7.7 Power

Power and Energy Transfer

Power tells you how fast work gets done or energy gets transferred. Two machines can do the same total work, but the one that does it in less time has greater power. This concept shows up everywhere, from comparing car engines to estimating your electricity bill.

Rate of Energy Transfer

Power ($P$) is defined as work done (or energy transferred) divided by time:

$P = \frac{W}{t}$

where $W$ is work (in joules) and $t$ is time (in seconds). The SI unit for power is the watt (W), which equals one joule per second (J/s).

Since work itself depends on force and displacement, you can also think of power in terms of those quantities. Recall that work is:

$W = Fd\cos\theta$

where $F$ is the applied force, $d$ is the displacement, and $\theta$ is the angle between the force and displacement vectors. When force and motion point the same direction, $\theta = 0°$ and $\cos\theta = 1$, so $W = Fd$.

Energy transfer can take many forms: mechanical (motion), electrical (current flow), and thermal (heat), among others. Regardless of the form, power always measures the rate of that transfer.

Example 1: A 100 N force moves an object 5 m in 2 s (force and displacement in the same direction):

$P = \frac{(100 \text{ N})(5 \text{ m})}{2 \text{ s}} = 250 \text{ W}$

Example 2: If 1000 J of energy is transferred in 10 s:

$P = \frac{1000 \text{ J}}{10 \text{ s}} = 100 \text{ W}$

Cost of Electrical Energy Usage

For electrical circuits, power is calculated by multiplying voltage and current:

$P = VI$

where $V$ is voltage (in volts) and $I$ is current (in amperes).

Energy consumption over time is then:

$E = Pt$

where $E$ is energy in joules, $P$ is power in watts, and $t$ is time in seconds.

In practice, utility companies don't bill you in joules. They use the kilowatt-hour (kWh), which is the energy consumed when 1000 W of power runs for 1 hour. Converting to SI units: 1 kWh = 3.6 × 10⁶ J.

To estimate cost, multiply energy consumption (in kWh) by the price per kWh:

$\text{Cost} = \text{Energy (kWh)} \times \text{Cost per kWh}$

Example: A 100 W light bulb runs for 10 hours at $0.12 per kWh.

1. Find energy consumed: $(100 \text{ W})(10 \text{ h}) = 1000 \text{ Wh} = 1 \text{ kWh}$
2. Find cost: $(1 \text{ kWh})(\$0.12/\text{kWh}) = \$0.12$

Keep in mind that no device converts energy with 100% efficiency. Some energy is always lost (usually as heat), so the actual power drawn from the outlet is higher than the useful power output. This affects both total consumption and cost.

Power Output Comparisons

It helps to have a sense of scale for power across different systems. Here are some approximate values:

One unit you'll see for engines is horsepower (hp). 1 hp ≈ 746 W. So a 200 hp car engine produces roughly 149,200 W, or about 149 kW.

Notice the enormous range: the Sun's output is roughly 10²⁴ times greater than a car engine. These comparisons give you a feel for what "a lot of power" actually means in different contexts.

Mechanical Power and Energy

A few additional connections worth knowing:

• Torque and rotation: For rotating machinery (motors, turbines), power depends on torque and rotational speed. You'll encounter this more in later courses, but the core idea is the same: power measures how fast energy is being transferred through rotation.
• Mechanical advantage: Simple machines (levers, pulleys) can increase the force you apply, but they don't increase power. You trade force for distance (or speed), so the rate of energy transfer stays the same (minus friction losses).
• Kinetic energy: A moving object carries kinetic energy ($KE = \frac{1}{2}mv^2$). The power required to accelerate that object depends on how quickly you change its kinetic energy. Faster acceleration demands more power, even if the final speed is the same.