Power

A weak motor can eventually do the same total work as a strong one — it just takes longer. Power captures the rate at which work is done. It measures how fast energy is transferred.

Concept

Average power over time $\Delta t$:

$P_{\text{avg}} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}$

Instantaneous power:

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

SI unit: watt (W), where $1 \, \text{W} = 1 \, \text{J/s}$.

Other units: 1 horsepower (hp) $\approx 746 \, \text{W}$.

Power is a scalar (dot product of two vectors).

Derivation

Starting from $dW = \vec{F}\cdot d\vec{r}$,

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

So instantaneous power equals force dotted with velocity. If force and velocity are parallel:

$P = Fv$

For constant force and constant velocity, $P$ is constant and equals $W/\Delta t$.

For variable force, total work equals the integral of power over time:

$W = \int P(t)\,dt$

Worked Example

A car of mass 1200 kg accelerates uniformly from rest to 20 m/s in 10 s. Assuming no friction, find (a) the work done, (b) average power, (c) instantaneous power at the end.

Solution:

(a) $W = \Delta K = \tfrac{1}{2}(1200)(20^2) = 2.4 \times 10^5 \, \text{J}$.

(b) $P_{\text{avg}} = W/\Delta t = 2.4 \times 10^5 / 10 = 24\,000 \, \text{W} = 24 \, \text{kW}$.

(c) Acceleration $a = 2 \, \text{m/s}^2$, force $F = ma = 2400 \, \text{N}$. Final velocity $v = 20 \, \text{m/s}$. Instantaneous power:

$P = Fv = 2400 \times 20 = 48\,000 \, \text{W} = 48 \, \text{kW}$

The instantaneous power at the end is twice the average — typical for uniform acceleration from rest.

Common Confusions

• Power $\neq$ energy. Power is the rate. A 1 kW heater used for 1 s delivers 1 kJ; for 1 hr it delivers 3600 kJ.
• kWh is energy, not power. $1 \, \text{kWh} = 3.6 \times 10^6 \, \text{J}$.
• Negative power. Power is negative if force opposes velocity (e.g., brakes).
• Constant velocity ≠ zero power. A car moving at constant speed against air drag delivers positive power (engine) and absorbs equal negative power (drag); these cancel in net work, but the engine is doing real work.

Key Takeaways

• Average power: $P_{\text{avg}} = W/\Delta t$.
• Instantaneous power: $P = \vec{F}\cdot\vec{v}$.
• SI unit: watt (W); 1 hp ≈ 746 W.
• $W = \int P\,dt$ is the total work.
• $P$ is a scalar; can be positive (driving) or negative (braking).