Energy

1. Overview

Energy is a fundamental concept in physics, describing the "capacity to do work." In any process, energy is never created or destroyed; instead, it shifts between different stores through various transfer pathways. Understanding how energy changes form allows us to calculate the behavior of everything from falling objects to national power grids.

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Key Definitions

• Energy: The capacity of a system to do work, measured in Joules (J).
• System: An object or a group of objects being studied.
• Conservation of Energy: A fundamental law stating that the total energy of a closed system remains constant.
• Work Done: The amount of energy transferred by a force or an electrical current.

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Core Content

Energy Stores

Energy is held by objects in different "stores." You must be able to identify these:

• Kinetic: Energy of a moving object.
• Gravitational Potential (GPE): Energy stored by an object due to its position in a gravitational field (height).
• Chemical: Energy stored in bonds (e.g., food, fuel, batteries).
• Elastic (Strain): Energy stored when an object is stretched or compressed.
• Nuclear: Energy stored in the nucleus of an atom.
• Electrostatic: Energy stored when like charges are pushed together or unlike charges are pulled apart.
• Internal (Thermal): The total kinetic and potential energy of the particles within an object (experienced as temperature).

Energy Transfers

Energy moves between stores via four main pathways:

1. Mechanical Work: A force moving an object through a distance.
2. Electrical Work: Charges moving due to a potential difference (current).
3. Heating: Due to a temperature difference.
4. Waves: Energy transferred by electromagnetic radiation (light), sound, or water waves.

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The Principle of Conservation of Energy

Energy cannot be created or destroyed. In any event: Total Energy Input = Total Energy Output

Worked Example (Simple Flow): An electric fan transfers 100J of electrical energy. 80J is transferred to the kinetic store of the air, and 20J is transferred to the thermal store of the motor.

• Conservation Check: 80J + 20J = 100J.

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Extended Content (Extended Only)

Kinetic Energy ($E_k$)

The energy of a moving object depends on its mass and the square of its speed.

• Equation: $E_k = \frac{1}{2}mv^2$

Change in Gravitational Potential Energy ($\Delta E_p$)

When an object is raised or lowered, its GPE changes.

• Equation: $\Delta E_p = mg\Delta h$

Worked Example (Complex Transfer): A 2 kg block is dropped from a height of 5m. Calculate its speed just before it hits the floor (assume no air resistance, $g = 9.8 m/s^2$).

1. Calculate GPE at the top: $\Delta E_p = 2 \times 9.8 \times 5 = 98J$.
2. Apply Conservation of Energy: $GPE_{lost} = KE_{gained}$.
3. Set up KE equation: $98 = \frac{1}{2} \times 2 \times v^2$.
4. Solve for $v$: $98 = v^2 \rightarrow v = \sqrt{98} \approx 9.9 m/s$.

Sankey Diagrams

Sankey diagrams show energy efficiency. The width of the arrows represents the amount of energy.

• The horizontal arrow usually represents "useful" energy.
• The downward arrow represents "wasted" (dissipated) energy.

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Key Equations

Equation	Symbols	Units
$E_k = \frac{1}{2}mv^2$	$m$ = mass, $v$ = speed	$m$ (kg), $v$ (m/s), $E_k$ (J)
$\Delta E_p = mg\Delta h$	$m$ = mass, $g$ = gravitational field strength, $\Delta h$ = change in height	$m$ (kg), $g$ (N/kg), $\Delta h$ (m), $E_p$ (J)

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Common Mistakes to Avoid

• ❌ Wrong: Assuming the energy link between a turbine and a generator in a power station is "electrical."
• ✅ Right: The turbine-to-generator link is a physical, rotating axle; therefore, the energy is transferred mechanically.
• ❌ Wrong: Forgetting to square the velocity ($v$) when calculating Kinetic Energy.
• ✅ Right: Always calculate $v^2$ first before multiplying by $\frac{1}{2}m$.
• ❌ Wrong: Thinking total energy decreases as an object falls.
• ✅ Right: Only the GPE decreases; the total energy remains constant as it converts into Kinetic Energy.
• ❌ Wrong: Forgetting to include $g$ (gravitational field strength) in GPE calculations.
• ✅ Right: Always multiply mass by $g$ ($9.8$ or $10 N/kg$ as per paper instructions) and height.

