Power

⚡ Power

Part 1 of 7 — What Power Means

Topics in This Part

🔑 Key Concept: Work (or energy) tells you how much gets transferred. Power tells you how fast it gets transferred. Two machines can do the exact same job — the more powerful one just does it sooner.

Energy vs. Power

Imagine two elevators that each lift a $500\ \text{kg}$ load up the same $20\ \text{m}$ shaft.

• Both do the same work on the load — the same energy is transferred.
• But one takes $10\ \text{s}$ and the other takes $40\ \text{s}$.

The fast elevator isn't doing more work — it's doing the same work in less time. That "work per time" is exactly what power measures.

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

where $W$ is the work done (in joules, $\text{J}$) and $t$ is the time interval (in seconds, $\text{s}$).

💡 Because energy and work share the same unit (the joule), you'll also see power written as $P = \dfrac{\Delta E}{t}$ — energy transferred per unit time. They're the same formula.

Concept Check 🎯

The Watt

The SI unit of power is the watt ($\text{W}$), named for James Watt:

$1\ \text{watt} = 1\ \frac{\text{joule}}{\text{second}} = 1\ \frac{\text{J}}{\text{s}}$

So a $60\ \text{W}$ light bulb converts $60\ \text{J}$ of electrical energy every second.

⚠️ Watch the symbols: $W$ (italic) is work in joules, while $\text{W}$ (upright) is the unit watt. Context tells them apart — but never confuse "the work was $W$" with "the unit is the watt."

First Calculations 🧮

Use $P = \dfrac{W}{t}$.

1) A motor does $1200\ \text{J}$ of work in $4\ \text{s}$. Power $= \,?\ \text{W}$ 2) A pump transfers $9000\ \text{J}$ of energy in $30\ \text{s}$. Power $= \,?\ \text{W}$ 3) A $50\ \text{W}$ bulb runs for $20\ \text{s}$. Energy used $= \,?\ \text{J}$ (rearrange: $W = Pt$)

Spot the Units 🔽

Match each quantity to its correct SI unit.

What You've Got So Far

You can now (1) tell power apart from energy, (2) use $P = \dfrac{W}{t}$, and (3) read the watt as joules per second.

🔑 Carry this forward: Every power problem is just "find the energy transferred, then divide by the time." In Part 2 we'll connect $W$ to the work you already know — $W = Fd$ — to build the most-used power formulas in AP Physics 1.

⚡ Power

Part 2 of 7 — Power From Work: $P = \dfrac{Fd}{t}$

🔑 The Idea: Work done by a constant force along the motion is $W = Fd$. Drop that into $P = \dfrac{W}{t}$ and you get the second master formula for power.

Building the Formula

Start with the two definitions you know:

$W = Fd \qquad\text{and}\qquad P = \frac{W}{t}$

Substitute the first into the second:

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

Here $F$ is the force component along the direction of motion, $d$ is the displacement, and $t$ is the time.

Worked Example: Lifting a Box

You lift a $15\ \text{kg}$ box straight up $2.0\ \text{m}$ in $3.0\ \text{s}$ at constant speed. Find your power output.

Step 1 — Find the force. At constant speed the lifting force equals the weight: $F = mg = (15)(9.8) = 147\ \text{N}$

Step 2 — Find the work. $W = Fd = (147)(2.0) = 294\ \text{J}$

Step 3 — Find the power. $P = \frac{W}{t} = \frac{294}{3.0} = 98\ \text{W}$

💡 When you lift at constant velocity, the work you do equals the gravitational potential energy gained, $\Delta U = mgh$. That's why $W = mgh$ shows up so often in power problems.

Order the Steps 🔽

A $2.0\ \text{kg}$ object is lifted $5.0\ \text{m}$ in $4.0\ \text{s}$ at constant speed. Choose what happens at each stage. (Use $g = 9.8\ \text{m/s}^2$.)

Horizontal Pushes, Too

The formula isn't just for lifting. Any force doing work over a distance works:

Worked Example: Pushing a Crate

You push a crate with a steady $80\ \text{N}$ horizontal force, sliding it $12\ \text{m}$ across a floor in $6.0\ \text{s}$.

