Power

1. Overview

Power is a measure of how quickly work is being done or how rapidly energy is being transferred. In everyday terms, two machines might perform the exact same task (like lifting a crate), but the one that completes the task faster is considered more "powerful." Understanding power is essential for evaluating the performance of engines, motors, and electrical appliances.

Key Definitions

• Power: The rate at which work is done or the rate at which energy is transferred.
• Watt (W): The SI unit of power. One Watt is equal to one Joule of energy transferred per second ($1\text{ W} = 1\text{ J/s}$).

Core Content

Power is always about the relationship between energy and time. If you do work faster, your power output is higher.

Understanding the Equations There are two ways to express the same concept:

1. Work-based: $P = \frac{W}{t}$ (Used when a force moves an object).
2. Energy-based: $P = \frac{\Delta E}{t}$ (Used when energy changes from one form to another).

Power in Action

• If two motors lift the same weight to the same height, they do the same amount of work.
• However, if Motor A does it in 5 seconds and Motor B does it in 10 seconds, Motor A has twice the power of Motor B.

Worked Example: Lifting a Load A crane lifts a $200\text{ kg}$ crate to a height of $10\text{ m}$ in $4\text{ seconds}$. Calculate the power of the crane. (Take $g = 10\text{ m/s}^2$)

• Step 1: Calculate the Force (Weight) $F = m \times g = 200\text{ kg} \times 10\text{ m/s}^2 = 2000\text{ N}$
• Step 2: Calculate the Work Done $W = F \times d = 2000\text{ N} \times 10\text{ m} = 20,000\text{ J}$
• Step 3: Calculate Power $P = \frac{W}{t} = \frac{20,000\text{ J}}{4\text{ s}} = 5000\text{ W}$ (or $5\text{ kW}$)

Extended Content (Extended Only)

The IGCSE syllabus classifies the definition and calculation of Power under the Core curriculum. There is no additional calculation or theory required specifically for Extended students for this sub-topic.

Key Equations

Equation	Symbols	Units
$P = \frac{W}{t}$	$P$ = Power, $W$ = Work done, $t$ = time	$P$ (Watts), $W$ (Joules), $t$ (seconds)
$P = \frac{\Delta E}{t}$	$P$ = Power, $\Delta E$ = Energy transferred, $t$ = time	$P$ (Watts), $E$ (Joules), $t$ (seconds)

Common Mistakes to Avoid

• ❌ Wrong: Using minutes or hours for time in your calculation.
  • ✓ Right: Always convert time into seconds (e.g., $2\text{ mins} = 120\text{ s}$) before using the formula.
• ❌ Wrong: Using mass ($kg$) directly in the work equation ($W = F \times d$).
  • ✓ Right: You must convert mass to weight (Force) by multiplying by $g$ ($10\text{ m/s}^2$ or $9.8\text{ m/s}^2$).
• ❌ Wrong: Thinking the most powerful motor is the one that lifts the heaviest mass OR the highest distance regardless of time.
  • ✓ Right: Power is a combined effect. You must calculate $(m \times g \times h) / t$ to compare them fairly.
• ❌ Wrong: Rearranging the formula to $P = W \times t$ or $P = W / t^2$.
  • ✓ Right: Power is simply the rate: $P = \frac{\text{Work}}{\text{Time}}$. If you multiply Work by Time, the units no longer represent Power.

Exam Tips

1. Check your units first: If a question gives you power in kilowatts (kW), multiply it by $1000$ to get Watts before you start. If time is in minutes, multiply by $60$.
2. Two-step problems: Many exam questions won't give you "Work." They will give you mass, height, and time. You must remember to calculate Work ($m \times g \times h$) first, then divide by time to find Power.
3. The "Watt" Definition: If asked to define a Watt, remember it is "one Joule per second." This helps you remember the formula ($J / s$).

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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 crane lifts a steel beam vertically upwards. The beam has a weight of 5000 N and is lifted a distance of 12 m in a time of 60 s at a constant speed.

