Work-Energy Theorem Equation, Derivation & Rearrangements

Core idea

Overview

The Work-Energy Theorem states that the net work done by all forces acting on an object is equal to the change in the object's kinetic energy. This fundamental principle connects the concepts of force, displacement, and energy, providing an alternative method to analyze motion without directly using Newton's laws. It is particularly useful when forces are variable or when dealing with complex paths, as it only depends on the initial and final states of kinetic energy.

When to use: Apply this theorem when you need to find the net work done on an object, or when you know the net work and need to determine a change in speed. It's especially useful for problems where forces are not constant or paths are curved, as it avoids direct integration of force over distance.

Why it matters: This theorem is a powerful tool in physics and engineering, simplifying the analysis of motion and energy transfer. It's crucial for understanding energy conservation, designing efficient machines, analyzing collisions, and calculating the stopping distance of vehicles, providing insights into how energy transforms within a system.

Symbols

Variables

W_{\text{net}} = Net Work Done, KE_f = Final Kinetic Energy, KE_i = Initial Kinetic Energy, m = Mass, v_f = Final Velocity

W_{net}

Net Work Done

J

K E_{f}

Final Kinetic Energy

J

K E_{i}

Initial Kinetic Energy

J

m

Mass

k g

v_{f}

Final Velocity

m / s

v_{i}

Initial Velocity

m / s

Walkthrough

Derivation

Formula: Work-Energy Theorem

Derives the relationship between net work done and the change in an object's kinetic energy from Newton's Second Law.

The net force acts over a displacement.
The object is treated as a point mass.
The work done is by the net force.

1

Start with Newton's Second Law and Work Definition:

Begin with Newton's Second Law ( $F_{net} = ma$ ) and the definition of work done by a constant force ( $W = F_{net} d cos ϕ$ ). For simplicity, consider motion in one dimension where force and displacement are in the same direction, so $cos ϕ = 1$ and $W = F_{net} d$ .

F_{net} = ma and W = F_{net} d cos ϕ

2

Relate Acceleration to Velocity and Displacement:

From kinematics, the relationship between initial velocity ( $v_{i}$ ), final velocity ( $v_{f}$ ), acceleration ( $a$ ), and displacement ( $d$ ) is $v_{f}^{2} = v_{i}^{2} + 2 a d$ . Rearrange this to express acceleration $a$ in terms of velocities and displacement.

v_{f}^{2} = v_{i}^{2} + 2 a d ⟹ a = \frac{v _{f}^{2} - v _{i}^{2}}{2 d}

3

Substitute 'a' into Newton's Second Law:

Substitute the expression for acceleration into Newton's Second Law, $F_{net} = ma$ .

F_{net} = m (\frac{v _{f}^{2} - v _{i}^{2}}{2 d})

4

Substitute $F_{\text{net}}$ into Work Definition:

Now substitute this expression for $F_{net}$ into the work equation $W_{net} = F_{net} d$ .

W_{net} = (m \frac{v _{f}^{2} - v _{i}^{2}}{2 d}) d

5

Simplify to Work-Energy Theorem:

The displacement $d$ cancels out, and rearranging the terms yields the Work-Energy Theorem, showing that the net work done equals the change in kinetic energy ( $Δ K E = K E_{f} - K E_{i}$ ).

W_{net} = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}

Result

W_{net} = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}

Source: A-Level Physics Textbooks (e.g., AQA, Edexcel, OCR Energy modules), University Physics (e.g., Serway & Jewett, Halliday & Resnick)

Free formulas

Rearrangements

Solve for $m$

Work-Energy Theorem: Make m the subject

m = \frac{2 W _{net}}{v _{f}^{2} - v _{i}^{2}}

To make m (mass) the subject, multiply the net work by 2 and divide by the difference of the squared final and initial velocities.

Difficulty: 3/5

Solve for $v_{f}$

Work-Energy Theorem: Make $v_{f}$ the subject

v_{f} = \frac{2 W _{net} + m v _{i}^{2}}{m}

To make $v_{f}$ (final velocity) the subject, rearrange the equation to isolate $v_{f}^{2}$ , then take the square root of both sides.

Difficulty: 4/5

Solve for $v_{i}$

Work-Energy Theorem: Make $v_{i}$ the subject

v_{i} = \frac{m v _{f}^{2} - 2 W _{net}}{m}

To make $v_{i}$ (initial velocity) the subject, rearrange the equation to isolate $v_{i}^{2}$ , then take the square root of both sides.

Difficulty: 4/5

Solve for $K E_{f}$

Work-Energy Theorem: Make $K E_{f}$ the subject

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

To make $K E_{f}$ (final kinetic energy) the subject, add the initial kinetic energy ( $K E_{i}$ ) to the net work done ( $W_{net}$ ).