$W = Fd = (80)(12) = 960\ \text{J}$ $P = \frac{W}{t} = \frac{960}{6.0} = 160\ \text{W}$

⚠️ Force must be along the motion. If a force is perpendicular to the displacement (like the normal force on a flat floor, or gravity on a horizontal slide), it does zero work and contributes zero power. Only the component of force in the direction of motion counts.

Power From Force and Distance 🧮

Use $P = \dfrac{Fd}{t}$. (Use $g = 9.8\ \text{m/s}^2$ where needed.)

1) A $200\ \text{N}$ force pushes a box $10\ \text{m}$ in $5.0\ \text{s}$. Power $= \,?\ \text{W}$ 2) A crane lifts a $100\ \text{kg}$ beam $6.0\ \text{m}$ in $12\ \text{s}$ at constant speed. Power $= \,?\ \text{W}$ 3) A $400\ \text{N}$ force does work over $15\ \text{m}$ in $10\ \text{s}$. Power $= \,?\ \text{W}$

Concept Check 🎯

⚡ Power

Part 3 of 7 — Power From Speed: $P = Fv$

🔑 The Shortcut: When something moves at a steady speed against a steady force, you don't need to track distance and time separately — you can get power straight from force and velocity: $P = Fv$.

Deriving $P = Fv$

Start from $P = \dfrac{Fd}{t}$ and group the $d$ and $t$ together:

$P = \frac{Fd}{t} = F \cdot \frac{d}{t}$

But $\dfrac{d}{t}$ is just the speed $v$. So:

$\boxed{P = Fv}$

This is instantaneous power when $v$ is the speed at a single instant, and average power when $v$ is the average speed.

💡 Check the units: $\text{N}\cdot\dfrac{\text{m}}{\text{s}} = \dfrac{\text{N}\cdot\text{m}}{\text{s}} = \dfrac{\text{J}}{\text{s}} = \text{W}$. ✓ Everything is consistent.

Concept Check 🎯

Worked Examples

Example 1 — Cruising Car

A car's engine pushes it forward with $1500\ \text{N}$ while it cruises at a steady $25\ \text{m/s}$.

$P = Fv = (1500)(25) = 37{,}500\ \text{W} = 37.5\ \text{kW}$

Example 2 — Solve for Force

A $20{,}000\ \text{W}$ motor drives a conveyor belt at $4.0\ \text{m/s}$. What force does it apply?

Rearrange $P = Fv$ to $F = \dfrac{P}{v}$:

$F = \frac{20{,}000}{4.0} = 5000\ \text{N}$

⚠️ Slower means stronger. For a fixed power, $F = \dfrac{P}{v}$ means force and speed trade off. That's why cars shift to low gear (low speed, high force) to climb a steep hill — the engine power is capped, so the only way to get more force is to go slower.

Using $P = Fv$ 🧮

1) A motor applies $600\ \text{N}$ to move a load at $3.0\ \text{m/s}$. Power $= \,?\ \text{W}$ 2) A $4500\ \text{W}$ engine moves a cart at $9.0\ \text{m/s}$. Force $= \,?\ \text{N}$ (use $F = P/v$) 3) A $2400\ \text{W}$ winch pulls with $800\ \text{N}$. Speed $= \,?\ \text{m/s}$ (use $v = P/F$)

Pick the Right Formula 🔽

For each situation, choose the most direct power formula.

⚡ Power

Part 4 of 7 — Power and Kinematics

🔑 The Connection: When an object speeds up, work goes into kinetic energy. Average power is then the change in kinetic energy divided by the time it took.

Power From Kinetic Energy

By the work-energy theorem, the net work done on an object equals its change in kinetic energy:

$W_{\text{net}} = \Delta KE = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$

Divide by the time and you get average power:

$\boxed{P_{\text{avg}} = \frac{\Delta KE}{t}}$

Worked Example: Accelerating a Cart

A $4.0\ \text{kg}$ cart starts from rest and reaches $6.0\ \text{m/s}$ in $3.0\ \text{s}$. Find the average power delivered.

