Power Revision Notes for Edexcel A-Level Further Mathematics

14.1.4 Power

Introduction

Power measures the rate of doing work or the rate at which energy is transferred. For example, a more powerful engine can do the same amount of work faster or perform more work in the same amount of time.

Key Formulae for Power

Average Power:

\text{Power} = \frac{\text{Work done}}{\text{Time taken}} = \frac{Fd}{t}

where:

$F$ is the force ( $\text{N}$ )
$d$ is the displacement ( $\text{m}$ )
$t$ is the time ( $\text{s}$ )

Instantaneous Power:

\text{Power} = \text{Force} \times \text{Velocity}

where:

$F$ is the force acting along the motion ( $\text{N}$ )
$v$ is the velocity of the object ( $\text{ms}^{-1}$ )

Units of Power

The unit of power is the watt ( $W$ ), where:

1 \, \text{W} = 1 \, \text{J/s}

Other units include:

Kilowatt ( $kW$ ): $1 \, \text{kW} = 1000 \, \text{W}$

Horsepower ( $hp$ ): $1 \, \text{hp} \approx 746 \, \text{W}$

Worked Examples

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Example 1: Calculating Average Power

Problem

An engine applies a constant force of $1400 \, \text{N}$ to move a vehicle over $375 \, \text{m}$ in $15 \, \text{s}$

Find the average power developed by the engine.

Step 1: Use the formula for average power:

\text{Power} = \frac{Fd}{t}

Step 2: Substitute the values:

\text{Power} = \frac{1400 \times 375}{15}

Step 3: Calculate:

\text{Power} = \frac{525,000}{15} = 35,000 \, \text{W}

Convert to kilowatts:

\text{Power} = \frac{35,000}{1000} = 35 \, \text{kW}

Final Answer:

The average power developed is 35 kW

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Example 2: Calculating Resistance Force and Acceleration

Problem

A driver is operating a car with a mass of $1300 \, \text{kg}$ on a horizontal road. The car travels at a constant speed of $72 \, \text{km/h}$ using a power output of $20 \, \text{kW}$

(a) Find the resistance force acting on the car.

(b) If the engine power increases to $25 \, \text{kW}$ , find the resulting acceleration of the car, assuming the resistance remains constant.

(a) Find the Resistance Force

At a constant speed, the tractive force provided by the engine equals the resistance force.

Step 1: Convert the speed to metres per second:

72 \, \text{km/h} = \frac{72 \times 1000}{3600} = 20 \, \text{ms}^{-1}

Step 2: Use the formula for instantaneous power:

\text{Power} = Fv \quad \Rightarrow \quad F = \frac{\text{Power}}{v}

Step 3: Substitute the values:

F = \frac{20,000}{20} = 1000 \, \text{N}

(b) Find the Acceleration

If the engine power increases, the net force is the difference between the tractive force and the resistance.

Step 1: Find the new tractive force at $25 \, \text{kW}$

F_\text{tractive} = \frac{\text{Power}}{v} = \frac{25,000}{20} = 1250 \, \text{N}

Step 2: Find the resultant force:

F_\text{net} = F_\text{tractive} - F_\text{resistance} = 1250 - 1000 = 250 \, \text{N}

Step 3: Use Newton's second law to find the acceleration:

F_\text{net} = ma \quad \Rightarrow \quad a = \frac{F_\text{net}}{m}

Step 4: Substitute the values:

a = \frac{250}{1300} \approx 0.192 \, \text{ms}^{-2}

Final Answer:

(a) The resistance force is 1000 N

(b) The acceleration of the car is 0.192 ms⁻²

Note Summary

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Common Mistakes

Incorrect unit conversions: Always convert speeds to $\text{ms}^{-1}$ before calculations.
Ignoring resistive forces: At constant speed, resistance equals the tractive force.
Confusing average and instantaneous power: Use $P = Fd/t$ for average power and $P = Fv$ for instantaneous power.

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Key Formulas

Average Power:

P = \frac{Fd}{t}

Instantaneous Power:

P = Fv

Force from Power:

F = \frac{P}{v}

Newton's Second Law:

F_\text{net} = ma \quad \Rightarrow \quad a = \frac{F_\text{net}}{m}

Power (Edexcel A-Level Further Mathematics): Revision Notes

14.1.4 Power

Introduction

Key Formulae for Power

Average Power:

Instantaneous Power:

Units of Power

Worked Examples

Example 1: Calculating Average Power

Example 2: Calculating Resistance Force and Acceleration

Note Summary

Common Mistakes

Key Formulas

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