(a) Define power. [1]

(b) Calculate the work done in lifting the steel beam. [2]

(c) Calculate the power output of the crane. [2]

Worked Solution:

(a)

1. Power is the rate at which energy is transferred. Power is the work done per unit time. [Definition of power]

How to earn full marks:

• State "rate at which energy is transferred" OR "work done per unit time"

(b)

1. Work done is equal to the force multiplied by the distance. $W = F \times d$ [Recall work done equation]
2. Substitute the force (weight) and distance to find the work done. $W = 5000 \text{ N} \times 12 \text{ m} = 60000 \text{ J}$ [Calculation of work done]

How to earn full marks:

• Correct formula, even if rearranged
• Correct substitution of 5000 N and 12 m
• Correct answer: $\boxed{60000 \text{ J}}$

(c)

1. Power is work done divided by time. $P = \frac{W}{t}$ [Recall power equation]
2. Substitute the work done and time to find the power. $P = \frac{60000 \text{ J}}{60 \text{ s}} = 1000 \text{ W}$ [Calculation of power]

How to earn full marks:

• Correct formula, even if rearranged
• Correct substitution of 60000 J and 60 s (ECF from (b) allowed)
• Correct answer: $\boxed{1000 \text{ W}}$

Common Pitfall: Make sure you remember the definition of power. Also, double-check that you're using the correct units (Joules for work done, seconds for time) to get the power in Watts.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

A small electric motor is used to lift a mass of 0.20 kg vertically through a height of 1.5 m. The motor is connected to a 6.0 V supply, and a current of 0.10 A flows through it. The time taken to lift the mass is 5.0 s.

(a) Calculate the work done in lifting the mass. Take $g = 10 \text{ N/kg}$. [2]

(b) Calculate the electrical power input to the motor. [2]

(c) Calculate the efficiency of the motor. [2]

Worked Solution:

(a)

1. The force required to lift the mass is its weight. $F = mg = 0.20 \text{ kg} \times 10 \text{ N/kg} = 2.0 \text{ N}$ [Calculate the weight of the mass]
2. Work done is equal to the force multiplied by the distance. $W = F \times d = 2.0 \text{ N} \times 1.5 \text{ m} = 3.0 \text{ J}$ [Calculate work done]

How to earn full marks:

• Calculate weight: $2.0 \text{ N}$
• Correct final answer: $\boxed{3.0 \text{ J}}$

(b)

1. Electrical power is equal to voltage multiplied by current. $P = VI = 6.0 \text{ V} \times 0.10 \text{ A} = 0.60 \text{ W}$ [Calculate electrical power]

How to earn full marks:

• Correct substitution of 6.0 V and 0.10 A
• Correct answer: $\boxed{0.60 \text{ W}}$

(c)

1. Efficiency is the ratio of useful power output to total power input. First, calculate the power output from the work done per unit time. $P_{output} = \frac{W}{t} = \frac{3.0 \text{ J}}{5.0 \text{ s}} = 0.60 \text{ W}$ [Calculate the output power]
2. Calculate the efficiency as output power / input power $\text{Efficiency} = \frac{P_{output}}{P_{input}} = \frac{0.60 \text{ W}}{0.60 \text{ W}} = 1.0$ [Calculate efficiency]
3. Express as a percentage $\text{Efficiency} = 1.0 \times 100% = 100%$

How to earn full marks:

• Calculate the output power (0.60 W)
• Correct substitution into efficiency equation (ECF from (a) and (b) allowed)
• Correct answer: $\boxed{100%}$

Common Pitfall: Don't forget to calculate the weight of the object using $W=mg$ before calculating the work done. Also, efficiency is a ratio, so it should be a decimal or a percentage.

Exam-Style Question 3 — Extended Response [8 marks]

Question:

A cyclist is riding up a hill. The hill is 500 m long and has a vertical height of 25 m. The cyclist and bicycle have a combined mass of 80 kg. The cyclist reaches the top of the hill in 100 s. Assume the cyclist is working at a constant rate.

