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°.