Energy conservation tells us how much energy is transferred in a process, but not how fast. Power measures the rate at which energy is transferred or dissipated. It is the answer to "how many watts does this device use?" or "how much power can an engine produce?"

Definition of power

Power is the rate of energy transfer:

\[ P = \frac{dE}{dt} \]

where E is energy [J] and t is time [s].

SI unit: watt (W)

\[ 1 \text{ W} = 1 \text{ J/s} = 1 \text{ N} \cdot \text{m/s} \]

Average power over a time interval \( \Delta t \):

\[ P_{\text{avg}} = \frac{\Delta E}{\Delta t} \]

Instantaneous power at a moment:

\[ P = \frac{dE}{dt} \]

Power for a force doing work

When a force \( \mathbf{F} \) does work on an object, the instantaneous power is:

\[ P = \mathbf{F} \cdot \mathbf{v} = |\mathbf{F}| |\mathbf{v}| \cos\theta \]

where v is the instantaneous velocity of the object and \( \theta \) is the angle between force and velocity.

Why? Recall from §7.1 that work is \( W = \mathbf{F} \cdot \Delta \mathbf{r} \). Over a tiny time \( dt \), the displacement is \( d\mathbf{r} = \mathbf{v} \, dt \), so:

\[ dW = \mathbf{F} \cdot d\mathbf{r} = \mathbf{F} \cdot (\mathbf{v} \, dt) \]

\[ P = \frac{dW}{dt} = \mathbf{F} \cdot \mathbf{v} \]

Example: Pushing a car

A person pushes a car with force 500 N horizontally. The car moves at 2 m/s.

\[ P = F \cos(0°) \times v = 500 \times 1 \times 2 = 1000 \text{ W} = 1 \text{ kW} \]

The person delivers 1 kilowatt of power to the car.

Example: Force at an angle

A tractor pulls a plow with force 800 N at 30° above the horizontal. The plow moves at 1.5 m/s horizontally.

\[ P = F \cos(30°) \times v = 800 \times 0.866 \times 1.5 \approx 1039 \text{ W} \]

Only the horizontal component of force contributes to power.

Example: Climbing stairs

A 70 kg person climbs 10 meters vertically in 5 seconds.

Work done (against gravity): \[ W = mgh = 70 \times 9.8 \times 10 = 6860 \text{ J} \]

Average power: \[ P_{\text{avg}} = \frac{6860}{5} = 1372 \text{ W} \approx 1.4 \text{ kW} \]

The person's muscles deliver about 1.4 kilowatts during the climb. (In reality, the muscles are only ~25% efficient, so the metabolic power is 4 times higher.)

Efficiency

Efficiency is the ratio of useful output power to input power:

\[ \eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% \]

Most machines waste energy as heat due to friction and other inefficiencies.

Common efficiencies

Device Efficiency
Electric motor 85–95%
Car engine (internal combustion) 25–35%
Steam engine 30–40%
Solar cell 15–22%
Incandescent lightbulb ~5%
LED lightbulb ~30%
Bicycle + human ~25%

Example: Car engine efficiency

A car engine burns fuel and converts chemical energy into mechanical work. Suppose the engine:

\[ \eta = \frac{15}{50} \times 100\% = 30\% \]

The remaining 35 kW is dissipated as:

Power and energy rate

Example: Battery charge and discharge

A smartphone battery stores 15 Wh (watt-hours) of energy.

If charged in 1 hour: \[ P = \frac{15 \text{ Wh}}{1 \text{ h}} = 15 \text{ W} \]

If discharged (used) in 8 hours of typical use: \[ P = \frac{15 \text{ Wh}}{8 \text{ h}} = 1.875 \text{ W} \]

If fast-charged in 30 minutes: \[ P = \frac{15 \text{ Wh}}{0.5 \text{ h}} = 30 \text{ W} \]

Same energy, but delivered at different rates (different powers).

Power and kinetic energy

If an object accelerates from rest, the power delivered relates to kinetic energy:

\[ P = \frac{dE_{\text{kinetic}}}{dt} = \frac{d(\frac{1}{2}mv^2)}{dt} = m v \frac{dv}{dt} = m v a = \mathbf{F} \cdot \mathbf{v} \]

This confirms the formula \( P = \mathbf{F} \cdot \mathbf{v} \).

Example: Accelerating a car (revisited)

A 1500 kg car accelerates from 0 to 25 m/s via an engine force of 3000 N.

At v = 0 (starting): \[ P = F \times v = 3000 \times 0 = 0 \text{ W} \]

No power is needed (the car hasn't started moving yet).

At v = 10 m/s (partway through): \[ P = 3000 \times 10 = 30{,}000 \text{ W} = 30 \text{ kW} \]

The engine delivers 30 kW at this speed.

At v = 25 m/s (reaching max speed): \[ P = 3000 \times 25 = 75{,}000 \text{ W} = 75 \text{ kW} \]

If the engine has a maximum power output (say, 60 kW), it cannot maintain the 3000 N force at this speed — the force must reduce as speed increases to stay within the power limit.

Kilowatt-hour: a unit of energy

In daily life, we use the kilowatt-hour (kWh) as an energy unit (on electric bills):

\[ 1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J} = 3.6 \text{ MJ} \]

Example: A 1.5 kW space heater running for 8 hours consumes: \[ E = 1.5 \text{ kW} \times 8 \text{ h} = 12 \text{ kWh} = 43.2 \text{ MJ} \]

If electricity costs \$0.12 per kWh: \[ \text{Cost} = 12 \text{ kWh} \times \$0.12/\text{kWh} = \$1.44 \]

Connection to prior sections

What you should be able to do

After this section, you should be able to:

Summary of Chapter 7

Chapter 7 develops the concept of energy from fundamentals to applications:

  1. §7.1–§7.2: Work is energy transfer by a force over a displacement.
  2. §7.3: The work–energy theorem relates work to kinetic energy change.
  3. §7.4: Potential energy "stores" energy in the configuration of a system.
  4. §7.5: Mechanical energy is conserved when only conservative forces act.
  5. §7.6: Power is the rate of energy transfer.

Unifying principle: Energy is conserved; it can be transformed between kinetic, potential, and thermal forms, but the total is constant (in an isolated system). Power measures the rate of these transformations.

📚 See also: Halliday Vol 1, Ch 7, §7.7 — Power. 📖 Open reference: OpenStax University Physics Vol 1 — Chapter 7.5: Power. 🎓 Real-world context: IEA (International Energy Agency) — World Energy Statistics (global power consumption trends).

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