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Power
The Rate of Doing Work

Understand how power quantifies the rate at which energy is transferred or work is done, and learn to apply \( P = \frac{W}{t} \) across real-world scenarios.

AP Physics 1

What is Power?

Power is the rate at which work is done or energy is transferred. Two machines can do the same total work, but the one that does it faster has greater power.

Definition
\( P = \frac{W}{t} \)
SI Unit
Watt (W), \( 1 \;\text{W} = 1 \;\text{J/s} \)
Symbol
\( P \) (uppercase italic)

Why Power Matters

Work tells you how much energy is transferred. Power tells you how quickly. This distinction is critical in engineering, biology, and everyday life.

Car Engines
A \( 300 \;\text{hp} \) engine delivers energy to the wheels faster than a \( 150 \;\text{hp} \) engine, enabling quicker acceleration.
Light Bulbs
A \( 100 \;\text{W} \) bulb converts electrical energy to light and heat at twice the rate of a \( 50 \;\text{W} \) bulb.

Power as Energy Transfer Rate

Time (s) Work (J) 0 2 4 6 8 0 400 800 High Power Low Power
\( \text{Slope} = P \)

The Power Equation

\( P \)
Power (watts, W)
\( W \)
Work done (joules, J)
\( t \)
Time interval (seconds, s)

Since work equals energy transferred, power also equals energy change per unit time: \( P = \frac{\Delta E}{t} \). This applies to any form of energy: kinetic, potential, thermal, or electrical.

Alternative Form: \( P = Fv \)

1
Start with \( P = \frac{W}{t} \)
2
Substitute \( W = Fd\cos\theta \), so \( P = \frac{Fd\cos\theta}{t} \)
3
Recognize \( \frac{d}{t} = v \), giving \( P = Fv\cos\theta \)

When force is parallel to velocity (\( \theta = 0^\circ \)), this simplifies to \( P = Fv \). This form is especially useful for objects moving at constant velocity.

Additional worked example

The following is a separate example from the long-form lesson. Its values may differ from the example already on this slide.

An elevator of mass 1200 kg carries passengers with a combined mass of 300 kg. The elevator motor lifts the system at a constant velocity of 2.5 m/s. Determine the power output of the motor. Then calculate the total energy the motor delivers over a 12-second ride.

Show answer and explanation

Step 1 — Identify Given Values: Total mass: m = 1200 kg + 300 kg = 1500 kg. Constant velocity: v = 2.5 m/s (upward). Acceleration due to gravity: g = 9.8 m/s². Time of ride: Δt = 12 s. Because the elevator moves at constant velocity, the net force is zero, so the motor's upward tension force equals the total weight. m = 1500 kg, v = 2.5 m/s, g = 9.8 m/s², Δt = 12 s Step 2 — Calculate the Required Force: At constant velocity, the applied force must balance gravity: F = mg = 1500 kg × 9.8 m/s² = 14,700 N. This is the tension in the cable (assuming no friction or cable mass). F = 14,700 N Step 3 — Apply P = Fv: Since the force and velocity are both directed upward (θ = 0°, cos 0° = 1): P = Fv = 14,700 N × 2.5 m/s = 36,750 W. P = 36,750 W ≈ 36.8 kW Step 4 — Calculate Total Energy Delivered: Rearranging P = W/Δt gives W = P × Δt = 36,750 W × 12 s = 441,000 J. This equals the gravitational potential energy gained: ΔPE = mgh = 1500 × 9.8 × (2.5 × 12) = 1500 × 9.8 × 30 = 441,000 J. ✓ The energy balance confirms our result. W = 441,000 J = 441 kJ Step 5 — Verify Units and Reasonableness: Unit check: [kg][m/s²][m/s] = kg·m²/s³ = W ✓. A 36.8 kW motor is roughly 49 hp, which is reasonable for a commercial elevator. The 441 kJ of energy delivered equals the gravitational potential energy gained over a 30-meter rise, consistent with a 12-story building.

Equal 1500 kg loads rise 30 m and each gain 441 kJ. At 2.5 m/s the lift takes 12 s and delivers 36.75 kW; at 1.25 m/s it takes 24 s and delivers 18.375 kW. This is a comparison illustration in the separate Physics C car lesson.
Equal 1500 kg loads rise 30 m and each gain 441 kJ. At 2.5 m/s the lift takes 12 s and delivers 36.75 kW; at 1.25 m/s it takes 24 s and delivers 18.375 kW. This is a comparison illustration in the separate Physics C car lesson.

