Energy cannot be created or destroyed ā only transferred or transformed between stores
What you will learn
ā” State the law of conservation of energy in your own words
š Identify energy stores and the pathways between them
š Calculate useful and wasted energy in a system
š„ Explain why energy is always dissipated to thermal stores
š” Apply conservation of energy to real-world scenarios
š§® Use energy equations involving kinetic, gravitational, and elastic stores
ā” The Law of Conservation of Energy
Law of Conservation of Energy: Energy cannot be created or destroyed. It can only be transferred from one object to another or transformed from one store to another. The total energy in a closed system always remains constant.
This is one of the most fundamental laws in all of physics. It applies everywhere ā from a bouncing ball to a nuclear reactor. No matter what process takes place, if you add up all the energy in every store before and after, the total is always the same.
We talk about energy being stored in different energy stores. Energy moves between these stores through pathways (sometimes called mechanisms of transfer).
Total energy input = Total energy output (useful + wasted)
The word "wasted" does not mean the energy disappears ā it simply means that some energy is transferred to a store that is not useful for the intended purpose. Most commonly this is the thermal (heat) store of the surroundings.
šļø Energy Stores and Pathways
AQA requires you to know the following energy stores:
Energy Store
Example
Kinetic
A moving car, a spinning top
Gravitational potential
A book on a shelf, a roller coaster at the top of a loop
Elastic potential
A stretched spring, a compressed rubber ball
Chemical
Food, fuel, batteries
Thermal (internal)
A hot cup of tea, a warm radiator
Nuclear
Uranium fuel rods in a reactor
Magnetic
A permanent magnet attracting iron filings
Electrostatic
A charged Van de Graaff generator
Energy is transferred between stores through pathways:
Pathway
Example
Mechanically (by a force)
Pushing a box up a ramp
Electrically
Current flowing through a wire
By heating
A hot plate warming a pan
By radiation (waves)
Sunlight reaching Earth
š§® Key Energy Equations
To apply conservation of energy quantitatively, you need to calculate the energy in different stores.
Kinetic Energy Store:
Ek = Ā½ Ć m Ć vĀ²
Symbol
Quantity
Unit
Ek
Kinetic energy
Joules (J)
m
Mass
Kilograms (kg)
v
Speed/velocity
metres per second (m/s)
Gravitational Potential Energy Store:
Ep = m Ć g Ć h
Symbol
Quantity
Unit
Ep
Gravitational potential energy
Joules (J)
m
Mass
Kilograms (kg)
g
Gravitational field strength
N/kg (use 10 N/kg)
h
Height above reference point
Metres (m)
Elastic Potential Energy Store:
Ee = Ā½ Ć k Ć eĀ²
Symbol
Quantity
Unit
Ee
Elastic potential energy
Joules (J)
k
Spring constant
N/m
e
Extension or compression
Metres (m)
š„ Dissipation ā Where Does "Wasted" Energy Go?
In any real energy transfer, some energy is always dissipated to the thermal store of the surroundings. This happens because of friction, air resistance, or electrical resistance. This energy spreads out and becomes less useful ā it is harder to use again for a specific purpose.
Dissipation: The process by which energy is spread out into the surroundings (usually as thermal energy), making it less useful and harder to recover.
Common causes of dissipation:
Friction between moving surfaces ā thermal energy store of the surfaces
Air resistance ā thermal energy store of the air
Electrical resistance in wires ā thermal energy store of the wire
Sound produced by vibrations ā thermal energy store of the surroundings
Even when energy is "wasted," the total energy in the system and surroundings is always the same. The law of conservation of energy is never broken.
š Sankey Diagrams and Energy Calculations
A Sankey diagram is a visual way to show how energy is transferred in a system. The width of each arrow represents the amount of energy. The main arrow going into the system represents the total input energy. Arrows branching off represent useful and wasted outputs.
From conservation of energy:
Energy input = Useful energy output + Wasted energy output
You can rearrange this to find any unknown:
Wasted energy = Energy input ā Useful energy output
Useful energy = Energy input ā Wasted energy
Efficiency tells us what fraction of the input energy is usefully transferred:
Efficiency = Useful energy output Ć· Total energy input
Efficiency can be expressed as a decimal (0 to 1) or as a percentage (multiply by 100). No real device can have an efficiency of 1 (or 100%) because some energy is always dissipated. Conservation of energy tells us that efficiency can never exceed 100%.
A higher efficiency means less energy is wasted. We can reduce dissipation by using lubrication, better insulation, and more efficient components.
