7.6 Power and Rates of Energy Transfer

Same Work, Different Engines

Two cars climb the same hill. Same mass, same height, same gravitational potential energy change. By the work-energy theorem, both engines do the same amount of work against gravity. But Car A reaches the top in 30 seconds, and Car B reaches it in 10 seconds.

If the work is the same, why does Car B need a bigger engine?

Because Car B transfers the same energy in one-third the time. The total energy transferred is not the whole story. The rate of energy transfer --- how many joules per second the engine delivers --- is what determines whether the engine can actually do the job. A small engine can eventually haul any load up any hill, given enough time. But if you need to get there in 10 seconds, you need an engine that can deliver energy fast enough.

This is the concept of power: the rate at which work is done, the rate at which energy is transferred. It is the quantity that separates a lawnmower engine from a rocket engine, even when both could, in principle, do the same total work.

Prediction

Before you read on: An elevator motor lifts a 1000 kg load at a constant speed of 2 m/s. The motor must exert a force equal to the load's weight to maintain constant velocity.

(a) What power must the motor deliver?

(b) Now the same motor lifts the same load at 4 m/s (double the speed). Does the required power double, more than double, or less than double?

Commit to numerical answers for both parts before continuing.

The Guiding Question

How quickly is energy being transferred, and why does the rate matter in real systems?

Work (Section 7.1) told us how much energy is transferred by a force. The work-energy theorem (Section 7.3) connected that transfer to changes in kinetic energy. But neither concept addresses timing. A crane that lifts a steel beam in one minute and a crane that lifts the same beam in one hour do the same work. The difference between them is not energy --- it is power. And in the real world, power is often the binding constraint. Motors have power ratings. Electrical circuits have wattage limits. Your muscles can exert a large force, but only for a short time before they fatigue --- which is fundamentally a power limitation.

Exploration: The Motor and the Load

[Interactive: Power Lab. A motor lifts a load vertically at constant speed. Two sliders control the mass of the load (100--5000 kg) and the lifting speed (0.5--10 m/s). The display shows:

The force the motor must exert ($F = mg$, since the speed is constant).
The power delivered ($P = Fv = mgv$), updated in real time as the sliders move.
A power-vs-time graph that builds as the motor operates. For constant speed and constant load, the graph is a horizontal line --- power is constant.

Guided prompts:

"Set the mass to 1000 kg and the speed to 2 m/s. What power does the motor deliver? Now double the speed to 4 m/s. What happened to the power?"
"Set the speed to 5 m/s. Now double the mass from 500 kg to 1000 kg. What happened to the power?"
"Which has a bigger effect on power requirements: doubling the speed or doubling the mass?"
"What limits the speed of the elevator --- the motor's maximum force or its maximum power?"]

Spend time with this. The key observation is that power depends on both force and velocity. A motor with a fixed power output faces a tradeoff: it can exert a large force at low speed, or a small force at high speed, but not both at once. This is why a car in low gear (large force, low speed) and high gear (small force, high speed) can use the same engine. The engine's power output is the constraint, and the gearbox trades force for speed within that constraint.

Pause and think: A car engine produces a maximum power of 150 kW. At 30 m/s (about 67 mph), what is the maximum force the engine can deliver to the wheels? At 60 m/s? What does this tell you about acceleration at high speed?

Concept Reveal: Power as a Rate

Average power

If a force does work $W$ over a time interval $\Delta t$, the average power is:

$$P_{\text{avg}} = \frac{W}{\Delta t}$$

This is exactly analogous to average velocity being displacement over time. It tells you the overall rate of energy transfer, smoothed over the interval.

Instantaneous power

Just as instantaneous velocity is the derivative of position, instantaneous power is the derivative of work with respect to time:

$$P = \frac{dW}{dt}$$

This gives the rate of energy transfer at a specific moment. When you floor the accelerator and the engine roars, the instantaneous power spikes. When you coast on a flat road, the instantaneous power from the engine drops to near zero.

The force-velocity form

Here is where the formula becomes especially useful. Recall that work done by a force over an infinitesimal displacement is $dW = \vec{F} \cdot d\vec{r}$. Dividing both sides by $dt$:

$$P = \frac{dW}{dt} = \vec{F} \cdot \frac{d\vec{r}}{dt} = \vec{F} \cdot \vec{v}$$

This gives us the central result:

$$\boxed{P = \vec{F} \cdot \vec{v}}$$

Power is the dot product of force and velocity. If the force is along the direction of motion, this simplifies to $P = Fv$. If the force is at an angle $\theta$ to the velocity, then $P = Fv\cos\theta$ --- exactly the component of force along the velocity, multiplied by the speed.

