14.1.3 Conservation of Mechanical Energy

Introduction

The principle of conservation of mechanical energy states that when only conservative forces (e.g., gravity) act on a system, the total mechanical energy remains constant. This means that energy can transform between kinetic energy (KE) and potential energy (PE), but the total energy does not change.

This note focuses on:

Understanding the conservation of mechanical energy.
Solving problems involving motion under gravity, including objects moving on inclined planes.

Conservation of Mechanical Energy

If only conservative forces (e.g., gravity) act on an object:

KE_\text{initial} + PE_\text{initial} = KE_\text{final} + PE_\text{final}

This principle allows us to solve problems without directly calculating the forces involved.

Worked Examples

infoNote

Example 1: Object Falling Freely Under Gravity

Problem

A ball of mass 2 kg is dropped from a height of 10 m.

Ignoring air resistance, find:

The speed of the ball just before it hits the ground.
The kinetic energy at that point.

Part 1: Speed at the Ground

Step 1: Use conservation of mechanical energy:

KE_\text{initial} + PE_\text{initial} = KE_\text{final} + PE_\text{final}

Step 2: At the top:

KE_\text{initial} = 0 \quad (\text{ball starts from rest})

PE_\text{initial} = mgh = 2 \times 9.8 \times 10 = :highlight[196 J]

Step 3: At the bottom:

PE_\text{final} = 0 \quad (\text{height is 0})

KE_\text{final} = \frac{1}{2}mv^2

Step 4: Equate total energy at the top and bottom:

196 = \frac{1}{2}(2)v^2

Step 5: Solve for $v$ :

196 = v^2 \quad \Rightarrow \quad v = \sqrt{196} = :success[14 ms⁻¹]

Part 2: Kinetic Energy at the Ground

Step 1: Use the kinetic energy formula:

KE = \frac{1}{2}mv^2

Step 2: Substitute $m = 2 \, \text{kg}$ and $v = 14 \, \text{ms}^{-1}$

KE = \frac{1}{2}(2)(14)^2 = 1 \times 196 = :highlight[196 J]

Final Answer:

Speed: 14 ms⁻¹

Kinetic Energy: 196 J

infoNote

Example 2: Block Sliding Down a Smooth Incline

Problem

A block of mass 3 kg slides down a smooth incline of length 5 m that makes an angle of 30° with the horizontal.

The block starts from rest at the top. Find:

The speed of the block at the bottom.
The total mechanical energy at the top and bottom.

Part 1: Speed at the Bottom

Step 1: Use conservation of mechanical energy:

KE_\text{initial} + PE_\text{initial} = KE_\text{final} + PE_\text{final}

Step 2: At the top:

KE_\text{initial} = 0 \quad (\text{block starts from rest})

PE_\text{initial} = mgh

where $h$ is the height of the incline:

h = d \sin \theta = 5 \times \sin 30^\circ = 5 \times 0.5 = :highlight[2.5 m]

PE_\text{initial} = 3 \times 9.8 \times 2.5 = :highlight[73.5 J]

Step 3: At the bottom:

PE_\text{final} = 0 \quad (\text{height is 0})

KE_\text{final} = \frac{1}{2}mv^2

Step 4: Equate total energy:

73.5 = \frac{1}{2}(3)v^2

Step 5: Solve for vv:

73.5 = 1.5v^2 \quad \Rightarrow \quad v^2 = \frac{73.5}{1.5} = 49

v = \sqrt{49} = :success[7 ms⁻¹]

Part 2: Total Mechanical Energy

At the top:

E_\text{mechanical} = KE_\text{initial} + PE_\text{initial} = 0 + 73.5 = :highlight[73.5 J]

At the bottom:

E_\text{mechanical} = KE_\text{final} + PE_\text{final} = 73.5 + 0 = :highlight[73.5 J]

Final Answer:

Speed: 7 ms⁻¹

Total Mechanical Energy: 73.5 J (at both the top and bottom)

Note Summary

infoNote

Common Mistakes

Forgetting to calculate height correctly: Use $h = d \sin \theta$ for inclined planes.
Mixing energy types: Ensure you correctly identify initial and final KE and PE.
Neglecting conservation principles: Remember that total mechanical energy remains constant if no external work is done.
Omitting units: Always express energy in joules (J) and speeds in meters per second (ms⁻¹).
Ignoring starting conditions: Include initial velocity in calculations when relevant.

infoNote

Key Formulas

Kinetic Energy:

KE = \frac{1}{2}mv^2

Potential Energy:

PE = mgh

Conservation of Mechanical Energy:

KE_\text{initial} + PE_\text{initial} = KE_\text{final} + PE_\text{final}

Height on an Incline:

h = d \sin \theta

Conservation of Mechanical Energy Simplified Revision Notes for A-Level Edexcel Further Maths Further Mechanics 1

14.1.3 Conservation of Mechanical Energy

Introduction

Conservation of Mechanical Energy

Worked Examples

Example 1: Object Falling Freely Under Gravity

Example 2: Block Sliding Down a Smooth Incline

Note Summary

Common Mistakes

Key Formulas

500K+ Students Use These Powerful Tools to Master Conservation of Mechanical Energy For their A-Level Exams.

Other Revision Notes related to Conservation of Mechanical Energy you should explore