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Power

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Power - Complete Interactive Lesson

Part 1: What Power Means

⚡ Power

Part 1 of 7 — What Power Means

Topics in This Part

Section
Energy vs. Power — the Key Distinction
The Definition: $P = \dfrac{W}{t}$
The Watt and Its Size

🔑 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.

Concept Check 🎯

The Watt

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

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

First Calculations 🧮

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

1) A motor does $1200\ \text{J}$ of work in $4\ \text{s}$ . Power A pump transfers of energy in . Power A bulb runs for . Energy used

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 — $W = F$ — to build the most-used power formulas in AP Physics 1.

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

⚡ 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 and you get the second master formula for power.

Part 3: Power From Speed: $P = Fv$

⚡ 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 $P = Fv$

Part 4: Power and Kinematics

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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:

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

Part 5: Efficiency & Wasted Power

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Part 5 of 7 — Efficiency & Wasted Power

🔑 The Reality: No real machine turns 100% of its input power into useful output. Some always leaks away as heat, sound, or friction. Efficiency measures how much actually makes it through.

Efficiency

Efficiency is the ratio of useful output power to total input power:

$\boxed{\text{Efficiency} = \frac{P_{\text{useful out}}}{P_{\text{total in}}}}$

Part 6: Real-World Power: Units, Horsepower & the Kilowatt-Hour

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Part 6 of 7 — Real-World Power: Units, Horsepower & the Kilowatt-Hour

🔑 Beyond the Watt: Real machines and electricity bills use bigger units — the kilowatt, the horsepower, and the kilowatt-hour. Knowing how they relate to the watt and the joule is a classic AP-style skill.

The Power Unit Family

Unit	Equals	Notes
$1\ \text{kW}$	$1000\ \text{W}$	kilowatt — everyday machines

Part 7: Mixed Mastery & Exit Quiz

⚡ 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 , (4) get power from , (5) handle efficiency and wasted power, and (6) convert between watts, horsepower, and kilowatt-hours. Time to put it all together.

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

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.

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

Device	Typical power
LED bulb	$\sim 10\ \text{W}$
Human at rest	$\sim 100\ \text{W}$
Microwave oven	$\sim 1000\ \text{W} = 1\ \text{kW}$
Compact car engine	$\sim 75{,}000\ \text{W} = 75\ \text{kW}$

⚠️ 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."

(rearrange: $W = Pt$ )

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 in at constant speed. Power A force does work over in . Power

Concept Check 🎯

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:

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

Symbol	Meaning	Unit
$P$	power	$\text{W}$
$F$	force along the motion	$\text{N}$
$v$	speed	$\text{m/s}$

💡 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.

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 $5.0\ \text{s}$ . $\Delta KE = \,?\ \text{J}$ Same ball: average power A car goes from rest to in . Average power

Average vs. Instantaneous Power

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

	Average power	Instantaneous power
Formula	$P_{\text{avg}} = \dfrac{W}{t} = \dfrac{\Delta KE}{t}$	$P = Fv$ (with $v$ at one instant)
Time	over a whole interval	at a single moment
Use when	total energy & total time known	force & current speed known

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.

It's a fraction between $0$ and $1$ , usually written as a percentage. Multiply by $100\%$ to get the percent.

Worked Example: An Electric Motor

A motor draws $500\ \text{W}$ of electrical power but delivers only $400\ \text{W}$ of useful mechanical power.

$\text{Efficiency} = \frac{400}{500} = 0.80 = 80\%$

The other $100\ \text{W}$ is wasted power — lost mostly as heat:

$P_{\text{wasted}} = P_{\text{in}} - P_{\text{out}} = 500 - 400 = 100\ \text{W}$

💡 Input always equals useful output plus wasted: $P_{\text{in}} = P_{\text{out}} + P_{\text{wasted}}$ . Energy is conserved — the "lost" power isn't destroyed, it just becomes a form you didn't want (usually heat).

Efficiency Logic 🔽

A machine takes in $1000\ \text{W}$ and delivers $750\ \text{W}$ of useful power.

Working Backward From Efficiency

Worked Example: Sizing a Pump

A water pump must deliver $600\ \text{W}$ of useful power, but it's only $75\%$ efficient. How much electrical power must it draw?

Rearrange the efficiency formula to solve for input:

$P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{600}{0.75} = 800\ \text{W}$

So you must supply $800\ \text{W}$ to get $600\ \text{W}$ of useful work — the extra $200\ \text{W}$ is lost to heat and friction.