Example 1: A 2 kg ball is dropped from a height of 5 m. Assuming no air resistance, calculate the speed of the ball just before it hits the ground. (g = 10 N/kg)
1 Identify what is happening. The ball starts at height h = 5 m with zero kinetic energy. All energy is in the gravitational potential store. As it falls, this is transferred entirely to the kinetic store (no air resistance means no dissipation).
2 Calculate the gravitational potential energy at the top:
Ep = m Ć g Ć h = 2 Ć 10 Ć 5 = 100 J
3 By conservation of energy, all GPE converts to KE at the bottom:
Ek = 100 J
4 Use the kinetic energy equation to find speed:
Ek = Ā½ Ć m Ć vĀ²
100 = Ā½ Ć 2 Ć vĀ²
100 = vĀ²
v = ā100 = 10 m/s
The ball hits the ground at 10 m/s. Total energy is conserved: 100 J of GPE ā 100 J of KE.
Example 2: A light bulb is supplied with 500 J of electrical energy. It produces 125 J of light energy and the rest is wasted as thermal energy. Calculate (a) the wasted energy and (b) the efficiency of the bulb.
1 Write down the known values:
Total energy input = 500 J
Useful energy output (light) = 125 J
2 Apply conservation of energy to find wasted energy:
Wasted energy = Energy input ā Useful energy output
Wasted energy = 500 ā 125 = 375 J
3 Calculate the efficiency:
Efficiency = Useful energy output Ć· Total energy input
Efficiency = 125 Ć· 500 = 0.25
4 Express as a percentage:
Efficiency = 0.25 Ć 100 = 25%
(a) Wasted energy = 375 J. (b) Efficiency = 25%. Note: 500 J in = 125 J (light) + 375 J (thermal) = 500 J total. Energy is conserved.
Example 3: A spring with spring constant k = 400 N/m is compressed by 0.05 m. When released, it fires a 0.1 kg ball horizontally. Assuming all elastic potential energy converts to kinetic energy, find the speed of the ball.
1 Calculate the elastic potential energy stored in the compressed spring:
Ee = Ā½ Ć k Ć eĀ²
Ee = Ā½ Ć 400 Ć (0.05)Ā²
Ee = Ā½ Ć 400 Ć 0.0025
Ee = 0.5 J
2 By conservation of energy, all elastic PE converts to KE of the ball:
Ek = 0.5 J
3 Use the kinetic energy equation:
Ek = Ā½ Ć m Ć vĀ²
0.5 = Ā½ Ć 0.1 Ć vĀ²
0.5 = 0.05 Ć vĀ²
vĀ² = 0.5 Ć· 0.05 = 10
v = ā10 ā 3.16 m/s
The ball is launched at approximately 3.16 m/s. Elastic PE store (0.5 J) ā Kinetic store (0.5 J). Energy is conserved.
Example 4: A 500 g ball rolls down a ramp from a height of 2 m. At the bottom it has a speed of 5.5 m/s. Calculate the energy dissipated by friction during the descent. (g = 10 N/kg)
1 Convert mass to kg: m = 500 g = 0.5 kg
2 Calculate initial GPE at the top of the ramp:
Ep = m Ć g Ć h = 0.5 Ć 10 Ć 2 = 10 J
3 Calculate KE at the bottom of the ramp:
Ek = Ā½ Ć m Ć vĀ² = Ā½ Ć 0.5 Ć (5.5)Ā²
Ek = Ā½ Ć 0.5 Ć 30.25 = 7.56 J
4 Apply conservation of energy. The "missing" energy was dissipated by friction:
Energy dissipated = Ep ā Ek
Energy dissipated = 10 ā 7.56 = 2.44 J
Energy dissipated by friction = 2.44 J. This went to the thermal store of the ball and ramp. Total: 10 J = 7.56 J (KE) + 2.44 J (thermal). Energy is conserved.
Question 1: Which of the following best describes the law of conservation of energy?
Question 2: A 3 kg object is lifted to a height of 4 m. What is the gravitational potential energy stored? (g = 10 N/kg)
Question 3: A motor transfers 800 J of electrical energy. 600 J is transferred usefully as kinetic energy. Calculate the wasted energy in joules.
Question 4: Which energy store is energy most commonly dissipated into?
Question 5: A 1 kg ball is dropped from rest and hits the ground at 6 m/s. Calculate the kinetic energy of the ball just before impact. Give your answer in joules.
Challenge 1: A roller coaster car of mass 800 kg starts from rest at the top of a 30 m high hill. Assuming no friction or air resistance, calculate:
(a) The gravitational potential energy at the top.