Step 1 — Change in kinetic energy (starts from rest, so $KE_i = 0$): $\Delta KE = \tfrac{1}{2}(4.0)(6.0)^2 - 0 = \tfrac{1}{2}(4.0)(36) = 72\ \text{J}$

Step 2 — Average power: $P_{\text{avg}} = \frac{\Delta KE}{t} = \frac{72}{3.0} = 24\ \text{W}$

💡 This is average power. The instantaneous power at the end (when $v = 6.0\ \text{m/s}$) is larger, because at higher speed $P = Fv$ is bigger even though the force is the same.

Power From $\Delta KE$ 🧮

Use $P_{\text{avg}} = \dfrac{\Delta KE}{t}$ with $\Delta KE = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$.

1) A $2.0\ \text{kg}$ ball accelerates from rest to $10\ \text{m/s}$ in $5.0\ \text{s}$.    $\Delta KE = \,?\ \text{J}$ 2) Same ball: average power $= \,?\ \text{W}$ 3) A $1000\ \text{kg}$ car goes from rest to $20\ \text{m/s}$ in $8.0\ \text{s}$. Average power $= \,?\ \text{W}$

Average vs. Instantaneous Power

This trips students up, so let's nail it down:

Worked Example: Both at Once

A constant $50\ \text{N}$ force accelerates a cart. At the moment the cart is moving at $2.0\ \text{m/s}$, the instantaneous power is:

$P = Fv = (50)(2.0) = 100\ \text{W}$

Later, when the cart reaches $8.0\ \text{m/s}$, the same force delivers:

$P = Fv = (50)(8.0) = 400\ \text{W}$

⚠️ Same force, four times the power — because power depends on speed, not just force. A constant force does not mean constant power.

Concept Check 🎯

Work Through the KE Power 🔽

A $5.0\ \text{kg}$ cart starts from rest and reaches $4.0\ \text{m/s}$ in $2.0\ \text{s}$. Fill in each stage.

⚡ Power

Part 5 of 7 — Energy Graphs and Negative Power

🔑 The Big Idea: Power is the rate at which a system's energy changes (EK 3.5.A.1). That makes it the slope of an energy-versus-time graph — and it means power can be negative when a force takes energy out of a system.

Power as a Slope

Because $P = \dfrac{\Delta E}{\Delta t}$, a graph of a system's energy versus time encodes the power:

Average vs. Instantaneous

• Average power over an interval is the total energy change divided by the time: the slope of the line joining the two endpoints (EK 3.5.A.2).
• Instantaneous power is the slope of the tangent at one instant. For a constant force along the motion it equals $Fv$ (EK 3.5.A.4).

Worked Example: Reading a Graph

A cart's kinetic energy rises from $0$ to $90\ \text{J}$ in the first $3.0\ \text{s}$, then stays at $90\ \text{J}$.

• From $0$ to $3.0\ \text{s}$: $P_{\text{avg}} = \dfrac{90 - 0}{3.0} = 30\ \text{W}$.
• After $3.0\ \text{s}$: the graph is flat, so the net power delivered to the cart is $0$ — it moves at constant speed.

Read the Graph 🔽

A block's kinetic energy is graphed against time. Choose what each feature tells you.

Negative Power

Work can be positive, negative or zero (EK 3.2.A.2), so power can be too. The power delivered by a force is

$P = Fv\cos\theta$

where $\theta$ is the angle between the force and the velocity.

Worked Example: Braking

A car moves at $20\ \text{m/s}$ while its brakes exert a $6000\ \text{N}$ friction force backward. At that instant the brakes deliver

$P = (6000)(20)\cos 180° = -120{,}000\ \text{W}$

The car's kinetic energy is dropping at $120\ \text{kW}$; that energy becomes thermal energy in the brakes.

Worked Example: Gravity on a Falling Ball

A $0.50\ \text{kg}$ ball falls at $6.0\ \text{m/s}$. Gravity points along the velocity, so gravity delivers $P = mgv = (0.50)(9.8)(6.0) = 29.4\ \text{W}$ at that instant. On the way up at the same speed, gravity delivers $-29.4\ \text{W}$: kinetic energy is being converted into gravitational potential energy.