This formula is remarkably revealing. It says two things at once:

For a fixed force, power grows linearly with speed. This is why it takes more engine power to maintain 120 km/h than 60 km/h, even if the resistive forces were the same. (In reality, drag grows with $v^2$, making it even worse.)
For a fixed power, force and speed are inversely related. $F = P/v$. A motor with fixed power output delivers less force at higher speed. This is the fundamental reason cars have gearboxes.

Returning to the prediction

The elevator motor lifts a 1000 kg load at constant speed. The force is $F = mg = (1000)(9.8) = 9800$ N.

At $v = 2$ m/s: $P = Fv = (9800)(2) = 19{,}600$ W $= 19.6$ kW.

At $v = 4$ m/s: $P = Fv = (9800)(4) = 39{,}200$ W $= 39.2$ kW.

The power exactly doubles. This should now make complete sense: the motor does the same work per meter of height gained, but it covers those meters twice as fast, so it must deliver energy twice as fast. Power is linear in speed (when force is constant).

Units of Power

The SI unit of power is the watt (W), named after James Watt:

$$1 \text{ W} = 1 \text{ J/s} = 1 \text{ kg} \cdot \text{m}^2/\text{s}^3$$

One watt is one joule of energy transferred per second. Some useful benchmarks:

System	Approximate power
Human body at rest (metabolism)	80 W
A person climbing stairs briskly	200--500 W
Professional cyclist, sustained	300--400 W
Household microwave oven	1000 W (1 kW)
Small car engine	75 kW (about 100 hp)
Sports car engine	300 kW (about 400 hp)
Commercial jet engine (one)	30 MW
Large power plant	1 GW

Horsepower

The unit horsepower (hp) predates the watt and remains common in engineering:

$$1 \text{ hp} = 746 \text{ W} \approx 750 \text{ W}$$

James Watt defined it by estimating the sustained power output of a draft horse. The conversion is worth knowing because motor specifications are still given in horsepower --- your car's engine, your air conditioner, your vacuum cleaner.

The kilowatt-hour

One oddity of power units trips up many students: the kilowatt-hour (kWh) is not a unit of power. It is a unit of energy.

$$1 \text{ kWh} = (1000 \text{ W})(3600 \text{ s}) = 3.6 \times 10^6 \text{ J} = 3.6 \text{ MJ}$$

A kilowatt-hour is the amount of energy delivered by a source running at 1 kW for 1 hour. Your electricity bill is measured in kWh because the electric company is selling you energy, not power. The power rating of your appliances tells you the rate at which they consume that energy.

Pause and think: Your electricity bill says you used 900 kWh last month. If electricity costs \$0.12 per kWh, what was your bill? More importantly: a 100 W light bulb left on 24/7 for a month --- how many kWh does it use?

Check your answer

Bill: $900 \times 0.12 = \$108$. Light bulb: $100 \text{ W} \times 24 \text{ h/day} \times 30 \text{ days} = 72{,}000 \text{ Wh} = 72 \text{ kWh}$. At \$0.12/kWh, that is \$8.64 per month. A single light bulb is not trivial.

Connection: Rates Everywhere

The relationship between work and power is structurally identical to relationships you have already internalized:

Quantity	Its rate
Position $x$	Velocity $v = dx/dt$
Velocity $v$	Acceleration $a = dv/dt$
Work $W$	Power $P = dW/dt$

In each case, the "rate" version is a derivative. In each case, the total quantity is the integral of its rate:

$$W = \int_{t_1}^{t_2} P \, dt$$

Just as total displacement is the area under the velocity-vs-time curve, total work is the area under the power-vs-time curve. If you are comfortable with velocity as the rate of position change, you already have the conceptual framework for power as the rate of energy transfer.

[Video: Rates as a Unifying Idea. A side-by-side animation. On the left: a car moves along a road, with a $v(t)$ graph below it. The area under the curve fills in, showing displacement accumulating. On the right: a motor lifts a load, with a $P(t)$ graph below it. The area under the curve fills in, showing work accumulating. The visual parallel makes the structural analogy unmistakable. Narration emphasizes: "Velocity tells you how fast position changes. Power tells you how fast energy changes. The mathematics is the same."]