Visualizing \( P = Fv \)

\( m \)
\( F \)
\( v \)
\( P = Fv \)

Power vs. Work vs. Energy

Property Power Work Energy
What it measures Rate of energy transfer Amount of energy transferred Capacity to do work
Equation \( P = W/t \) \( W = Fd\cos\theta \) \( KE = \tfrac{1}{2}mv^2 \)
SI Unit Watt (W) Joule (J) Joule (J)
Scalar/Vector Scalar Scalar Scalar

Average vs. Instantaneous Power

Average Power
Total work divided by total time: \( P_{\text{avg}} = \frac{W_{\text{total}}}{\Delta t} \). Useful when the rate varies over an interval.
Instantaneous Power
Power at a single moment: \( P = Fv\cos\theta \). Useful for analyzing a specific instant in time.

How to Solve Power Problems

1
Identify what is given: force, distance, time, velocity, mass, or height.
2
Choose the right equation: Use \( P = W/t \) if you know total work and time. Use \( P = Fv \) if you know force and velocity.
3
Calculate work if needed: \( W = Fd\cos\theta \) or \( W = \Delta KE \) or \( W = mgh \) for lifting against gravity.
4
Plug in and solve. Check units: force in N, distance in m, time in s gives watts.
5
Verify your answer: Does the magnitude make sense? (A human \( \approx 100 \;\text{W} \), a car engine \( \approx 100{,}000 \;\text{W} \).)

Example 1: Lifting a Box

20 kg
\( h = 3 \) m
\( t = 5 \) s

Example 2: Car at Constant Speed

\( F = 500 \) N
\( v = 20 \) m/s

Example 3: Accelerating Object

A \( 1{,}200 \;\text{kg} \) car accelerates from rest to \( 30 \;\text{m/s} \) in \( 8 \;\text{s} \). Find the average power delivered by the engine (ignore friction).

1
Find work via kinetic energy: \( W = \Delta KE = \frac{1}{2}mv^2 - 0 = \frac{1}{2}(1200)(30)^2 = \) 540{,}000 J
2
Average power: \( P_{\text{avg}} = \frac{W}{t} = \frac{540{,}000}{8} = \) 67{,}500 W = 67.5 kW

Units of Power

Unit Symbol Equivalent
Watt W \( 1 \;\text{J/s} \)
Kilowatt kW \( 1{,}000 \;\text{W} \)
Horsepower hp \( 746 \;\text{W} \)
Kilowatt-hour kWh \( 3.6 \times 10^6 \;\text{J} \) (energy, not power)

Common Mistakes

✕ Confusing Power and Work
More work does not mean more power. Power depends on how fast the work is done.
✕ Forgetting the Angle
When force and velocity are not parallel, you must use \( P = Fv\cos\theta \), not \( P = Fv \).
✕ Unit Mix-ups
Using km/h instead of m/s, or kJ instead of J. Always convert to SI before calculating.

Quick Reference Summary

Formula When to Use Key Variables
\( P = \frac{W}{t} \) Total work and time known \( W \) in J, \( t \) in s
\( P = \frac{\Delta E}{t} \) Energy change and time known Any \( \Delta E \) (KE, PE, etc.)
\( P = Fv\cos\theta \) Force and velocity known at an instant \( F \) in N, \( v \) in m/s
\( P = Fv \) Force parallel to velocity Simplified case (\( \theta = 0 \))

Question 1 of 5

A motor does \( 6{,}000 \;\text{J} \) of work in \( 30 \;\text{s} \). What is the power output of the motor?

Question 2 of 5

A \( 60 \;\text{kg} \) student runs up a \( 5 \;\text{m} \) tall staircase in \( 4 \;\text{s} \). What is the student's average power output against gravity? (Use \( g = 10 \;\text{m/s}^2 \).)

Question 3 of 5

A truck engine exerts a force of \( 4{,}000 \;\text{N} \) on a trailer while moving at a constant \( 15 \;\text{m/s} \). What is the engine's power output?

Question 4 of 5

Two cranes lift identical \( 500 \;\text{kg} \) loads to the same height. Crane A takes \( 10 \;\text{s} \) and Crane B takes \( 20 \;\text{s} \). Which statement is correct?

Question 5 of 5

A \( 900 \;\text{kg} \) car accelerates from rest to \( 20 \;\text{m/s} \) in \( 10 \;\text{s} \). What is the average power delivered by the engine? (Ignore friction.)