Work-Energy Theorem

Equivalent forms

W_{\text{net}} = \frac{1}{2}m(v_f^2 - v_i^2)

\int \vec{F} \cdot d\vec{s} = \Delta KE

Bridges Newton's force-based mechanics with the energy perspective — the gateway to Lagrangian mechanics.

Symbol	Name	SI unit	Dimension	Range
$W_net$	Net Workoutput Total work done by all forces on the object	J	ML²T⁻²	-10000 – 10000
$m$	Mass Mass of the object	kg	M	0.1 – 2000
$v_f$	Final Velocity Velocity of the object after work is done	m/s	LT⁻¹	0 – 100
$v_i$	Initial Velocity Velocity of the object before work is done	m/s	LT⁻¹	0 – 100

Unit systems

Symbol	SI	CGS	Imperial
$W_net$	J	erg	ft·lbf
$m$	kg	g	slug
$v_f$	m/s	cm/s	ft/s
$v_i$	m/s	cm/s	ft/s

Where it holds

Valid for non-relativistic speeds (v << 299792458 m/s) and point particles or rigid bodies. Does not account for internal energy changes (heat, deformation).

Dimensional analysis

W \to [ML^{2}T^{-2}]

\tfrac{1}{2}m(v_f^{2} - v_i^{2}) \to [M]\cdot [LT^{-1}]^{2} = [ML^{2}T^{-2}]

Both sides are

[ML^{2}T^{-2}]

= Joule.

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Discovery

Gaspard-Gustave de Coriolis · 1829

Coriolis formalized the concept of 'work' in Du calcul de l'effet des machines, linking force over distance to energy change.

Try this

A car brakes from 30 m/s to rest. How does braking force relate to stopping distance?

W_net = ΔKE = 0 - ½mv². For a 1500 kg car at 30 m/s: ΔKE = -675,000 J. With F_brake = 10,000 N, stopping distance d = 67.5 m.

Research status: stable

Real-world applications

Vehicle braking distance estimation
Roller coaster energy budgets
Impact engineering
Sports performance analysis

Common misconceptions

Work done by individual forces can be positive, negative, or zero — only the net matters
Normal force on a flat surface does zero work (perpendicular to motion)
Friction always does negative work on the sliding object

Experimental verification

Air track experiments with photogates, crash test energy absorption measurements.

Derivation

Start with

F_net = ma

.

Multiply both sides by ds:

F_net\cdot ds = ma\cdot ds = m(dv/dt)\cdot ds = mv\cdot dv

.

Integrate from v_i to v_f:

W_net = \int mv\cdot dv = \tfrac{1}{2}mv_f^{2} - \tfrac{1}{2}mv_i^{2}

.

Limiting cases

v_f = v_i

⟶

W_net = 0

No change in speed means zero net work done.

v_i = 0

⟶

W_net = \tfrac{1}{2}mv_f^{2}

Starting from rest, net work equals final kinetic energy.

v_f = 0

⟶

W_net = -\tfrac{1}{2}mv_i^{2}

Bringing to rest requires negative work (energy removed by friction/braking).

What if…

What if braking force doubles?

Stopping distance halves. With 20,000 N of braking force: $d = 675000/20000 = 33.75\,\mathrm{m}$ .

What if the road is inclined upward?

Gravity does negative work too, assisting braking. W_gravity $= -mgh$ . The car stops in a shorter distance.

What about friction doing work?

Friction converts KE to thermal energy. W_friction $= -\mu mgd$ . This is why brakes heat up during hard stops.

1

Car braking distance

Given ·

m:: 1500
v i:: 30
v f:: 0
F brake:: 10000

Find · stopping distance d

Steps

$\Delta KE = \tfrac{1}{2}m(v_f^{2} - v_i^{2}) = \tfrac{1}{2}(1500)(0 - 900) = -675$ ,000 J
Work by brakes: $W = -F\cdot d$ (negative because force opposes motion)
675, $000 = 10$ , $000 \times d \to d = 67.5\,\mathrm{m}$

Result ·

W = \Delta KE \to F\cdot d = \tfrac{1}{2}mv^{2} \to d = mv^{2}/(2F) = 1500\times 900/(2\times 10000) = 67.5\,\mathrm{m}

2

Accelerating a sprinter

Given ·

m:: 75
v i:: 0
v f:: 10

Find · W_net

Steps

Starting from rest: $v_i = 0$
Final speed: $v_f = 10\,\mathrm{m/s}$
$W_net = \tfrac{1}{2}mv_f^{2} - \tfrac{1}{2}mv_i^{2} = \tfrac{1}{2}(75)(100) - 0 = 3750\,\mathrm{J}$
This energy comes from the sprinter's muscles

Result ·

W = \tfrac{1}{2}(75)(10^{2}) - 0 = 3750\,\mathrm{J}

Kinetic Energy→Newton's Second Law→Hooke's Law→