Work (Section 7.1) measured the total energy transfer by a force over a displacement. Power measures how fast that transfer happens. Both are needed. Work tells you how much; power tells you how quickly. Just as knowing a car traveled 300 km is incomplete without knowing it took 3 hours (which gives you the speed), knowing that 50 kJ of work was done is incomplete without knowing it took 5 seconds (which gives you the power).

Average vs. Instantaneous Power

The distinction between average and instantaneous power mirrors the distinction between average and instantaneous velocity from Chapter 2.

Average power is the total work divided by the total time:

$$P_{\text{avg}} = \frac{W_{\text{total}}}{\Delta t}$$

This is useful for overall performance questions. "How powerful an engine do I need to lift this load to that height in this time?" Average power answers that.

Instantaneous power is the rate of work at a specific moment:

$$P(t) = \vec{F}(t) \cdot \vec{v}(t)$$

This is useful for understanding how power demands vary during a process. A car accelerating from rest has low instantaneous power at first (large force but low velocity) and high instantaneous power later (moderate force but high velocity). The engine is working hardest --- in terms of energy delivery rate --- when $Fv$ is at its peak, not necessarily when the force is largest.

Pause and think: A sprinter leaves the blocks. At $t = 0$, the force from her legs is enormous but her velocity is zero. At top speed, her leg force has decreased but her velocity is high. At what point during the sprint is the instantaneous power greatest: at the start, in the middle, or at top speed?

Check your answer

At the start, $P = Fv \approx (\text{large})(0) = 0$. At top speed, if force has dropped to just match air resistance, $P$ may be moderate. The instantaneous power is greatest somewhere in the middle of the acceleration phase, when *both* force and velocity are substantial. The exact moment depends on how $F$ decreases and $v$ increases, but the qualitative answer is clear: maximum power is not at maximum force nor at maximum speed, but at a point between them where their product peaks.

Spaced Retrieval

Before moving to practice, recall concepts from earlier in this chapter.

Recall prompt 1: What is the definition of work done by a constant force? What role does the angle between force and displacement play? (Section 7.1)

Recall prompt 2: State the work-kinetic energy theorem. What does "net work" mean in that theorem? (Section 7.3)

Recall prompt 3: On an energy diagram, how do you find turning points? What determines whether an equilibrium is stable or unstable? (Section 7.5)

Practice Layers

Layer 1: Concrete --- Compute Power from Force and Velocity

Problem 1. A tugboat pulls a barge with a constant horizontal force of $5.0 \times 10^4$ N. The barge moves at a steady 3.0 m/s.

(a) What power does the tugboat engine deliver to the tow cable?

(b) How much work does the tugboat do in 10 minutes?

Check your answer

**(a)** The force is along the direction of motion, so $P = Fv = (5.0 \times 10^4)(3.0) = 1.5 \times 10^5$ W $= 150$ kW. That is about 200 hp. **(b)** At constant power, $W = P \cdot \Delta t = (1.5 \times 10^5)(600) = 9.0 \times 10^7$ J $= 90$ MJ. Alternatively: $W = F \cdot d = (5.0 \times 10^4)(3.0 \times 600) = (5.0 \times 10^4)(1800) = 9.0 \times 10^7$ J. Same answer, as it must be.

Problem 2. A person pushes a shopping cart with a force of 40 N at an angle of 30 degrees below the horizontal. The cart moves horizontally at 1.5 m/s.

(a) What power does the person deliver to the cart?

(b) Why is the answer not $40 \times 1.5 = 60$ W?

Check your answer

**(a)** The velocity is horizontal. The force is 30 degrees below horizontal. The angle between $\vec{F}$ and $\vec{v}$ is $30°$. $$P = Fv\cos\theta = (40)(1.5)\cos 30° = 60 \times 0.866 = 52 \text{ W}$$ **(b)** Only the component of force along the direction of motion does work. The downward component of the push presses the cart into the floor (increasing the normal force) but does not contribute to the cart's horizontal motion. Power is $\vec{F} \cdot \vec{v}$, not $|\vec{F}||\vec{v}|$.

