Unit 2 · Lesson 4a

4aWork and Power

Explore how forces transfer energy through work and how power measures the rate of energy transfer.

Work and power are the bridge between forces and energy — understanding them is essential for analyzing engines, machines, and any system that transfers energy.

Lesson Overview

In physics, work is done on an object when a force causes a displacement. The formula W = Fd cosθ accounts for the angle between the force and displacement vectors. Work can be positive (force aids motion), negative (force opposes motion), or zero (force perpendicular to motion). Power is the rate of doing work: P = W/t = Fv. The SI unit of work is the joule (J) and of power is the watt (W).

Key Concepts

Work Formula

W = Fd cosθ — only the component of force parallel to displacement does work.

Positive Work

Force has a component in the direction of motion (0° ≤ θ < 90°); energy is transferred to the object.

Negative Work

Force opposes motion (90° < θ ≤ 180°); energy is removed from the object (e.g., friction).

Zero Work

Force is perpendicular to displacement (θ = 90°); no energy transfer (e.g., normal force on level surface).

Power

P = W/t = Fv — the rate at which work is done. 1 W = 1 J/s.

Units

Work: joule (J = N·m). Power: watt (W = J/s). Also: 1 horsepower ≈ 746 W.

Example 1

A person pushes a box with a force of 50 N horizontally over a distance of 8 m. How much work is done?

Answer:W = Fd cosθ = 50 × 8 × cos0° = 50 × 8 × 1 = 400 J.
Example 2

A person pulls a suitcase with a force of 40 N at 30° above the horizontal for 10 m. How much work does the pulling force do?

Answer:W = Fd cosθ = 40 × 10 × cos30° = 400 × 0.866 ≈ 346 J.
Example 3

Friction exerts a 20 N force opposing a box sliding 5 m. How much work does friction do?

Answer:The friction force is opposite to displacement, so θ = 180°. W = 20 × 5 × cos180° = 100 × (−1) = −100 J.
Example 4

A motor lifts a 200 N load 15 m in 10 s. Find (a) the work done and (b) the power output.

Answer:(a) W = Fd = 200 × 15 = 3000 J. (b) P = W/t = 3000/10 = 300 W.
Example 5

A car engine exerts a constant force of 4000 N while the car travels at 25 m/s. What is the engine's power output?

Answer:P = Fv = 4000 × 25 = 100,000 W = 100 kW.
Guided Problem 1

A 60 N force is applied at 45° to the horizontal to push a crate 12 m. How much work is done?

Hint: Use W = Fd cosθ with θ = 45°. cos45° ≈ 0.707.

Guided Problem 2

A weightlifter holds a 500 N barbell stationary overhead for 5 seconds. How much work does the weightlifter do on the barbell?

Hint: Work requires displacement. Is the barbell moving?

Guided Problem 3

A 1200 W motor runs for 30 seconds. How much work does it do?

Hint: Rearrange P = W/t to find W.

Guided Problem 4

A horse pulls a cart with 800 N of force at 2 m/s. What is the horse's power output in watts and in horsepower?

Hint: Use P = Fv, then convert: 1 hp ≈ 746 W.

Guided Problem 5

A normal force acts on a box sliding along a horizontal floor. How much work does the normal force do? Explain.

Hint: What is the angle between the normal force (upward) and the displacement (horizontal)?

Key Vocabulary

Work

The energy transferred to or from an object by a force acting over a displacement; W = Fd cosθ.

Example: Pushing a box 5 m with a 10 N horizontal force does 50 J of work.

Joule

The SI unit of work and energy; 1 J = 1 N·m = 1 kg·m²/s².

Example: Lifting a 1 N apple 1 m upward requires 1 J of work.

Power

The rate at which work is done; P = W/t = Fv. SI unit: watt (W = J/s).

Example: A 100 W light bulb uses 100 J of energy every second.

Watt

The SI unit of power; 1 W = 1 J/s. Named after James Watt.

Example: A typical human can sustain about 75–100 W of mechanical power output.

Interactive Practice — 5 Questions

1

A 30 N force moves an object 4 m in the direction of the force. How much work is done?

2

A force is applied perpendicular to an object's displacement. How much work is done?

3

A machine does 6000 J of work in 2 minutes. What is its power output?

4

Friction does negative work on a sliding box. This means:

5

A car travels at constant velocity of 20 m/s with a driving force of 500 N. What is the engine's power?

Independent Practice

1

A 75 N force is applied at 60° above the horizontal to push a box 20 m along the floor. Find (a) the work done by the applied force and (b) the work done by the normal force.

2

A 500 W motor lifts boxes. How long does it take to do 15,000 J of work? How high can it lift a 100 N box in 30 s?

3

Explain why carrying a heavy backpack horizontally across a flat floor does zero work on the backpack (in the physics sense), even though it feels tiring.

4

A 1500 kg car accelerates from rest to 20 m/s in 8 s. The engine exerts a constant net force. Find (a) the net force, (b) the work done, and (c) the average power.

5

★ A 60 kg person climbs a 10 m staircase in 15 s, then descends the same staircase in 10 s. Find the power output going up and the work done by gravity going down. Explain the sign of each answer.

Challenge
⚠️

Common Mistakes

Using the full force magnitude in W = Fd when the force is at an angle.

Only the component of force parallel to displacement does work: W = Fd cosθ. Always include the cosθ factor.

Thinking that holding a heavy object stationary requires work in the physics sense.

Work requires displacement. W = Fd cosθ = 0 when d = 0, regardless of how large the force is.

Confusing power and work — saying "the motor has more power so it does more work."

Power is the rate of doing work. A higher-power motor does the same work faster, but total work depends on both power and time: W = Pt.

💡

Math Tips

📌

Remember the three cases: θ = 0° → W = Fd (maximum, positive); θ = 90° → W = 0; θ = 180° → W = −Fd (maximum negative). For any other angle, use W = Fd cosθ.

📌

Power can be calculated two ways: P = W/t (when you know total work and time) or P = Fv (when you know force and instantaneous speed). Both give watts.