Defining Work & Work Done

Introduction

Key Concepts

1. Definition of Work

In physics, work is defined as the process of energy transfer when a force is applied to an object causing it to move. The concept of work quantifies the amount of energy required to perform physical tasks and is a scalar quantity, meaning it has magnitude but no direction.

The fundamental equation for work ($W$) is given by: $$ W = F \cdot d \cdot \cos(\theta) $$ where:

• $F$: The magnitude of the force applied (in Newtons, N).
• $d$: The displacement of the object in the direction of the force (in meters, m).
• $\theta$: The angle between the force vector and the direction of displacement.

This equation highlights that only the component of the force that acts in the direction of the displacement does work.

2. Units of Work

The SI unit of work is the Joule (J), where: $$ 1 \, \text{Joule} = 1 \, \text{Newton} \cdot \text{meter} = 1 \, \text{N} \cdot \text{m} $$ This unit signifies the work done when a force of one Newton displaces an object by one meter in the direction of the force.

3. Work Done by a Variable Force

When a force varies over the displacement, work cannot be calculated simply by multiplying force and displacement. Instead, calculus is employed to integrate the force over the path of displacement: $$ W = \int_{a}^{b} F(x) \, dx $$ where $F(x)$ represents the force as a function of position $x$, and the integration bounds $a$ and $b$ denote the initial and final positions.

For example, consider a force that increases linearly with displacement: $F(x) = kx$, where $k$ is a constant. The work done from $x = 0$ to $x = d$ would be: $$ W = \int_{0}^{d} kx \, dx = \frac{1}{2} k d^2 $$

4. Positive and Negative Work

Work can be positive or negative depending on the angle $\theta$ between the force and the displacement:

• Positive Work: Occurs when $0^\circ \leq \theta < 90^\circ$, meaning the force has a component in the direction of displacement. For instance, lifting a weight against gravity involves positive work.
• Negative Work: Occurs when $90^\circ < \theta \leq 180^\circ$, meaning the force has a component opposite to the direction of displacement. An example is the work done by friction when slowing down a moving object.

5. Work-Energy Theorem

The Work-Energy Theorem states that the net work done on an object is equal to the change in its kinetic energy ($\Delta KE$): $$ W_{\text{net}} = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 $$ where:

• $m$: Mass of the object (in kilograms, kg).
• $v_f$: Final velocity of the object (in meters per second, m/s).
• $v_i$: Initial velocity of the object (in meters per second, m/s).

This theorem illustrates the direct relationship between work and the energy of an object.

6. Power

Power is the rate at which work is done or energy is transferred. It is a measure of how quickly work can be performed and is given by: $$ P = \frac{W}{t} $$ where:

• $P$: Power (in Watts, W).
• $W$: Work done (in Joules, J).
• $t$: Time taken to do the work (in seconds, s).

Alternatively, power can be expressed in terms of force and velocity: $$ P = F \cdot v $$ where $v$ is the velocity in the direction of the force.

Understanding power is essential when analyzing how quickly energy is utilized or transferred in various physical systems.

7. Practical Examples of Work in Physics

• Lifting Objects: When you lift a book from the floor to a shelf, you apply an upward force against gravity, resulting in positive work.
• Falling Objects: As an object falls, gravity does positive work on it, increasing its kinetic energy.
• Friction: Sliding a box across the floor involves friction doing negative work, dissipating energy as heat.
• Pulling a Sled: When pulling a sled at a constant velocity, the applied force does positive work while friction does negative work, resulting in net zero work if forces balance out.

8. Calculating Work in Different Scenarios

Let’s explore a couple of scenarios to illustrate work calculations:

• Scenario 1: Pushing a shopping cart with a constant force of 50 N over a distance of 20 meters on a flat surface. Assuming the force is applied in the direction of displacement, the work done is: $$ W = F \cdot d \cdot \cos(0^\circ) = 50 \, \text{N} \cdot 20 \, \text{m} \cdot 1 = 1000 \, \text{J} $$
• Scenario 2: Lowering a weight slowly, applying a force of 100 N upwards over a displacement of 2 meters downward. The angle $\theta = 180^\circ$. $$ W = 100 \, \text{N} \cdot 2 \, \text{m} \cdot \cos(180^\circ) = 100 \cdot 2 \cdot (-1) = -200 \, \text{J} $$ Negative work indicates energy is being transferred from the system.

9. Caution in Calculations

When calculating work, it is crucial to:

• Identify the correct force to consider.
• Determine the displacement in the direction of the force.
• Accurately measure the angle between force and displacement vectors.
• Use consistent units to avoid calculation errors.

Comparison Table

Summary and Key Takeaways