Layer 2: Pattern --- Compare Power Requirements

Problem 3. A car drives at constant speed on a level road. At this speed, the total resistive force (drag plus rolling friction) is $F_{\text{resist}}$.

(a) If the car doubles its speed and the resistive force doubles (a rough approximation at moderate speeds), by what factor does the required engine power increase?

(b) If instead drag is the dominant resistance and it follows $f_D = \frac{1}{2}C_D \rho A v^2$, by what factor does the required power increase when speed doubles?

Check your answer

**(a)** If $F \to 2F$ and $v \to 2v$, then $P = Fv \to (2F)(2v) = 4Fv$. Power increases by a factor of 4. **(b)** If $F_{\text{drag}} \propto v^2$, then doubling $v$ quadruples the drag force. The required power is $P = F_{\text{drag}} \cdot v \propto v^2 \cdot v = v^3$. Doubling the speed increases the required power by a factor of $2^3 = 8$. **(c)** Engine power --- and therefore fuel consumption rate --- scales roughly as $v^3$ at speeds where aerodynamic drag dominates. Going from 55 mph to 75 mph (a factor of 1.36 in speed) increases the power demand by a factor of $1.36^3 \approx 2.5$. The engine burns fuel 2.5 times faster, but you cover distance only 1.36 times faster, so fuel economy (miles per gallon) drops significantly. This is the fundamental physics behind the speed-versus-efficiency tradeoff.

Problem 4. Three elevators lift identical 800 kg loads to the same height (20 m).

Elevator A takes 40 s.
Elevator B takes 20 s.
Elevator C takes 10 s.

(a) Compute the work done by each elevator motor.

(b) Compute the average power for each.

Check your answer

**(a)** All three do the same work: $W = mgh = (800)(9.8)(20) = 156{,}800$ J $\approx 157$ kJ. Work depends on force and displacement, not on time. **(b)** Average power: - A: $P = 156{,}800 / 40 = 3{,}920$ W $\approx 3.9$ kW - B: $P = 156{,}800 / 20 = 7{,}840$ W $\approx 7.8$ kW - C: $P = 156{,}800 / 10 = 15{,}680$ W $\approx 15.7$ kW **(c)** Elevator C's motor must be rated at four times the power of Elevator A's motor, even though both accomplish the same task. Power is the performance specification; work is the task specification.

Layer 3: Structure --- Why $P = Fv$ Is So Revealing

Problem 5. The formula $P = Fv$ says that at constant power, force and velocity are inversely related: $F = P/v$.

(a) Explain physically why a car can accelerate more strongly at low speed than at high speed, even if the engine delivers the same power throughout.

(b) A car engine delivers a constant 120 kW. Plot (or describe) the maximum force available at the wheels as a function of speed from $v = 10$ m/s to $v = 60$ m/s.

(c) How does this explain the purpose of a gearbox?

Check your answer

**(a)** At constant power $P$, the maximum force the engine can deliver is $F = P/v$. At low speed, $v$ is small, so $F$ can be large --- the engine can push hard. At high speed, $v$ is large, so $F$ must be small --- the same power output is "spread over" more distance per second. Since acceleration is $a = F/m$, the car accelerates more strongly at low speed even though the engine is working just as hard (delivering energy at the same rate). **(b)** $F = 120{,}000/v$. - At $v = 10$ m/s: $F = 12{,}000$ N - At $v = 20$ m/s: $F = 6{,}000$ N - At $v = 30$ m/s: $F = 4{,}000$ N - At $v = 40$ m/s: $F = 3{,}000$ N - At $v = 60$ m/s: $F = 2{,}000$ N This is a hyperbola ($F \propto 1/v$). The force drops rapidly at first and then levels off. **(c)** A gearbox changes the mechanical advantage between the engine and the wheels. In a low gear, the engine turns many times for each wheel revolution --- this trades speed for force, keeping the engine in its power-efficient RPM range while delivering large force at low wheel speed. In a high gear, fewer engine turns per wheel revolution --- this trades force for speed. The gearbox lets the engine operate near its peak power output across a wide range of vehicle speeds, rather than being limited to one force-speed combination.

Layer 4: Transfer --- Power in Everyday Technology

Problem 6. A rooftop solar panel delivers an average power of 300 W over the course of a sunny day.

(a) How many joules of energy does it deliver in 8 hours of sunlight?

