Instantaneous Power

Derivation

Instantaneous power is the rate of doing work at a specific instant:

$P = \frac{dW}{dt}$

Work done in a small displacement $d\vec{s}$:

$dW = \vec{F} \cdot d\vec{s}$

Divide by $dt$:

$P = \frac{dW}{dt} = \vec{F} \cdot \frac{d\vec{s}}{dt} = \vec{F} \cdot \vec{v}$

$\boxed{P = \vec{F} \cdot \vec{v} = Fv\cos\theta}$

where $\theta$ is the angle between the force vector and the velocity vector.

What this means

Power is the dot product of force and velocity. Only the component of force along the velocity contributes to power.

• Force parallel to velocity ($\theta = 0$): $P = Fv$ — maximum power
• Force perpendicular to velocity ($\theta = 90°$): $P = 0$ — no power (centripetal force, normal force)
• Force antiparallel to velocity ($\theta = 180°$): $P = -Fv$ — negative power (braking)

How power varies during motion

Even if force is constant, power $P = Fv\cos\theta$ varies as velocity and direction change.

Car accelerating: As speed increases, the engine delivers increasing power (at constant force).

Body sliding down incline at constant speed: $F_{net} = 0$, but gravity does work at rate $P = mgv\sin\theta$ and friction dissipates at rate $P = f_k v$. They are equal (body moves at constant speed).

Finding velocity when power and force are given

$P = Fv \implies v = \frac{P}{F}$

Maximum speed of a vehicle: At maximum speed, the driving force equals resistance (constant velocity, $a = 0$):

$P_{engine} = F_{resistance} \cdot v_{max}$

$v_{max} = \frac{P_{engine}}{F_{resistance}}$

Example: Engine power 60 kW, road resistance 2000 N:

$v_{max} = \frac{60000}{2000} = 30 \text{ m/s} = 108 \text{ km/h}$

Units

$[P] = [F][v] = \text{N} \cdot \text{m/s} = \text{J/s} = \text{W}$ — consistent with the average power formula.