14.1.2 Work-Energy Principle

Introduction

The work-energy principle relates the work done on an object to the change in its kinetic energy. It provides a powerful way to solve problems involving forces and motion, without relying on the equations of motion.

This note focuses on:

The work done by constant forces.
Applying the principle to problems involving changes in kinetic energy and motion on inclined planes.

Work Done by a Constant Force

Work ( $W$ ) is the energy transferred by a constant force acting over a displacement. It is given by:

W = Fd\cos\theta

where:

$F$ is the magnitude of the constant force ( $\text{N}$ )
$d$ is the displacement ( $\text{m}$ )
$\theta$ is the angle between the force and displacement. If the force is in the same direction as the displacement ( $\theta = 0^\circ$ ):

W = Fd

If the force is perpendicular to the displacement ( $\theta = 90^\circ$ ):

W = 0

Work-Energy Principle

The work-energy principle states that the net work done on an object is equal to the change in its kinetic energy:

W_\text{net} = \Delta KE = KE_\text{final} - KE_\text{initial}

This principle applies to problems involving:

Forces acting along straight-line motion.
Motion on inclined planes with constant forces (e.g., gravity or friction).

Worked Examples

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Example 1: Box on a Smooth Horizontal Surface

Problem

A box of mass 3 kg is pushed along a smooth horizontal surface with a constant force of 10 N over a distance of 5 m. The box starts from rest. Find:

The work done on the box.
The final speed of the box.

Step 1: Use the work formula:

W = Fd\cos\theta

Since the force acts along the displacement ( $\theta = 0^\circ, \cos 0^\circ = 1$ ):

W = Fd = 10 \times 5 = :success[50 J]

Step 2: Use the work-energy principle:

W_\text{net} = \Delta KE = KE_\text{final} - KE_\text{initial}

Since $KE_\text{initial} = 0$ (the box starts from rest):

W_\text{net} = KE_\text{final}

Step 3: Write $KE_\text{final}$ in terms of the final velocity $v$ :

W_\text{net} = \frac{1}{2}mv^2

Step 4: Solve for $v$ :

50 = \frac{1}{2}(3)v^2 \quad \Rightarrow \quad v^2 = \frac{50 \times 2}{3} = 33.33

v = \sqrt{33.33} \approx :success[5.77 ms⁻¹]

Final Answer:

Work done: 50 J
Final speed: 5.77 ms⁻¹

infoNote

Example 2: Object Sliding Down an Inclined Plane

Problem

A block of mass 4 kg slides 6 m down a smooth incline of 30°, starting from rest. Find:

The work done by gravity.
The block's speed at the bottom of the incline.

Step 1: Find the component of gravitational force along the incline:

F_\text{gravity} = mg \sin\theta = 4 \times 9.8 \times \sin 30^\circ

F_\text{gravity} = 4 \cdot 9.8 \times 0.5 = :success[19.6 N]

Step 2: Calculate the work done:

W = F_\text{gravity} \times d = 19.6 \times 6 = :success[117.6 J]

Step 3: Use the work-energy principle:

W_\text{net} = \Delta KE = KE_\text{final} - KE_\text{initial}

Since $KE_\text{initial} = 0$

W_\text{net} = KE_\text{final}

Step 4: Write $KE_\text{final}$ in terms of $v$ :

117.6 = \frac{1}{2}mv^2 = \frac{1}{2}(4)v^2

Step 5: Solve for $v$ :

117.6 = 2v^2 \quad \Rightarrow \quad v^2 = \frac{117.6}{2} = 58.8

v = \sqrt{58.8} \approx :success[7.67 ms⁻¹]

Final Answer:

Work done: 117.6 J

Final speed: 7.67 ms⁻¹

Note Summary

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Common Mistakes

Forgetting to resolve forces: For inclined planes, resolve gravity into components parallel and perpendicular to the incline.
Mixing units: Ensure consistent units, especially for force ( $N$ ), displacement ( $m$ ), and work ( $j$ ).
Ignoring the angle $\theta$ : Work depends on the cosine of the angle between force and displacement.
Misapplying work-energy principle: Use only the net work done by all forces, not just a single force.
Omitting starting conditions: Account for initial velocity in $KE_\text{initial}$

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Key Formulas

Work Done by a Constant Force:

W = Fd\cos\theta

Kinetic Energy:

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

Work-Energy Principle:

W_\text{net} = \Delta KE = KE_\text{final} - KE_\text{initial}

Gravitational Work on an Incline:

W_\text{gravity} = F_\text{gravity} \cdot d, \quad F_\text{gravity} = mg\sin\theta

Work-Energy Principle Simplified Revision Notes for A-Level Edexcel Further Maths Further Mechanics 1

14.1.2 Work-Energy Principle

Introduction

Work Done by a Constant Force

Work-Energy Principle

Worked Examples

Example 1: Box on a Smooth Horizontal Surface

Example 2: Object Sliding Down an Inclined Plane

Note Summary

Common Mistakes

Key Formulas

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