Mechanical Power

Equivalent forms

P = \vec{F} \cdot \vec{v}

P = \frac{dW}{dt}

P = \tau\omega

The dimension of intensity — how the same work feels very different at different speeds.

Symbol	Name	SI unit	Dimension	Range
$P$	Poweroutput Rate of doing work or transferring energy	W	ML^2T^-3	0 – 100000
$F$	Force Applied force in the direction of motion	N	MLT^-2	0 – 5000
$v$	Velocity Velocity along the force direction	m/s	LT^-1	0 – 100

Unit systems

Symbol	SI	CGS	Imperial
$P$	W	erg/s	hp
$F$	N	dyn	lbf
$v$	m/s	cm/s	ft/s

Where it holds

Valid for any classical force-velocity pair. For rotational power

use P = \tau \omega

. Relativistic corrections needed when v approaches 299792458 m/s.

Dimensional analysis

P \to [ML^{2}T^{-3}]

F\cdot v \to [MLT^{-2}]\cdot [LT^{-1}] = [ML^{2}T^{-3}]

Both sides are watts.

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Discovery

James Watt · 1782

Watt coined 'horsepower' (550 ft·lbf/s) to market his steam engines against draft horses, giving us a unit that still rates car engines today.

Try this

How many 100-watt humans does it take to power a 100 hp car?

Convert 100 horsepower to watts and divide by typical sustained human output (~100 W). Apply P = Fv to understand engine output scales.

Research status: stable

Real-world applications

Car engine specifications (kW and hp)
Electric vehicle range calculations
Cyclist FTP (Functional Threshold Power) training
Wind turbine output rating

Common misconceptions

Horsepower (746 W) is not the same as 1 hp of human output $(\sim 100\,\mathrm{W}$ sustained)
Power and energy are different: a 100 W bulb for 10 s uses 1000 J of energy
Holding a heavy weight is exhausting but delivers zero mechanical power

Experimental verification

Dynamometers measure engine torque and shaft RPM, then compute

P = \tau \omega

. Cyclist power meters use strain gauges to measure

F\cdot v

in real time.

Derivation

Work over a small displacement:

dW = F\cdot dx

.

Divide by dt:

P = dW/dt = F\cdot (dx/dt) = F\cdot v

.

For variable force or angle,

P = F\cdot v \cos \theta = F

⃗

\cdot v

⃗.

Limiting cases

v \to 0

⟶

P \to 0

Static force does no work — holding a weight in place delivers zero mechanical power.

F \to 0

⟶

P \to 0

No force, no power, regardless of speed.

F ⊥ v⟶

P \to 0

Perpendicular force does no work (e.g., centripetal force on a circular orbit).

What if…

What if velocity doubles at constant force?

Power doubles. To maintain a car's speed against quadrupled drag at $2\times$ speed, the engine must output $8\times$ more power.

What if force is at angle

\theta to

velocity?

$P = Fv \cos \theta$ . Tangential component does work; perpendicular component does not.

What if the system rotates instead of translates?

$Use P = \tau \omega$ : torque times angular velocity. A $200 N\cdot m$ engine at 3000 RPM delivers about 63 kW.

1

100 hp car ≈ how many humans?

Given ·

P car hp:: 100
P human:: 100

Find · ratio

Steps

Convert: $100\,\mathrm{hp} \times 746\,\mathrm{W/hp} = 74600\,\mathrm{W}$
Sustained human cyclist output $\approx 100\,\mathrm{W}$
Ratio $= 74600 / 100 = 746$ — a car engine outputs nearly 750 human-power steady

Result ·

100\,\mathrm{hp} \times 746\,\mathrm{W/hp} = 74600\,\mathrm{W}

;

74600 \div 100 \approx 746

humans

2

Cyclist climbing a hill

Given ·

F:: 80
v:: 5

Find · P

Steps

Force overcoming gravity + drag along the slope: 80 N
Speed up the hill: 5 m/s
$P = 80 \times 5 = 400\,\mathrm{W}$ — elite Tour de France climbing zone

Result ·

P = Fv = 80 \times 5 = 400\,\mathrm{W}

Work-Energy Theorem→Kinetic Energy→Newton's Second Law→