Power measured how fast energy is transferred. Two students climbing the same stairs do the same Work by a Constant Force, but the one who runs delivers it faster.
Key idea
Power is the rate of energy transfer: $P = \dfrac{W}{\Delta t}$, in watts ($1\ \text{W} = 1\ \text{J/s}$). For a force pushing along the velocity, $P = Fv$.
Why it works
Work divided by time is a rate, like speed is distance divided by time. If a force $F$ pushes along the motion, in a short time $\Delta t$ the object moves $v\Delta t$, so the work is $Fv\Delta t$ and the power is $Fv$. That's why a car's engine works hardest at highway speed: drag is large and so is $v$.
Worked examples
Worked example: Climbing stairs
A $60\ \text{kg}$ hiker climbs $12\ \text{m}$ of stairs in $15\ \text{s}$. What power does the climb take?
- Work against gravity: $mgh = (60)(9.8)(12) = 7{,}056\ \text{J}$.
- $P = \dfrac{7056}{15} \approx 470\ \text{W}$.
Answer: about $470\ \text{W}$
Worked example: A car at cruising speed
A car cruises at $25\ \text{m/s}$ against $600\ \text{N}$ of drag and rolling friction. What power reaches the wheels?
- At steady speed the forward force equals $600\ \text{N}$, along the velocity.
- $$P = Fv = (600)(25) = 15{,}000\ \text{W} = 15\ \text{kW}.$$
Answer: $15\ \text{kW}$
Worked example: Energy from power
A $1{,}500\ \text{W}$ hair dryer runs for $5$ minutes. How much energy does it use?
- $5\ \text{min} = 300\ \text{s}$.
- $E = P\,\Delta t = 1500 \times 300 = 450{,}000\ \text{J}$.
Answer: $450{,}000\ \text{J}$
Worked example: Kilowatt-hours
An electric bill counts energy in kilowatt-hours. How many joules is $1\ \text{kW}\cdot\text{h}$?
- $1\ \text{kW} = 1000\ \text{J/s}$ and $1\ \text{h} = 3600\ \text{s}$.
- $1\ \text{kW}\cdot\text{h} = 1000 \times 3600 = 3.6 \times 10^{6}\ \text{J}$, an amount of energy, not a power.
Answer: $3.6 \times 10^{6}\ \text{J}$
Common mistake
Mixing up power with energy (watts are joules per second; a kilowatt-hour is an amount of energy), or using $P = Fv$ with a force that isn't along the velocity. Check the units: power must come out in J/s. If the force is at an angle, use only its part along the velocity: $P = Fv\cos\theta$.