulled-back catapult, a compressed spring in a pen, trampoline springs, car suspension springs, a drawn archery bow and mattress springs. In archery, energy stored in the bent bow limbs converts to kinetic energy of the arrow.
Extension is the increase in length of a spring when a force is applied, measured from its natural length in metres. Extension is not the total length: a spring with a natural length of 10 cm stretched to 15 cm has an extension of 5 cm, or 0.05 m.
To find the spring constant, rearrange Ee = ½ke² to give k = 2 × Ee ÷ e². A spring storing 2.5 J when extended by 5.0 cm (0.050 m) has k = 5.0 ÷ 0.0025 = 2000 N/m. The factor of ½ becomes a factor of 2 when rearranging.
The limit of proportionality is the point beyond which a spring no longer obeys Hooke's Law, so extension is no longer proportional to force. The elastic limit is the point beyond which permanent deformation begins and the spring stops returning to its original shape.
In the elastic potential energy equation Ee = ½ke², k is the spring constant. The spring constant measures how stiff a spring is and is defined as the force needed per unit extension, measured in newtons per metre (N/m).
The half in Ee = ½ke² appears because the force increases linearly with extension (Hooke's Law), so the average force during stretching is half the final force. The energy stored is this average force multiplied by the distance.
Elastic potential energy quadruples when the extension doubles, because energy depends on extension squared in Ee = ½ke². Small increases in stretch therefore create large increases in stored energy.
The elastic potential energy equation only applies when Hooke's law holds, so force is proportional to extension. Beyond the limit of proportionality the force–extension graph stops being a straight line, and beyond the elastic limit the spring is permanently deformed.
A stiff spring has a high spring constant, so it needs more force to stretch it the same distance as a weak spring with a low spring constant. Spring constant is measured in N/m.
The energy stored in a stretched spring is found with Ee = ½ke² after converting the extension to metres. A spring with k = 200 N/m stretched by 8.0 cm has e = 0.080 m, so Ee = ½ × 200 × 0.0064 = 0.64 J.