(a) Calculate the gain in gravitational potential energy of the cyclist and bicycle. Take $g = 10 \text{ N/kg}$. [3]

(b) Calculate the average power output of the cyclist as they cycle up the hill. [2]

(c) The cyclist eats an energy bar that provides 500 kJ of energy. Assuming the cyclist converts all of this energy into useful work, calculate the distance they could cycle along a flat road against a constant frictional force of 40 N. [3]

Worked Solution:

(a)

1. The gain in gravitational potential energy (GPE) is given by $mgh$. $\Delta GPE = mgh$ [Recall the GPE formula]
2. Substitute the mass, gravitational field strength, and height. $\Delta GPE = 80 \text{ kg} \times 10 \text{ N/kg} \times 25 \text{ m} = 20000 \text{ J}$ [Calculate the GPE]

How to earn full marks:

• Correct formula, even if rearranged
• Correct substitution of 80 kg, 10 N/kg, and 25 m
• Correct answer: $\boxed{20000 \text{ J}}$

(b)

1. Power is the energy transferred per unit time. $P = \frac{\Delta GPE}{t} = \frac{20000 \text{ J}}{100 \text{ s}} = 200 \text{ W}$ [Calculate power using the GPE calculated in (a)]

How to earn full marks:

• Correct formula, even if rearranged
• Correct substitution of 20000 J and 100 s (ECF from (a) allowed)
• Correct answer: $\boxed{200 \text{ W}}$

(c)

1. Work done is equal to the energy transferred. $W = 500 \text{ kJ} = 500000 \text{ J}$ [Convert kJ to J]
2. Work done is also equal to force multiplied by distance. $W = F \times d$ [Recall work done formula]
3. Rearrange the formula to find the distance. $d = \frac{W}{F} = \frac{500000 \text{ J}}{40 \text{ N}} = 12500 \text{ m}$ [Calculate the distance]

How to earn full marks:

• Correct conversion of 500 kJ to 500000 J
• Correct rearrangement of formula
• Correct answer: $\boxed{12500 \text{ m}}$

Common Pitfall: Remember to convert kilojoules (kJ) to joules (J) before using the energy value in calculations. Also, make sure you understand the relationship between work done, force, and distance.

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

Question:

A small hydroelectric power station uses water falling from a height to generate electricity. Water flows from a reservoir at a rate of 500 kg/s. The water falls a vertical distance of 40 m to a turbine. The turbine converts the kinetic energy of the water into electrical energy with an efficiency of 80%. Take $g = 10 \text{ N/kg}$.

(a) Calculate the loss in gravitational potential energy of the water each second. [3]

(b) Calculate the kinetic energy of the water just before it reaches the turbine (assuming all the potential energy is converted to kinetic energy). [1]

(c) Calculate the electrical power output of the power station. [3]

(d) State one reason why, in a real power station, not all of the gravitational potential energy of the water is converted to electrical energy. [2]

Worked Solution:

(a)

1. The loss in gravitational potential energy (GPE) is given by $mgh$. $\Delta GPE = mgh$ [Recall the GPE formula]
2. The mass of water falling per second is 500 kg.
3. Substitute the mass, gravitational field strength, and height. $\Delta GPE = 500 \text{ kg} \times 10 \text{ N/kg} \times 40 \text{ m} = 200000 \text{ J}$ [Calculate the GPE]

How to earn full marks:

• Correct formula, even if rearranged
• Correct substitution of 500 kg, 10 N/kg, and 40 m
• Correct answer: $\boxed{200000 \text{ J}}$

(b)

1. The kinetic energy (KE) gained is equal to the loss in GPE. $KE = \Delta GPE = 200000 \text{ J}$ [State that all GPE is converted to KE]

How to earn full marks:

• Recognizing that the KE is equal to the GPE lost in (a)
• Correct answer: $\boxed{200000 \text{ J}}$

(c)

1. The electrical power output is 80% of the kinetic energy per second. $P_{output} = \text{Efficiency} \times \frac{KE}{t} = 0.80 \times \frac{200000 \text{ J}}{1 \text{ s}} = 160000 \text{ W}$ [Calculate the power output]

How to earn full marks:

• Correctly calculating 80% of the KE from part (b)
• Correct answer: $\boxed{160000 \text{ W}}$

(d)

1. Some energy is lost due to friction in the turbine bearings. [Friction will convert some energy to heat]
2. Some energy is lost due to the sound produced by the turbine. [Sound also carries away energy]

How to earn full marks:

• Any valid reason, such as friction in the turbine, sound produced, or heat generated in the generator.
• Explanation of the energy loss (e.g., friction converts energy to heat)

Common Pitfall: Remember that efficiency is a percentage of the total energy converted. Also, when explaining energy losses, be specific about how the energy is lost (e.g., friction converts kinetic energy into thermal energy).