(b) The speed of the car at the bottom of the hill.
(c) Explain, using the law of conservation of energy, why the car cannot rise above 30 m on the next hill if friction is present.
(a) Ep = m Ć g Ć h = 800 Ć 10 Ć 30 = 240,000 J (240 kJ)
(b) All GPE ā KE at bottom (no friction):
Ek = 240,000 J
Ā½ Ć 800 Ć vĀ² = 240,000
vĀ² = 240,000 Ć· 400 = 600
v = ā600 ā 24.5 m/s
(c) With friction present, some energy is dissipated to the thermal stores of the track and wheels. This means not all 240,000 J is available as GPE on the next hill. Since Ep = mgh, with less energy available the car can only reach a height less than 30 m. Conservation of energy tells us the car cannot gain more energy than it started with, so it can never go higher than the original starting point.
Challenge 2: A spring launcher (k = 800 N/m) is compressed by 0.08 m to launch a 0.2 kg ball vertically upward. Using conservation of energy:
(a) Calculate the elastic potential energy stored in the spring.
(b) Calculate the maximum height reached by the ball (assume no air resistance).
(c) In reality the ball only reaches 1.8 m. Calculate the energy dissipated against air resistance.
(b) All elastic PE ā GPE at maximum height:
m Ć g Ć h = 2.56
0.2 Ć 10 Ć h = 2.56
h = 2.56 Ć· 2 = 1.28 m
(c) GPE gained at 1.8 m = 0.2 Ć 10 Ć 1.8 = 3.6 J
But wait ā 3.6 J > 2.56 J. Let's recheck: the ball only reaches 1.8 m, so GPE = 0.2 Ć 10 Ć 1.8 = 3.6 J. This is more than the stored energy, which is impossible. Re-reading: the ball reaches 1.8 m instead of 1.28 m (theoretical maximum ā this scenario is not physically consistent). Correction for exam use: If theoretical max is 1.28 m but actual height is only less than this, say 1.0 m:
GPE at 1.0 m = 0.2 Ć 10 Ć 1.0 = 2.0 J
Energy dissipated = 2.56 ā 2.0 = 0.56 J
(Your teacher may give a specific actual height ā always use: dissipated = Ee ā Ep at actual height.)
Challenge 3: A student claims that a machine with 92% efficiency is "almost perfect" and that only 8% of energy is "lost." Write a paragraph evaluating this claim using your knowledge of conservation of energy and energy dissipation. Include a numerical example using 5000 J input energy.
Model Answer:
The student is partially correct ā 92% efficiency is high, and the machine transfers most of its input energy usefully. With 5000 J input: useful output = 0.92 Ć 5000 = 4600 J; wasted energy = 5000 ā 4600 = 400 J.
However, the claim that energy is "lost" is incorrect. The law of conservation of energy states that energy cannot be destroyed. The 400 J is not lost ā it is dissipated to the thermal store of the surroundings, most likely through friction or electrical resistance. This energy is simply spread out and becomes harder to use for a specific purpose.
Furthermore, even a small percentage of wasted energy matters in large-scale applications. A power station processing millions of joules per second would waste enormous amounts of energy even at 92% efficiency. Engineers always aim to reduce dissipation further through lubrication, better insulation, and improved materials.
Challenge 4 (Extended): A 0.5 kg ball is thrown horizontally from the top of a 20 m cliff with an initial speed of 8 m/s. (g = 10 N/kg)
(a) Calculate the total mechanical energy of the ball at the top of the cliff (KE + GPE, taking the base as reference).
(b) Using conservation of energy, calculate the speed of the ball just before it hits the ground (ignore air resistance).
(c) Explain why the horizontal component of velocity remains 8 m/s throughout the fall.
(a)
KE at top = Ā½ Ć 0.5 Ć 8Ā² = Ā½ Ć 0.5 Ć 64 = 16 J
GPE at top = 0.5 Ć 10 Ć 20 = 100 J
Total mechanical energy = 16 + 100 = 116 J
(b) At the bottom, all energy is KE (GPE = 0 at ground level):
Ā½ Ć 0.5 Ć vĀ² = 116
0.25 Ć vĀ² = 116
vĀ² = 464
v = ā464 ā 21.5 m/s
(c) The horizontal component of velocity stays at 8 m/s because there is no horizontal force acting on the ball (air resistance is ignored). Only gravity acts, and gravity acts vertically downward. By Newton's first law, with no horizontal force, the horizontal velocity does not change. Energy is conserved overall ā the GPE converts to additional vertical KE, while the horizontal KE remains constant.