Difficulty: 1/5

Solve for $K E_{i}$

Work-Energy Theorem: Make $K E_{i}$ the subject

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

To make $K E_{i}$ (initial kinetic energy) the subject, subtract the net work done ( $W_{net}$ ) from the final kinetic energy ( $K E_{f}$ ).

Difficulty: 1/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Visual intuition

Graph

Graph unavailable for this formula.

The graph is a straight line with a slope of one, featuring a y-intercept at negative initial kinetic energy that increases steadily as final kinetic energy grows. This linear relationship means that for every unit increase in final kinetic energy, the net work done increases by the exact same amount. Large values on the x-axis represent a state of high final energy, while small values indicate that the object has lost energy or remains at a lower energy state. The most important feature is the constant slope, whic

Graph type: linear

Why it behaves this way

Intuition

Visualize an object changing its speed: the total push or pull (net work) directly accounts for how much its energy of motion increases or decreases.

W_{net}

Net work done on the object by all forces.

The total energy transferred to or from the object by external actions; positive work increases the object's speed, negative work decreases it.

m

Mass of the object.

A measure of an object's inertia; a larger mass requires more work to achieve the same change in speed.

v_{f}

Final speed of the object.

How fast the object is moving after the net work has been done.

v_{i}

Initial speed of the object.

How fast the object is moving before the net work begins.

\frac{1}{2} m v_{f}^{2}

Final kinetic energy of the object.

The energy of motion the object possesses at the end; it depends quadratically on speed, meaning doubling speed quadruples this energy.

\frac{1}{2} m v_{i}^{2}

Initial kinetic energy of the object.

The energy of motion the object possesses at the beginning.

Signs and relationships

\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2: This subtraction represents the change in kinetic energy (ΔK). If the net work ( $W_{n}$ et) is positive, the final kinetic energy is greater than the initial, indicating the object gained speed.

Free study cues

Insight

Canonical usage

This equation requires consistent units for work and kinetic energy, typically using SI units where work is in joules, mass in kilograms, and velocity in meters per second.

Common confusion

A common mistake is mixing unit systems, such as using mass in grams with velocity in meters per second, which will lead to incorrect energy units.

Unit systems

$W_{net}$ Joule (J) · Represents the net work done on the object. In CGS, the unit is erg.

$m$ kilogram (kg) · Represents the mass of the object. In CGS, the unit is gram (g).

$v_{f}$ meter per second (m/s) · Represents the final speed of the object. In CGS, the unit is centimeter per second (cm/s).

$v_{i}$ meter per second (m/s) · Represents the initial speed of the object. In CGS, the unit is centimeter per second (cm/s).

One free problem

Practice Problem

A 2 kg object accelerates from rest to a speed of 5 m/s. Calculate the net work done on the object.

Mass2 kg

Initial Velocity0 m/s

Final Velocity5 m/s

Solve for: $W_{t} e x t n e t$

Hint: Calculate initial and final kinetic energies first, then find their difference.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

Calculating the work done by brakes to stop a car, or the speed of a roller coaster at the bottom of a hill.

Study smarter

Tips

Remember that $W_{net}$ is the work done by the *net* force, not just one force.
Kinetic energy is always positive, but the change in kinetic energy ( $Δ K E$ ) can be negative if the object slows down.
Ensure all units are consistent (Joules for work/energy, kg for mass, m/s for velocity).
The theorem applies regardless of the path taken, only depending on initial and final speeds.

Avoid these traps

Common Mistakes

Forgetting to use the *net* work, or including work done by non-conservative forces incorrectly.
Confusing initial and final velocities, leading to incorrect sign for $Δ K E$ .
Not squaring the velocities when calculating kinetic energy.

Keep going

Related Formulas

Common questions

Frequently Asked Questions

Derives the relationship between net work done and the change in an object's kinetic energy from Newton's Second Law.

Apply this theorem when you need to find the net work done on an object, or when you know the net work and need to determine a change in speed. It's especially useful for problems where forces are not constant or paths are curved, as it avoids direct integration of force over distance.

This theorem is a powerful tool in physics and engineering, simplifying the analysis of motion and energy transfer. It's crucial for understanding energy conservation, designing efficient machines, analyzing collisions, and calculating the stopping distance of vehicles, providing insights into how energy transforms within a system.

Forgetting to use the *net* work, or including work done by non-conservative forces incorrectly. Confusing initial and final velocities, leading to incorrect sign for $\Delta KE$. Not squaring the velocities when calculating kinetic energy.