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Exam Tips

1. Trace the Source: Remember that for hydroelectric, wind, and fossil fuels, the Sun is the original source of energy (driving the water cycle, weather patterns, and photosynthesis).
2. Units Matter: Always convert mass into kilograms (kg) and height into meters (m) before using the $E_k$ or $E_p$ formulas.
3. Sankey Width: When drawing or interpreting Sankey diagrams, use a ruler. If 1cm width equals 10J, then a 50J input must be exactly 5cm wide.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A small electric motor lifts a mass of 0.25 kg vertically through a height of 0.80 m.

(a) State the energy store that increases as the mass is lifted. [1]

(b) Calculate the increase in this energy store. The gravitational field strength, $g$, is 9.8 N/kg. [3]

(c) Suggest one reason why the electrical energy supplied to the motor will be greater than your answer to (b). [1]

Worked Solution:

(a)

1. The mass is being lifted against gravity, so it gains gravitational potential energy. Gravitational potential energy

How to earn full marks:

• State "gravitational potential energy" or "GPE".

(b)

1. Recall the formula for the change in gravitational potential energy. $\Delta E = mg\Delta h$ Formula must be correct.

2. Substitute the given values into the formula. $\Delta E = 0.25 \times 9.8 \times 0.80$ Correct substitution of all three values.

3. Calculate the change in gravitational potential energy, including the correct unit. $\Delta E = 1.96 \ J$ Correct numerical value, including the unit.

How to earn full marks:

• Correct formula for GPE.
• Correct substitution of all values.
• Correct final answer with the correct unit (J).

(c)

1. The motor will not be 100% efficient; some energy will be transferred to other forms. Some energy is transferred as thermal energy in the motor.

How to earn full marks:

• Suggest that some energy is transferred to the surroundings as thermal energy (heat) due to friction in the motor.

Common Pitfall: Don't forget that real-world systems are never perfectly efficient. When energy is transferred, some of it is usually 'lost' as heat due to friction or other processes, meaning the useful output energy is always less than the total input energy.

Exam-Style Question 2 — Extended Response [8 marks]

Question:

A student investigates the energy transfers involved when a small ball is dropped onto a spring. The ball has a mass of 0.050 kg.

(a) The ball is held 0.45 m above the top of the spring. Calculate the gravitational potential energy (GPE) of the ball relative to the top of the spring. State the formula used. $g = 9.8 \ N/kg$. [3]

(b) The ball is dropped. Just before it hits the spring, it has a speed of 2.97 m/s. Calculate the kinetic energy (KE) of the ball at this point. State the formula used. [3]

(c) Explain why the kinetic energy of the ball just before it hits the spring is less than the gravitational potential energy it had before it was dropped. [2]

Worked Solution:

(a)

1. Recall the formula for gravitational potential energy. $GPE = mgh$ Formula must be correct.

2. Substitute the given values into the formula. $GPE = 0.050 \times 9.8 \times 0.45$ Correct substitution of all three values.

3. Calculate the gravitational potential energy, including the correct unit. $GPE = 0.2205 \ J \approx \boxed{0.22 \ J}$ Correct numerical value, including the unit.

How to earn full marks:

• Correct formula for GPE.
• Correct substitution of all values.
• Correct final answer with the correct unit (J).

(b)

1. Recall the formula for kinetic energy. $KE = \frac{1}{2}mv^2$ Formula must be correct.

2. Substitute the given values into the formula. $KE = \frac{1}{2} \times 0.050 \times (2.97)^2$ Correct substitution of all values.

3. Calculate the kinetic energy, including the correct unit. $KE = 0.2205 \ J \approx \boxed{0.22 \ J}$ Correct numerical value, including the unit.

How to earn full marks:

• Correct formula for KE.
• Correct substitution of all values.
• Correct final answer with the correct unit (J).

(c)

1. Some of the GPE is not converted into KE. Air resistance does work on the ball as it falls. [1]

2. Therefore, some energy is transferred to the surroundings. This work done by air resistance transfers energy to the thermal energy store of the air (heating it). [1]

How to earn full marks:

• State that air resistance is present (or friction).
• Explain that air resistance transfers energy to the thermal energy store (heats the air).