⚠️ Efficiency is never greater than $100\%$ . If you ever compute an efficiency above $1$ , you've flipped the ratio — useful output goes on top, total input on the bottom.

Efficiency Drills 🧮

Efficiency $= \dfrac{P_{\text{out}}}{P_{\text{in}}}$ , and $P_{\text{wasted}} = P_{\text{in}} - P_{\text{out}}$ .

1) Input $200\ \text{W}$ , useful output $150\ \text{W}$ . Efficiency as a percent $= \,?$ (enter just the number) 2) Same machine: wasted power $= \,?\ \text{W}$ 3) A machine is efficient and delivers of useful power. Input power

Concept Check 🎯

Kilowatt-Hour: a Unit of Energy, Not Power

Your electric bill is in kilowatt-hours ( $\text{kWh}$ ). Don't be fooled — this is energy, because it's power × time:

$1\ \text{kWh} = (1000\ \text{W})(3600\ \text{s}) = 3{,}600{,}000\ \text{J} = 3.6\ \text{MJ}$

⚠️ Trap: "kilowatt-hour" sounds like a rate, but it's the total energy used when you run $1\ \text{kW}$ for $1\ \text{hour}$ . A $2\ \text{kW}$ heater run for $3\ \text{hours}$ uses $2 \times 3 = 6\ \text{kWh}$ of energy.

Energy Bills, the Physics Way

Energy in kilowatt-hours is simply power (in kW) times time (in hours):

$E\,(\text{kWh}) = P\,(\text{kW}) \times t\,(\text{h})$

Worked Example: Running a Heater

A $1.5\ \text{kW}$ space heater runs for $4\ \text{hours}$ .

$E = (1.5\ \text{kW})(4\ \text{h}) = 6\ \text{kWh}$

In joules, that's $6 \times 3.6\ \text{MJ} = 21.6\ \text{MJ}$ — a big number, which is exactly why we use kilowatt-hours for billing instead of joules.

💡 To find the cost, multiply the energy by the price per kWh. At a rate of $0.20 per kWh, those $6\ \text{kWh} $cost$ 6 \times $0.20 = $1.20$.

Unit Conversions 🧮

Use $1\ \text{kW} = 1000\ \text{W}$ , $1\ \text{hp} \approx 746\ \text{W}$ , and $E(\text{kWh}) = P(\text{kW}) \times t(\text{h})$ .

1) Convert $5\ \text{kW}$ to watts. $= \,?\ \text{W}$ 2) A $2\ \text{kW}$ appliance runs for $5\ \text{hours}$ . Energy used $= ? kWh$ Convert to watts (use ).

Power or Energy? 🔽

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

Quick Reference

Goal	Formula
Power from work & time	$P = \dfrac{W}{t}$
Power from force, distance, time	$P = \dfrac{Fd}{t}$
Power from force & speed	$P = Fv$
Average power from speed change	$P_{\text{avg}} = \dfrac{\Delta KE}{t}$
Efficiency	$\dfrac{P_{\text{out}}}{P_{\text{in}}}$
Energy from power & time	$W = Pt$

⚠️ Top traps to avoid: (1) the watt is power but the joule and kWh are energy; (2) only force along the motion gives power; (3) constant force ≠ constant power (since $P = Fv$ grows with speed); (4) efficiency can't exceed $100\%$ .

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.

1\ \text{hp} \approx 746\ \text{W}

Power

Power - Complete Interactive Lesson

Part 1: What Power Means

⚡ Power

Topics in This Part

Energy vs. Power

The Watt

What You've Got So Far

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

⚡ Power

Part 3: Power From Speed: $P = Fv$

⚡ Power

Deriving P=FvP = FvP

Part 4: Power and Kinematics

⚡ Power

Power From Kinetic Energy

Part 5: Efficiency & Wasted Power

⚡ Power

Efficiency

Part 6: Real-World Power: Units, Horsepower & the Kilowatt-Hour

⚡ Power

The Power Unit Family

Part 7: Mixed Mastery & Exit Quiz

⚡ Power

Building the Formula

Worked Example: Lifting a Box

Horizontal Pushes, Too

Worked Example: Pushing a Crate

Worked Examples

Example 1 — Cruising Car

Example 2 — Solve for Force

Worked Example: Accelerating a Cart

Average vs. Instantaneous Power

Worked Example: Both at Once

Worked Example: An Electric Motor

Working Backward From Efficiency

Worked Example: Sizing a Pump

Kilowatt-Hour: a Unit of Energy, Not Power

Energy Bills, the Physics Way

Worked Example: Running a Heater

Quick Reference

Deriving $P = Fv$