(b) A home battery system stores 10 kWh. How many hours of sunlight does it take this single panel to fully charge the battery from empty?

(c) If a household uses an average of 1.2 kW continuously, how many of these panels are needed to cover that demand (during sunlight hours)?

Check your answer

**(a)** $E = P \cdot t = (300)(8 \times 3600) = 300 \times 28{,}800 = 8.64 \times 10^6$ J $= 8.64$ MJ. In more practical units: $300 \text{ W} \times 8 \text{ h} = 2400 \text{ Wh} = 2.4$ kWh. **(b)** The panel delivers 2.4 kWh per day (8 hours of sun). To store 10 kWh: $10 / 2.4 \approx 4.2$ days, or equivalently $10{,}000 / 300 \approx 33.3$ hours of sunlight. **(c)** The household needs 1.2 kW. Each panel delivers 0.3 kW. Number of panels: $1200/300 = 4$ panels (to match demand during sun hours; actual systems need more panels or batteries to cover nighttime and cloudy days). This is a power-budget calculation --- the same kind engineers do when sizing any energy system. The concepts of work and power turn vague questions ("Is one solar panel enough?") into precise arithmetic.

Problem 7. A cyclist produces a sustained power output of 250 W. She rides on a flat road at constant speed against a total resistive force of 25 N.

(a) What is her speed?

(b) She reaches a hill with a 5% grade (rise/run = 0.05, so $\sin\theta \approx 0.05$). If her mass plus bicycle is 70 kg and she maintains the same power output, what is her new constant speed uphill? (Assume the resistive force stays at 25 N.)

Check your answer

**(a)** At constant speed, $P = F_{\text{resist}} \cdot v$, so $v = P/F = 250/25 = 10$ m/s (about 36 km/h, a reasonable cycling speed on flat ground). **(b)** Uphill, the total force the cyclist must overcome is $F = F_{\text{resist}} + mg\sin\theta = 25 + (70)(9.8)(0.05) = 25 + 34.3 = 59.3$ N. At constant speed with the same power: $v = P/F = 250/59.3 = 4.2$ m/s (about 15 km/h). **(c)** On the flat, the cyclist fights only resistive forces. On the hill, she must also fight gravity, which adds $mg\sin\theta$ to the required force. Since her power is fixed, $v = P/F_{\text{total}}$, and the additional gravitational force at the denominator cuts her speed dramatically. A modest 5% grade more than doubled the required force, cutting her speed by more than half. This is the $P = Fv$ tradeoff in action: at fixed power, more force means less speed.

Reflection

Why do we need both "work" and "power" as separate concepts?

Work tells you the total energy transferred. Power tells you the rate of that transfer. They carry different information about the same physical process, just as displacement and velocity carry different information about motion.

Consider this: two batteries both store 1000 J of energy. One can discharge it in 1 second (1000 W). The other discharges it over 1000 seconds (1 W). They contain the same energy, but they are useful for entirely different purposes. The first can start a car engine. The second can run a small LED. The difference is not energy --- it is power.

In engineering and everyday life, power is often the more important quantity. We buy light bulbs by wattage, size electrical panels by amperage (which determines power capacity), and rate car engines in horsepower. All of these are power specifications. We care about work (total energy) mainly when we ask "how long will the fuel last?" or "how big a battery do I need?" --- and even those questions are answered by dividing stored energy by power demand.

Having both concepts lets you reason about both the total and the rate, which together tell the full story of energy transfer.

Looking Ahead

This section introduced power --- the rate of doing work --- and connected it to force and velocity through $P = \vec{F} \cdot \vec{v}$. You saw that power is the practical quantity that constrains real machines, and you computed power for motors, cars, cyclists, and solar panels.

With this section, you now have a complete energy toolkit: work as energy transfer (Section 7.1), the work-energy theorem connecting work to speed change (Section 7.3), potential energy and conservation for path-independent forces (Section 7.4), energy diagrams for qualitative motion analysis (Section 7.5), and power for rates of energy transfer (this section).

In Section 7.7, we step back and ask a strategic question: now that you have two complete methods for analyzing motion --- Newton's laws (force and acceleration) and energy methods (work and kinetic energy) --- how do you choose between them? When does each method shine, and when does it struggle? The answer will sharpen your problem-solving judgment and prepare you for the even richer toolkit that comes when momentum enters the picture in Chapter 8.