Prerequisites
Before studying this topic, review:
What is Power?
Power is the rate at which work is done or energy is transferred. While work tells us how much energy is transferred, power tells us how fast it happens.
Definition
$$\boxed{P = \frac{dW}{dt}}$$Power = Rate of work done = Rate of energy transfer
Average Power
Total work divided by total time:
$$\boxed{P_{avg} = \frac{W_{total}}{t_{total}} = \frac{\Delta E}{\Delta t}}$$Example
If a machine does 1000 J of work in 5 seconds:
$$P_{avg} = \frac{1000}{5} = 200 \text{ W}$$Instantaneous Power
Using Force and Velocity
$$\boxed{P = \vec{F} \cdot \vec{v} = Fv\cos\theta}$$where $\theta$ is the angle between force and velocity.
Derivation
$$P = \frac{dW}{dt} = \frac{\vec{F} \cdot d\vec{r}}{dt} = \vec{F} \cdot \frac{d\vec{r}}{dt} = \vec{F} \cdot \vec{v}$$Interactive Demo: Visualize Power and Work
Explore the relationship between power, force, and velocity.
Units of Power
SI Unit: Watt (W)
$$1 \text{ W} = 1 \text{ J/s} = 1 \text{ kg·m}^2/\text{s}^3$$Common Units
| Unit | Conversion |
|---|---|
| 1 kilowatt (kW) | 1000 W |
| 1 megawatt (MW) | $10^6$ W |
| 1 horsepower (hp) | 746 W |
| 1 kWh (energy) | $3.6 \times 10^6$ J |
kWh is a unit of energy, not power!
$$1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}$$Power in Common Scenarios
1. Lifting at Constant Velocity
Force = Weight = $mg$ (upward) Velocity = $v$ (upward)
$$P = mgv$$2. Vehicle Moving on Level Road
Against resistive force $f$ (like friction) at constant velocity:
$$P = fv$$3. Vehicle Moving Up an Incline
Against gravity component + friction:
$$P = (mg\sin\theta + f)v$$4. Maximum Velocity of a Vehicle
At maximum speed, all engine power goes to overcome resistance:
$$P = f_{max} \cdot v_{max}$$If resistance $f = kv^2$:
$$P = kv_{max}^3$$ $$v_{max} = \left(\frac{P}{k}\right)^{1/3}$$Motion Under Constant Power
When power $P$ is constant (compare with uniform acceleration where force is constant):
$$P = Fv = mav = m\frac{dv}{dt}v$$ $$P\,dt = mv\,dv$$Integrating from $v_0$ to $v$:
$$Pt = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2$$Starting from rest ($v_0 = 0$):
$$\boxed{v = \sqrt{\frac{2Pt}{m}}}$$So velocity increases as $\sqrt{t}$ under constant power.
Position under Constant Power
$$v = \frac{dx}{dt} = \sqrt{\frac{2Pt}{m}}$$ $$x = \int_0^t \sqrt{\frac{2P}{m}}\sqrt{t}\,dt = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3}t^{3/2}$$ $$\boxed{x = \frac{2}{3}\sqrt{\frac{2P}{m}}t^{3/2}}$$Position varies as $t^{3/2}$.
Worked Examples
Example 1: Power of a Pump
Solution: Work done:
$$W = mgh = 100 \times 10 \times 20 = 20000 \text{ J}$$Power:
$$P = \frac{W}{t} = \frac{20000}{50} = \boxed{400 \text{ W}}$$Example 2: Vehicle Power
Solution: At maximum speed:
$$P = f \cdot v_{max}$$ $$75000 = 1500 \times v_{max}$$ $$v_{max} = \frac{75000}{1500} = \boxed{50 \text{ m/s} = 180 \text{ km/h}}$$Example 3: Constant Power Motion
Solution:
Velocity:
$$v = \sqrt{\frac{2Pt}{m}} = \sqrt{\frac{2 \times 16 \times 4}{2}} = \sqrt{64} = \boxed{8 \text{ m/s}}$$Distance:
$$x = \frac{2}{3}\sqrt{\frac{2P}{m}}t^{3/2} = \frac{2}{3}\sqrt{\frac{32}{2}}(4)^{3/2}$$ $$x = \frac{2}{3}\sqrt{16} \times 8 = \frac{2}{3} \times 4 \times 8 = \boxed{\frac{64}{3} \approx 21.3 \text{ m}}$$Example 4: Horsepower Conversion
Solution:
$$P = 100 \times 746 = 74600 \text{ W} = \boxed{74.6 \text{ kW}}$$Power vs Time Graph
Area Under Power-Time Graph
$$\text{Area} = \int P\,dt = W = \text{Work done}$$If Power is Constant
Work = $P \times t$ = Area of rectangle
If Power Varies Linearly
Work = Area of trapezoid or triangle
Practice Problems
A crane lifts a 500 kg load at 2 m/s. Find the power developed.
A car of mass 1000 kg accelerates from rest to 20 m/s in 10 s. Find the average power delivered by the engine.
A motor operates at 5 kW for 2 hours. Calculate the energy consumed in kWh and Joules.
Under constant power 18 W, a 1 kg body starts from rest. Find its speed after traveling 12 m.