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Power

Lesson ~10 min read 8 MCQs

In simple terms: In simple terms, power is about how fast you do work or use energy. It's not just what you do, but how quickly you do it.

Why this matters

Imagine two friends, Priya and Marcus, are helping you move. You both need to carry an identical, heavy box of books up to your second-floor apartment. Priya, a track athlete, hustles up the stairs in 10 seconds flat. Marcus, who's been skipping the gym, takes a full 30 seconds.

Who did more work?

It's a trick question! Since the box, the mass, and the vertical height were the same, you both did the exact same amount of work against gravity. But you'd probably agree Priya was more "powerful." Why? Because she did the same work in less time.

That's the core of what we're talking about today: Power. It’s the physics way of measuring how quickly energy is transferred or work is done. We'll break down how to calculate it, the different formulas you'll need, and how it shows up in everything from lifting weights to driving a car.

Diagram

Instantaneous Power in Mechanical Systems A diagram showing a person pushing a lawnmower. It illustrates how an applied force at an angle relates to the power delivered. A force vector F is shown applied along the mower's handle, at an angle theta to the horizontal. The velocity vector v is horizontal. The component of the force parallel to the velocity, F cos(theta), is shown, and the formula P = Fv cos(theta) is displayed. F v F cos θ θ P = Fv cos θ P = (150 N)(1.2 m/s) cos(35°) P ≈ 147.4 W
This diagram shows a person pushing a lawnmower to illustrate the concept of instantaneous power. Vectors show the applied force at an angle, the horizontal velocity, and the parallel component of the force, visually explaining the formula P = Fv cos(theta).

Concept map

flowchart TD
    A[Energy (Joules)] -- "Change in energy is..." --> B(Work, W);
    B -- "Work done over time is..." --> C(Average Power, P_avg = W/Δt);
    A -- "Energy transferred over time is..." --> D(Average Power, P_avg = ΔE/Δt);
    E(Force, F) -- "applied at an angle θ to..." --> F(Velocity, v);
    F -- "results in..." --> G(Instantaneous Power, P = Fv cos θ);
    C -- "is a type of" --> H((Power));
    D -- "is a type of" --> H((Power));
    G -- "is a type of" --> H((Power));

Core explanation

Hello again! Let's dive into the idea of power. In everyday language, we use "power" to describe a strong athlete or a fast car. In physics, we have a very precise definition that captures that same intuitive idea of "doing a lot, quickly."

Power: The Rate of Doing Work

At its heart, power is the rate at which energy is used, transferred, or transformed. The key word here is rate. A rate always involves time.

Think about it this way: running a marathon and walking a marathon cover the same distance. But the runner completes it in a much shorter time. The runner is operating at a higher power level.

The fundamental equation for average power is:

P_avg = ΔE / Δt

Where:

  • P_avg is the average power.
  • ΔE is the change in energy (or energy transferred).
  • Δt is the time interval over which that change occurred.

The unit of power is the Watt (W), named after the 18th-century inventor James Watt. One Watt is equal to one Joule of energy transferred per second (1 W = 1 J/s). So if a light bulb has a power rating of 60 W, it means it converts 60 Joules of electrical energy into light and heat every single second.

Connecting Power and Work

Remember from our last lessons that doing work on a system is one way to change its energy (W = ΔE). Because of this direct link, we can substitute Work (W) for the change in energy (ΔE) in our power equation.

This gives us another essential formula for average power:

P_avg = W / Δt

This is incredibly useful. If you know the work done on an object and the time it took, you can find the average power delivered.

Let's go back to Priya and Marcus. Let's say the box has a mass of 15 kg and they carried it up a 4-meter-high flight of stairs. The work done against gravity is W = mgh = (15 kg)(9.8 m/s²)(4 m) = 588 J.

  • Priya's Power
    P = W / t = 588 J / 10 s = 58.8 W
  • Marcus's Power
    P = W / t = 588 J / 30 s = 19.6 W

They both did 588 Joules of work, but Priya's power output was three times greater because she did it three times faster.

Instantaneous Power

Average power is great, but what if the power isn't constant? A car accelerating from a stoplight uses a lot of power initially, but then less power to cruise at a constant speed. For these situations, we need instantaneous power—the power at a specific moment in time.

We can derive a handy formula for this. Let's start with our work-based power equation: P = W / t

And we know that for a constant force, W = F * d, where d is the displacement. P = (F * d) / t

Now, look closely at that d / t term. Displacement divided by time is velocity! P = F * v

This simple equation tells us that the instantaneous power delivered to an object is the product of the force being applied and the object's velocity.

What if it's not?

Power When Force is at an Angle

Think about pushing a lawnmower. You push down and forward on the handle, but the mower only moves forward horizontally. Only the part of your force that is parallel to the mower's motion is actually doing work and contributing to its power.