Signed Power Drills 🧮

Use $P = Fv\cos\theta$ and $g = 9.8\ \text{m/s}^2$. Include a minus sign when the power is negative.

1) Friction of $40\ \text{N}$ acts on a crate sliding at $3.0\ \text{m/s}$. Power delivered by friction $= \,?\ \text{W}$ 2) A $2.0\ \text{kg}$ rock falls at $5.0\ \text{m/s}$. Power delivered by gravity $= \,?\ \text{W}$ 3) A sled's kinetic energy drops from $500\ \text{J}$ to $200\ \text{J}$ in $6.0\ \text{s}$. Average power delivered to the sled $= \,?\ \text{W}$

Concept Check 🎯

⚡ Power

Part 6 of 7 — Power on Inclines and Against Friction

🔑 The Strategy: For a machine moving something at constant speed, the net force is zero. Use a free-body diagram to find the force the machine must supply, then multiply by the speed: $P = Fv$.

Step 1: Find the Force From Newton's Laws

At constant velocity, the applied force balances everything opposing the motion (EK 2.4.A.3).

Step 2: Multiply by the Speed

$P = Fv$

Worked Example: A Winch on a Ramp

A winch pulls a $200\ \text{kg}$ crate up a $30°$ ramp at a steady $0.50\ \text{m/s}$. The coefficient of kinetic friction is $0.20$. Use $g = 9.8\ \text{m/s}^2$.

• Gravity along the ramp: $mg\sin 30° = (200)(9.8)(0.50) = 980\ \text{N}$
• Friction: $\mu_k mg\cos 30° = (0.20)(200)(9.8)(0.866) \approx 339\ \text{N}$
• Winch force: $F = 980 + 339 = 1319\ \text{N}$
• Power: $P = Fv = (1319)(0.50) \approx 660\ \text{W}$

Force First, Then Power 🎯

Where Does the Power Go?

Energy conservation splits the machine's power into the rates at which each kind of energy changes (EK 3.4.B.2, 3.5.A.1). For the winch on the ramp:

💡 The rates add up to the winch's power, which is a good check on your arithmetic. If the winch suddenly delivered more power than this, the extra would go into kinetic energy and the crate would speed up.

Incline Power 🧮

Use $g = 9.8\ \text{m/s}^2$. A cyclist and bike ($80\ \text{kg}$ total) ride up a hill inclined at an angle whose $\sin\theta = 0.10$, at a steady $5.0\ \text{m/s}$. Ignore friction and air resistance.

1) Force needed along the slope $= \,?\ \text{N}$ 2) Power the cyclist must deliver $= \,?\ \text{W}$ 3) If the cyclist rides at $2.5\ \text{m/s}$ instead, the power needed $= \,?\ \text{W}$

Power or Energy? 🔽

Classify each quantity. Remember: power is a rate (joules per second), energy is a total (joules).

⚡ Power

Part 7 of 7 — Mixed Mastery & Exit Quiz

You can now (1) use $P = \dfrac{W}{t}$, (2) use $P = \dfrac{Fd}{t}$, (3) use $P = Fv$, (4) get power from $\Delta KE$, (5) read power as the slope of an energy–time graph, including negative power, and (6) find the force from Newton's laws before using $P = Fv$. Time to put it all together.

Quick Reference

⚠️ Top traps to avoid: (1) the watt is power but the joule is energy; (2) only force along the motion gives power; (3) constant force ≠ constant power (since $P = Fv$ grows with speed); (4) constant speed means zero net force, not zero applied force.

Mixed Practice 🎯

Putting It Together 🧮

(Use $g = 9.8\ \text{m/s}^2$ where needed.)

1) A $25\ \text{kg}$ load is lifted $4.0\ \text{m}$ in $5.0\ \text{s}$ at constant speed. Power $= \,?\ \text{W}$ 2) A $1200\ \text{W}$ motor drives a belt at $4.0\ \text{m/s}$. Force $= \,?\ \text{N}$ 3) A $2.0\ \text{kg}$ object goes from rest to $8.0\ \text{m/s}$ in $4.0\ \text{s}$. Average power $= \,?\ \text{W}$

Exit Quiz ✅

Answer all three to finish the lesson.