Understanding the concept of work done by a force is pivotal. It's not just about calculating a value but also about comprehending the physics behind the movement of objects. Work done is a measure of energy transferred by a force over a displacement, and this concept is central to many areas of physics and engineering.

Understanding Work Done by a Force

Work Done ($W$): Energy transferred when a force moves an object.

Force ($F$): Push or pull on an object.

Displacement ($d$): Distance over which force acts.

Formula for Work Done

The work done is given by the formula:

W = Fd cos(θ)

Where:

• W is the work done,
• F is the magnitude of the force,
• d is the displacement,
• θ (Theta) is the angle between the force and displacement vectors.

This formula accounts for the force component acting in the direction of displacement.

Image courtesy of Cuemath

Factors Affecting Work Done

1. Force Magnitude: More force = More work.

2. Displacement: Larger displacement = More work.

3. Angle of Application: Determines force component for work.

• At $0°$ (parallel): Maximum work.
• At $90°$ (perpendicular): No work $cos(90°)=0$.
• Other angles: Partial force contribution.

Work Done Examples:

Example 1: Horizontal Force

A force of 10 N is applied horizontally to move a box 5 meters. The task is to calculate the work done.

Solution:

Given Data:

• Force (F) = 10 N
• Displacement (d) = 5 m
• Angle (θ) = 0° (since the force is applied horizontally)

Calculating Work Done:

• The formula for work done (W) is $W = Fd \cos(\theta)$.
• In this case, $W = 10 \times 5 \times \cos(0°)$.
• Since $\cos(0°) = 1$,
• Therefore, $W = 10 \times 5 \times 1 = 50$Joules.

Conclusion: The work done in moving the box is 50 Joules.

Example 2: Force at an Angle

A force of 20 N is used to move an object 3 meters at an angle of 30° to the horizontal. The objective is to determine the work done.

Solution:

Given Data:

• Force (F) = 20 N
• Displacement (d) = 3 m
• Angle (θ) = 30°

Calculating Cosine of 30°:

• The cosine of 30° is $\cos(30°) \approx 0.866$ (rounded to three decimal places).

Calculating Work Done:

• The formula for work done is again $W = Fd \cos(\theta)$.
• Therefore, $W = 20 \times 3 \times 0.866$.
• Simplifying, $W \approx 51.96$ Joules.

Conclusion: The work done in this scenario is approximately 51.96 Joules.

Example 3: Varied Angles

Calculate the work done when a force of 15 N moves an object 4 meters at angles of 0°, 45°, and 90°.

Solution:

General Formula: $W = Fd \cos(\theta)$

• Where $F = 15 \, \text{N}, d = 4 \, \text{m},$ and $\theta$ is the angle between the force and displacement direction.

For $\theta = 0°$:

• $\cos(0°) = 1$
• $W = 15 \times 4 \times \cos(0°) = 60 \, \text{Joules}$

For $\theta = 45°:$

• $\cos(45°) \approx 0.707$
• $W = 15 \times 4 \times 0.707 \approx 42.43 \, \text{Joules}$

For $\theta = 90°$:

• $\cos(90°) = 0$
• $W = 15 \times 4 \times \cos(90°) = 0 \, \text{Joules}$

Conclusion: The angle at which the force is applied significantly affects the work done: 60 Joules at 0°, approximately 42.43 Joules at 45°, and 0 Joules at 90°.