(This is where the lesson's main visual of a person pushing a lawnmower would appear.)

As you can see in the diagram, if you apply a force F at an angle θ to the direction of motion, you have to break that force into components.

  • The component of the force parallel to the velocity is F_∥ = F cos θ.
  • The component perpendicular to the velocity is F_⊥ = F sin θ.

The perpendicular component (F sin θ) just pushes the mower into the ground. It doesn't cause it to move forward, so it does no work. Only the parallel component (F cos θ) does work.

So, to find the instantaneous power, we modify our equation:

P = F_∥ * v P = (F cos θ) * v

Or, as it's most commonly written:

P = Fv cos θ

This is the most general form of the equation for instantaneous power from a constant force.

  • If F and v are in the same direction, θ = 0°, and cos(0°) = 1. The equation becomes P = Fv.
  • If F and v are in opposite directions (like the force of friction on a moving car), θ = 180°, and cos(180°) = -1. The power is negative (P = -Fv), which means energy is being removed from the system.
  • If F and v are perpendicular, θ = 90°, and cos(90°) = 0. The power is zero. This is a critical concept: a force perpendicular to the direction of motion does no work and delivers no power!

And that's power in a nutshell! It's the bridge between energy, work, and time, telling us the rate at which the action happens.

Worked examples

Let's put these concepts into practice with a few examples.

Example 1

The Elevator

Problem: An elevator with a total mass of 800 kg is lifted by a motor from the ground floor to the 10th floor, a vertical distance of 30 meters. The journey takes 25 seconds at a constant velocity. What is the average power output of the motor? (Ignore friction).

Solution:

  1. 1
    Identify the goal
    We need to find the average power (P_avg). The best formula for this is P_avg = W / Δt.
  2. 2
    Find the work done (W)
    The motor is working against gravity to lift the elevator. The force it must exert is equal to the elevator's weight, F = mg. The work done is this force times the distance, W = Fd = mgd.
    • W = (800 kg) * (9.8 m/s²) * (30 m)
    • W = 235,200 J
  3. 3
    Calculate the power
    Now we can use the power formula with the given time, Δt = 25 s.
    • P_avg = W / Δt = 235,200 J / 25 s
    • P_avg = 9,408 W (or 9.4 kW)
Example 2

Cruising Down the Highway

Problem: A car is traveling at a constant velocity of 25 m/s (about 56 mph) on a flat, straight highway. The engine must overcome a total resistive force (air drag and friction) of 600 N. What is the instantaneous power being delivered by the engine to maintain this speed?

Solution:

  1. 1
    Identify the goal
    We need the instantaneous power (P) required to keep the car moving. Since we have a force and a velocity, P = Fv cos θ is the way to go.
  2. 2
    Analyze the forces and direction
    The engine provides a forward force to counteract the resistive force. To maintain a constant velocity, the engine's force must be equal in magnitude and opposite in direction to the resistive force. So, the force the engine applies is F = 600 N.
  3. 3
    Determine the angle (θ)
    The engine's force is pushing the car forward, and the car's velocity is also forward. The force and velocity are in the same direction, so the angle θ between them is 0°.
  4. 4
    Calculate the power
    • P = Fv cos θ
    • P = (600 N) * (25 m/s) * cos(0°)
    • Since cos(0°) = 1, this simplifies to P = (600 N) * (25 m/s) = 15,000 W (or 15 kW).
Example 3

Pushing the Lawnmower

Problem: Aaliyah is pushing a lawnmower across her yard with a constant velocity of 1.2 m/s. She applies a force of 150 N to the handle, which is angled at 35° below the horizontal. What is the instantaneous power she is supplying to the mower?

Solution:

  1. 1
    Identify the goal
    We need the instantaneous power, and we have force, velocity, and an angle. This is a perfect job for P = Fv cos θ.
  2. 2
    Assign the variables
    • F = 150 N
    • v = 1.2 m/s
    • θ = 35°
  3. 3
    Apply the formula
    The key is recognizing that θ is the angle between the force and the velocity. The velocity is horizontal, and the force is applied at 35° to the horizontal, so we can use the angle directly.
    • P = (150 N) * (1.2 m/s) * cos(35°)
    • P ≈ (150) * (1.2) * (0.819)
    • P ≈ 147.4 W

Try it yourself

Ready to try a couple on your own?

Problem 1

A 65 kg student, Jordan, runs up a 5.0-meter-tall flight of stairs in 6.2 seconds to get to class on time. What was Jordan's average power output during this climb?

Problem 2

A speedboat's engine provides 30,000 W (30 kW) of power to move the boat through the water at a constant speed of 15 m/s. What is the magnitude of the total resistive force (water drag) acting on the boat?

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