The Work-Energy Theorem

Hello! Welcome back to our exploration of work and energy.

Introduction

In our last lesson, we established a precise physical definition for work as the product of force and displacement, calculated by the formula $W = F d cos θ$ . We saw that work can be positive, negative, or zero depending on how a force acts relative to an object's motion.

But what does work actually accomplish? When you do work on an object, you are transferring energy to it. Today, we'll connect the concept of work directly to an object's motion. The learning outcome for this lesson is to relate the net work done on an object to the change in its kinetic energy, a principle known as the Work-Energy Theorem. This theorem is a powerful new tool in our physics toolkit, often simplifying problems that would be more complex to solve using Newton's laws alone.

1. Kinetic Energy and the Work-Energy Theorem

First, let's define the type of energy associated with motion. This is called kinetic energy (KE). It's the energy an object possesses because it is moving. The formula for kinetic energy is:

K E = \frac{1}{2} m v^{2}

where $m$ is the object's mass and $v$ is its velocity. Notice that since velocity is squared, kinetic energy is always a positive scalar quantity. An object either has it, or it doesn't.

So, how does work affect this energy? The fundamental idea is that positive work increases kinetic energy, and negative work decreases kinetic energy. Let's watch a video that builds the intuition for this relationship.

Work-energy theorem | Work & Energy | Physics | Khan Academy

This Khan Academy video provides an excellent conceptual explanation of how work changes an object's kinetic energy. It covers what happens when you do positive work, negative work, and what happens when multiple forces are acting.

Please watch from the beginning up to 8:02. Focus on how the examples (throwing a ball, catching a ball, a falling ball, pushing a cupboard) illustrate the connection between the work done and the object's speed.

The key idea from the video is that we must consider the net work—the sum of the work done by all forces acting on the object. This leads us to the formal statement of the Work-Energy Theorem.

Work-Energy Theorem Illustration — This diagram illustrates the core idea of the Work-Energy Theorem. The net work done ($W$) on the box causes its kinetic energy to change from an initial state ($KE_i$) to a final state ($KE_f$).

The theorem states:
The net work done on an object is equal to the change in its kinetic energy.

Mathematically, this is expressed as:

W_{net} = Δ K E = K E_{f} - K E_{i}

Where:

$W_{net}$ is the sum of the work done by all forces.
$K E_{f} = \frac{1}{2} m v_{f}^{2}$ is the final kinetic energy.
$K E_{i} = \frac{1}{2} m v_{i}^{2}$ is the initial kinetic energy.

If the net work is positive, $K E_{f} > K E_{i}$ , and the object speeds up. If the net work is negative, $K E_{f} < K E_{i}$ , and the object slows down. If the net work is zero, the kinetic energy (and thus the speed) remains constant.

2. Deriving the Theorem from First Principles

You've expressed an interest in understanding physics from "first principles," which is an excellent way to learn. The Work-Energy Theorem isn't a new, independent law of nature; it's a direct mathematical consequence of Newton's Second Law. Let's see how.

The derivation connects the definition of work with the kinematic equations you learned in our first module.

Work-energy theorem | Work & Energy | Physics | Khan Academy

Let's return to the Khan Academy video to watch the mathematical derivation. This section shows step-by-step how Newton's Second Law leads directly to the Work-Energy Theorem.

Watch the segment from 8:02 to 11:47. Follow the logic of how F=ma and v_f^2 = v_i^2 + 2as are combined to produce the final theorem.

To summarize the derivation for a constant net force:

Start with the definition of work done by a net force: $W_{net} = F_{net} \cdot s$ (using $s$ for displacement here).
Substitute Newton's Second Law, $F_{net} = ma$ : $W_{net} = (ma) s$ .
Recall the kinematic equation: $v_{f}^{2} = v_{i}^{2} + 2 a s$ . Rearranging this for the term $a s$ gives us $a s = \frac{v _{f}^{2} - v _{i}^{2}}{2}$ .
Substitute this expression for $a s$ into our work equation: $W_{net} = m (\frac{v _{f}^{2} - v _{i}^{2}}{2})$ .
Distribute the mass $m$ : $W_{net} = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}$ .
This is exactly $W_{net} = K E_{f} - K E_{i}$ , or $W_{net} = Δ K E$ .

