Unit 3.4 | Understanding and Applying Power

Track cyclist racing around the banked curve of a velodrome
  • 12 minutes
  • 2 formulas
  • 11 questions
  • 2 videos

Work tells you how much energy is transferred. Power tells you how fast. This final lesson of Unit 4 shows you how to calculate it and where it shows up in real life.

Unit 3 Breakdown

In this lesson: 

  • Learn the difference between energy and power
  • Calculate power from work and time
  • Use the [katex] P = Fv [/katex] shortcut for moving objects
  • Find work from a power vs. time graph
  • The 3 power problem types: work over time, steady speed, and power vs. time graphs

Introduction

This lesson covers power in physics: the difference between energy and power, how to calculate it, and where power in physics shows up in everyday life.

Two students carry identical boxes up the same flight of stairs. One walks. One sprints. Who did more work?

Neither! Same box, same height, so the same [katex] mgh [/katex]. Yet the sprinter is clearly doing something “more.” That something is power: not how much energy is transferred, but how fast.

What Is Power?

Power is the rate at which work is done, or the rate at which energy is transferred.

[katex] P = \frac{W}{t} = \frac{\Delta E}{t} [/katex]

Power is measured in watts (W): [katex] 1 \, W = 1 \, J/s [/katex]. A [katex] 60\,\text{W} [/katex] light bulb uses 60 joules every second. Car engines are often rated in horsepower, where [katex] 1 \, hp \approx 746 \, W [/katex].

Note – don’t mix up the letters. [katex] W [/katex] (italic) is work in joules; W (the unit) is watts.

Every power question on the AP exam is one of three types:

Problem typeThe clueThe setup
1. Work over timea lift, climb or push that takes a time[katex] P = \frac{W}{t} [/katex]
2. Steady speed“constant speed” against drag or friction[katex] P = Fv [/katex]
3. Power vs. time graphsa graph of [katex] P [/katex] against [katex] t [/katex]Area = work

Problem Type 1: Power from work and time

How to spot it: Something is lifted, climbs or is pushed, and you know (or need) how long it takes.

Setup: Find the work with the energy framework (often [katex] mgh [/katex]), then divide by the time.

A [katex] 60\,\text{kg} [/katex] student climbs a flight of stairs [katex] 4\,\text{m} [/katex] tall. What is the student’s average power if they walk up in [katex] 12\,\text{s} [/katex]? If they sprint up in [katex] 3\,\text{s} [/katex]?

1st step: Use the energy framework. The student gains height, so the work equals the gain in potential energy: [katex] W = mgh = (60)(9.8)(4) = 2352 \, J [/katex]

2nd step: Divide by the time: walking [katex] P = \frac{2352}{12} = 196 \, W [/katex], sprinting [katex] P = \frac{2352}{3} = 784 \, W [/katex].

The same \(2352\,\text{J}\) of work either way. Doing it in a quarter of the time takes four times the power.

AP trap: Thinking the faster climber does more work. Same mass, same height, same work; only the time changes.

A Shortcut: Power and Velocity

When a force pushes an object at a steady speed, there is a faster way to find power:

  1. Start with the definition: [katex] P = \frac{W}{t} [/katex].
  2. Replace work with force times distance: [katex] P = \frac{Fd}{t} [/katex].
  3. Distance over time is velocity: [katex] \frac{d}{t} = v [/katex].

[katex] P = Fv [/katex]

Like [katex] W = Fd [/katex], this only uses the part of the force along the velocity. It is perfect for anything moving at a steady speed against a resisting force: cars, boats, elevators, conveyor belts.

Problem Type 2: Steady speed: P = Fv

How to spot it: “Constant speed” or “steady speed” with drag, friction or a slope pushing back.

Setup: Constant speed means zero net force, so the engine’s force equals the resisting force. Then [katex] P = Fv [/katex].

A car cruises at a constant [katex] 25\,\text{m/s} [/katex]. Air resistance and friction push back with a combined [katex] 600\,\text{N} [/katex]. How much power must the engine deliver?

1st step: Constant speed means zero net force (Newton’s 1st law), so the engine pushes forward with [katex] 600\,\text{N} [/katex].

2nd step: [katex] P = Fv = (600)(25) = 15{,}000 \, W [/katex] ([katex] 15\,\text{kW} [/katex], about 20 horsepower).

Air resistance grows with speed (roughly with [katex] v^2 [/katex]), and so does [katex] v [/katex] itself, so power climbs with the cube of speed:

Power needed to hold a steady speed

Same car as above (\(600\,\text{N}\) of drag at \(25\,\text{m/s}\)). Going from \(25\) to \(35\,\text{m/s}\) nearly triples the power.

This is why driving fast burns so much more fuel.

AP trap: Using the net force in [katex] P = Fv [/katex]. At constant speed the net force is zero; use the force the engine or motor exerts.

Problem Type 3: Work from a power vs. time graph

How to spot it: A graph of power against time.

Setup: The area under the graph is the work done, since [katex] W = Pt [/katex].

Since [katex] W = Pt [/katex], the area under a power vs. time graph is the work done. Same trick as force vs. position in Unit 3.2.

Constant power for a time \(t\): the rectangle under the line is the work, \(W = Pt\).

AP trap: Reading the height of the graph as the work. The height is the power; the work is the area.

Recap

  1. Power is how fast energy is transferred. Work tells you how much, power tells you how fast.
  2. Use [katex] P = \frac{W}{t} [/katex] when you know the work (from the energy framework) and the time.
  3. Use [katex] P = Fv [/katex] when a force moves an object at a known speed.
  4. The area under a power vs. time graph is the work.

Unit 3 Recap

Read the Unit 3 Speed Recap to review everything in a few minutes, and lock in the 3 step energy framework with more problems in UBQ. Or watch the energy section of my speed review:

Ready for the whole unit? Take the unit review:

Unit 4 Preview

Next up is Unit 4: Momentum. Momentum is another conserved quantity, just like energy, and you will use both together to solve collisions and explosions.

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This article was written by Jason