Work, Energy and Power

Understanding energy transfer, conservation, and the rate of doing work

SPEC 3.4.1.7 • MECHANICS AND MATERIALS

What is Work?

Definition

Work is done when a force causes an object to move in the direction of the force. Work is a measure of energy transferred.

When force and displacement are in the same direction:

Formula

W = F s

Where: W = work done (J), F = force (N), s = displacement (m)

Unit

Work is a scalar quantity measured in joules (J). One joule = one newton-metre (1 J = 1 N m).

Work Done at an Angle

When the force is not in the same direction as the displacement, only the component of force parallel to the displacement does work.

General Formula

W = F s cos θ

Where θ = angle between force and displacement

Special Cases

θ = 0°: W = Fs (maximum)

θ = 90°: W = 0 (no work)

θ = 180°: W = −Fs (negative)

Important

Negative work means energy is transferred FROM the object (e.g., friction).

Work from F-s Graphs

When force varies with displacement, work done equals the area under the force-displacement graph.

Key Principle

W = Area under F-s graph = \int F d s

Common Shapes

Rectangle: W = F × s (constant force)

Triangle: W = ½ × base × height (e.g., springs where F = kx)

Trapezium: W = ½(F₁ + F₂) × s

01Worked Example - F-s Graph

The graph shows how force varies with displacement for an object. Calculate the total work done.

1

Split into two regions: Triangle (0-4m) + Rectangle (4-8m)

2

Triangle area: W₁ = ½ × base × height = ½ × 4 × 40 = 80 J

3

Rectangle area: W₂ = F × s = 40 × 4 = 160 J

4

Total work: W = W₁ + W₂ = 80 + 160 = 240 J

Answer: W = 240 J

Worked Example - Work at Angle02

A person pulls a sled with a force of 150 N at 30° above horizontal. The sled moves 20 m. Calculate the work done.

1

Identify: F = 150 N, θ = 30°, s = 20 m

2

Apply: W = Fs cos θ

3

W = 150 × 20 × cos 30° = 150 × 20 × 0.866

4

W = 2598 J

Answer: W = 2600 J (2 s.f.)

Kinetic Energy

Definition

Kinetic energy is the energy an object possesses due to its motion.

Formula

E_{k} = \frac{1}{2} m v^{2}

Where: E_k = kinetic energy (J), m = mass (kg), v = velocity (m s⁻¹)

Work-Energy Theorem

The net work done on an object equals its change in kinetic energy:

W_{n e t} = Δ E_{k} = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}

Worked Example - Kinetic Energy03

A 1200 kg car accelerates from 10 m s⁻¹ to 25 m s⁻¹. Calculate: (a) the change in KE, (b) the work done by the engine.

1

Initial KE: E_k1 = ½ × 1200 × 10² = 60,000 J

2

Final KE: E_k2 = ½ × 1200 × 25² = 375,000 J

3

(a) ΔE_k = 375,000 − 60,000 = 315,000 J

4

(b) By work-energy theorem: W = ΔE_k = 315,000 J

Answer: (a) ΔE_k = 315 kJ, (b) W = 315 kJ

Gravitational Potential Energy

Definition

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field.

Formula

E_{p} = m g h

Where: E_p = gravitational PE (J), m = mass (kg), g = 9.8 N kg⁻¹, h = height (m)

Important

GPE is measured relative to a chosen reference point. Only changes in GPE are physically meaningful.

Conservation of Energy

Principle

Energy cannot be created or destroyed, only transferred from one form to another. Total energy of a closed system remains constant.

Conservation Equation

E_{k} + E_{p} = constant

\frac{1}{2} m v_{1}^{2} + m g h_{1} = \frac{1}{2} m v_{2}^{2} + m g h_{2}

Worked Example - Conservation04

A 0.5 kg ball is dropped from 12 m. Calculate its speed just before hitting the ground (no air resistance).

1

At top: E_p = mgh = 0.5 × 9.8 × 12 = 58.8 J, E_k = 0

2

At bottom: E_p = 0, E_k = 58.8 J (conservation)

3

½mv² = 58.8

4

v² = 2 × 58.8 / 0.5 = 235.2

5

v = √235.2 = 15.3 m s⁻¹

Answer: v = 15.3 m s⁻¹

Power

Definition

Power is the rate of doing work, or the rate of energy transfer.