Seeing this derivation confirms that we are still operating under the same fundamental rules of motion, just looking at them through the lens of energy.

3. Applying the Theorem: Solved Examples

Now, let's see how this theorem is used to solve problems. It's particularly useful when you are asked for an object's speed and you know the forces and distance, without needing to calculate time or acceleration as intermediate steps.

First, a straightforward example.

Work-Energy Theorem: Statement, Examples, and Problems

This webpage provides several worked-out problems. Let's look at the first one to see a direct application of the theorem.

Read 'Problem 1' in the 'Solved Problems' section. Notice how the work is calculated first, and then the Work-Energy Theorem is used to find the final speed.

Now, let's look at a more realistic scenario involving friction. Remember, friction is a force that typically does negative work because it acts opposite to the direction of motion.

Work-Energy Theorem Car Braking Example — This image shows how to use the Work-Energy Theorem to find the stopping distance of a car. The initial kinetic energy is positive. The work done by friction is negative, which removes energy from the system, bringing the final kinetic energy to zero.

In this car example:

The initial kinetic energy ( $K E_{i}$ ) is known because the car has a speed.
The final kinetic energy ( $K E_{f}$ ) is zero because the car stops.
The net work is the work done by friction, $W_{friction} = F_{k} \cdot d \cdot cos (18 0^{\circ}) = - F_{k} d$ .
By setting $W_{net} = K E_{f} - K E_{i}$ , we get $- F_{k} d = 0 - K E_{i}$ . This allows us to solve for the stopping distance, $d$ .

4. Practice Problem

Time to try one on your own.

Problem: A 4.0 kg bowling ball is lifted from the floor to a height of 1.5 m. It is then dropped from rest. Using the Work-Energy Theorem, what is its speed just before it hits the floor? (Assume the only force acting on it as it falls is gravity, and use $g \approx 9.8 m/s^{2}$ ).

Click here for the solution

1. Identify the forces and calculate net work.

The only force doing work as the ball falls is gravity.
The force of gravity is $F_{g} = m g = (4.0 kg) (9.8 m/s^{2}) = 39.2 N$ .
The displacement is $d = 1.5 m$ downwards.
The force of gravity is also downwards, so the angle $θ$ between force and displacement is $0^{\circ}$ .
The net work is the work done by gravity: $W_{net} = W_{g} = F_{g} d cos (0^{\circ}) = (39.2 N) (1.5 m) (1) = 58.8 J$ .

2. Identify the initial and final kinetic energies.

The ball is "dropped from rest," so its initial velocity is $v_{i} = 0$ .
Therefore, the initial kinetic energy is $K E_{i} = \frac{1}{2} m v_{i}^{2} = 0$ .
The final kinetic energy is $K E_{f} = \frac{1}{2} m v_{f}^{2}$ , where $v_{f}$ is what we need to find.

3. Apply the Work-Energy Theorem.

W_{net} = K E_{f} - K E_{i}

58.8 J = \frac{1}{2} (4.0 kg) v_{f}^{2} - 0

58.8 = 2.0 \cdot v_{f}^{2}

4. Solve for the final velocity.

v_{f}^{2} = \frac{58.8}{2.0} = 29.4

v_{f} = 29.4 \approx 5.42 m/s

Answer: The ball's speed just before hitting the floor is approximately 5.42 m/s.

Conclusion

Today we established a critical link between forces and energy. This will fundamentally expand your ability to analyze physical systems.

Key Takeaways:

Kinetic Energy is the energy of motion, given by $K E = \frac{1}{2} m v^{2}$ .
The Work-Energy Theorem states that the net work done on an object equals the change in its kinetic energy: $W_{net} = Δ K E$ .
Positive net work increases speed, negative net work decreases speed, and zero net work means constant speed.
This theorem is not a new law, but a direct result of Newton's Second Law, providing an alternative and often simpler method for solving motion problems.

Preview of the Next Lesson:
In the braking car example, friction did negative work, and the car's kinetic energy decreased. But where did that energy go? It was converted into heat in the brakes and tires. This leads to a crucial question: is energy ever truly lost, or does it just change form? In our next lesson, we will explore this by distinguishing between conservative forces (like gravity), which store energy, and non-conservative forces (like friction), which dissipate it. This will set the stage for one of the most important principles in all of physics: the conservation of energy.