Common Pitfall: Remember to consider air resistance when an object is falling. Air resistance converts some of the gravitational potential energy into thermal energy, which is why the kinetic energy just before impact is less than the initial gravitational potential energy.

Exam-Style Question 3 — Short Answer [6 marks]

Question:

A cyclist rides along a flat road.

(a) State the main form of energy input that allows the cyclist to move. [1]

(b) As the cyclist pedals, the chemical energy store reduces. Describe two energy transfers that occur as the cyclist moves along the road. [4]

(c) The cyclist stops pedalling and slows down. Suggest one reason why the cyclist slows down. [1]

Worked Solution:

(a)

1. The cyclist uses chemical energy stored in their body. Chemical (energy)

How to earn full marks:

• State "chemical energy".

(b)

1. Chemical energy is transferred to kinetic energy. Chemical energy (in cyclist) is transferred to kinetic energy (of the cyclist and bicycle). [2]

2. Some energy is lost due to friction. Some chemical energy is transferred as thermal energy due to friction in the bicycle parts and between the tyres and the road. [2]

How to earn full marks:

• State chemical energy is transferred to kinetic energy.
• State that some energy is transferred as thermal energy due to friction.

(c)

1. Friction opposes the cyclist's motion. Friction/air resistance opposes the motion of the bicycle.

How to earn full marks:

• Suggest friction or air resistance.

Common Pitfall: When describing energy transfers, be specific about where the energy is going. For example, don't just say "energy is lost due to friction"; instead, say "energy is transferred as thermal energy due to friction between the tires and the road."

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A small hydroelectric power station uses water stored in a reservoir to generate electricity. The water falls through a vertical distance of 40 m to turn a turbine. The flow rate of the water is 500 kg/s.

(a) Calculate the change in gravitational potential energy (GPE) of the water each second. State the formula used. $g = 9.8 \ N/kg$. [3]

(b) Calculate the maximum possible power output of the power station. [2]

(c) The actual power output of the power station is 160 kW. Calculate the efficiency of the power station. [2]

(d) State two reasons why the power station is less than 100% efficient. [2]

Worked Solution:

(a)

1. Recall the formula for gravitational potential energy. $\Delta GPE = mg\Delta h$ Formula must be correct.

2. Substitute the given values into the formula. $\Delta GPE = 500 \times 9.8 \times 40$ Correct substitution of all three values.

3. Calculate the change in gravitational potential energy, including the correct unit. $\Delta GPE = \boxed{196000 \ J}$ Correct numerical value, including the unit.

How to earn full marks:

• Correct formula for GPE.
• Correct substitution of all values.
• Correct final answer with the correct unit (J).

(b)

1. The maximum power output is equal to the change in GPE per second. $Power = \frac{Energy}{time} = \frac{196000 \ J}{1 \ s}$ Understanding that power is energy transferred per second.

2. Calculate the power, including the correct unit. $Power = 196000 \ W = \boxed{196 \ kW}$ Correct numerical value, including the unit.

How to earn full marks:

• Recognise that power is energy/time.
• Correct final answer with the correct unit (W or kW).

(c)

1. Recall the formula for efficiency. $Efficiency = \frac{Useful \ power \ output}{Total \ power \ input}$ Formula must be correct.

2. Substitute the given values into the formula. $Efficiency = \frac{160000 \ W}{196000 \ W} = 0.8163$ Correct substitution of values.

3. Calculate the efficiency as a percentage. $Efficiency = 0.8163 \times 100 % \approx \boxed{81.6 %}$ Correct numerical value, including the percentage sign.

How to earn full marks:

• Correct formula for efficiency.
• Correct substitution of values.
• Correct final answer with the correct unit (%).

(d)

1. Friction in the turbine converts some energy to thermal energy. Friction in the turbine transfers some energy to the thermal energy store of the surroundings.

2. Sound energy is also produced. Some energy is transferred as sound energy.

How to earn full marks:

• State that friction is present in the turbine.
• State that some energy is transferred as thermal energy or sound energy.

Common Pitfall: When calculating efficiency, make sure you use the same units for both the input and output power. It's easy to make a mistake by using kW for one and W for the other. Also, remember to convert the final answer to a percentage by multiplying by 100.