Formulas

P = \frac{W}{t} = \frac{Δ E}{t}

Unit: watt (W), where 1 W = 1 J s⁻¹

Understanding power: Same work done in less time = more power

Alternative Formula

For constant force F on object moving at velocity v: P = Fv

Derivation: P = W/t = Fs/t = F(s/t) = Fv

Worked Example - Power05

A crane lifts a 500 kg load through 30 m in 20 s. Calculate: (a) work done, (b) power output.

1

(a) Work = gain in GPE = mgh

2

W = 500 × 9.8 × 30 = 147,000 J

3

(b) Power = W/t

4

P = 147,000 / 20 = 7350 W = 7.35 kW

Answer: (a) W = 147 kJ, (b) P = 7.35 kW

Worked Example - P = Fv06

A car engine produces 2400 N driving force at constant 25 m s⁻¹. Calculate: (a) power output, (b) resistive force.

1

(a) Power = Fv

2

P = 2400 × 25 = 60,000 W = 60 kW

3

(b) At constant speed, a = 0, so net force = 0

4

Resistive force = Driving force = 2400 N

Answer: (a) P = 60 kW, (b) Resistive force = 2400 N

Efficiency

Definition

Efficiency is the ratio of useful energy output to total energy input.

Formula

Efficiency = \frac{Useful output}{Total input} \times 100%

Important

Efficiency is always less than 100% for real machines - some energy is always dissipated as heat.

Worked Example - Efficiency07

An 800 W motor lifts a 50 kg mass at steady 0.6 m s⁻¹. Calculate the efficiency.

1

Useful output power = mg × v

2

P_out = 50 × 9.8 × 0.6 = 294 W

3

Efficiency = (P_out / Pᵢₙ) × 100%

4

Efficiency = (294/800) × 100% = 36.75%

Answer: Efficiency = 37% (2 s.f.)

Worked Example - Energy Dissipation08

A 60 kg skier starts from rest at 40 m height. At the bottom, speed is 20 m s⁻¹. Find: (a) initial GPE, (b) final KE, (c) energy dissipated.

1

(a) GPE = mgh = 60 × 9.8 × 40 = 23,520 J

2

(b) KE = ½mv² = ½ × 60 × 20² = 12,000 J

3

(c) Dissipated = Initial − Final

4

E_dissipated = 23,520 − 12,000 = 11,520 J

Answer: (a) 23.5 kJ, (b) 12 kJ, (c) 11.5 kJ

Practice

Question 1: A force of 80 N pushes a crate 15 m. The force is at 25° below horizontal. Calculate the work done.

1

θ = 25° (angle to displacement)

2

W = Fs cos θ = 80 × 15 × cos 25°

3

W = 80 × 15 × 0.906 = 1087 J

Answer: W = 1090 J (3 s.f.)

Practice

Question 2: A 800 kg roller coaster starts from rest at 50 m height. What is maximum speed at the bottom? Why would actual speed be less?

1

Conservation: mgh = ½mv² (mass cancels)

2

v = √(2gh) = √(2 × 9.8 × 50) = √980 = 31.3 m s⁻¹

3

Actual speed less due to friction and air resistance dissipating energy as heat.

Answer: v_max = 31.3 m s⁻¹; actual less due to dissipation

Practice

Question 3: A pump raises 200 kg of water per minute through 15 m height. With 60% efficiency, calculate input power required.

1

Useful output power = mgh/t = (200 × 9.8 × 15)/60 = 490 W

2

Pᵢₙ = P_out / Efficiency

3

Pᵢₙ = 490 / 0.60 = 817 W

Answer: Input power = 817 W (or 820 W)

Summary

Key Formulas

QuantityFormulaUnit

Work Done

W = F s

or

W = F s cos θ

or Area under F-s graphJ

Kinetic Energy

E_{k} = \frac{1}{2} m v^{2}

J

Gravitational PE

E_{p} = m g h

J

Power

P = \frac{W}{t} = F v

W

Efficiency

η = \frac{E _{o u t}}{E _{in}} \times 100%

%

Key Exam Skills

Use W = Fs when parallel; use cos θ when at an angle

Calculate work from area under F-s graphs (rectangle, triangle, trapezium)

Apply conservation: total E before = total E after

Use P = Fv for objects at constant velocity

Efficiency always less than 100% for real systems