Power defined power as work per time, $P = W/\Delta t$, and showed $P = Fv$ for a steady push. With Work by a Variable Force, power becomes the rate of work at each instant, even when the force and speed change.
Key idea
$$P = \dfrac{dW}{dt} = \vec{F}\cdot\vec{v}.$$ The power of the net force equals $\dfrac{dK}{dt}$. Units: watts (J/s). The total work over a time interval is the area under the $P$–$t$ graph, $W = \int P\,dt$.
Why it works
In a tiny time $dt$ the object moves $d\vec{r} = \vec{v}\,dt$, so the force does work $dW = \vec{F}\cdot d\vec{r} = \vec{F}\cdot\vec{v}\,dt$. Divide by $dt$. A force perpendicular to the velocity, like the tension on a ball whirled in a circle at steady speed, delivers zero power: it changes the direction, never the speed.
Worked example: Cruising against drag
A car cruises at $25\ \text{m/s}$, with air drag and rolling resistance adding up to $500\ \text{N}$. What power reaches the wheels?
- At steady speed the driving force equals the resistance, $500\ \text{N}$, along the velocity.
- $P = Fv = 500(25) = 12\,500\ \text{W} = 12.5\ \text{kW}$.
Answer: $12.5\ \text{kW}$.
Worked example: Constant power from rest
A $1000\ \text{kg}$ car's engine delivers a steady $50\ \text{kW}$ starting from rest. Ignoring losses, when does it reach $20\ \text{m/s}$?
- $\dfrac{dK}{dt} = P$ is constant, so $K = Pt$.
- $\tfrac{1}{2}(1000)(20)^2 = 200\,000\ \text{J} = 50\,000t$, so $t = 4\ \text{s}$.
- The force $P/v$ is enormous at first and falls as the car speeds up: the acceleration is not constant.
Answer: $4\ \text{s}$.
Common mistake
Using the average power when the force or speed is changing. For the cart in the picture, the average power over 10 s is $5\ \text{kW}$, but at the end the push delivers $10\ \text{kW}$, twice as much. The average $W/\Delta t$ tells you the total energy per time; the instantaneous $\vec{F}\cdot\vec{v}$ tells you the rate right now. Check: if $v$ changes during the interval, the two will differ.
The same idea carries over to turning: a torque $\tau$ on something spinning at $\omega$ delivers $P = \tau\omega$, which you'll derive in Work and Power in Rotation. An engine's power rating is a torque times a rotation rate.
Drag force grows like $v^2$, so the power to overcome it grows like $v^3$: driving 10% faster takes about $1.1^3 \approx 1.33$ times the power against drag. That is one reason fuel use rises quickly at highway speeds.