Power (AP Physics C: Mechanics 3.5)

Two students each carry a $10\ \text{kg}$ box up the same flight of stairs. They do the same work against gravity, but one takes $5\ \text{s}$ and the other a minute. What differs is how fast the energy moves, and that rate is power. It is why a car's engine is rated in kilowatts, not joules: the question is how quickly it can deliver energy.

Power is the rate of energy change

Power is the rate at which energy changes. That change can be a transfer into or out of a system (a motor doing work on an elevator, friction taking energy from a sliding box) or a conversion from one kind to another inside it (gravitational potential energy becoming kinetic energy as an object falls).

Because work is how a force changes an object's or a system's energy (Topic 3.2), average power is the total work divided by the time it took. Instantaneous power is the limit of that as the interval shrinks, exactly as instantaneous velocity is the limit of average velocity. The two agree at every moment only when the power is constant.

Running the definition backward gives the energy from the power: $W = \int P\,dt$ , the area under a graph of power versus time. A $1500\ \text{W}$ space heater running for $10$ minutes converts $(1500)(600) = 9.0\times10^5\ \text{J}$ of electrical energy to thermal energy, and a power that rises or falls over time needs the integral (or the area).

Power delivered by a force: P = F · v

For a force acting on a moving object, the work in a short time $dt$ is $dW = \vec{F}\cdot d\vec{r}$ . Divide by $dt$ , and since $\frac{d\vec{r}}{dt} = \vec{v}$ :

This form tells you power at an instant without knowing how far anything moved. Three consequences you'll use:

Holding a speed against resistance. At constant speed, the driving force equals the resistive force, so the power needed is that force times the speed. A boat pushing against $800\ \text{N}$ of water drag at $5.0\ \text{m/s}$ needs its engine to deliver $(800)(5.0) = 4.0\ \text{kW}$ to the water.
A constant force on an accelerating object delivers increasing power. Push a $3.0\ \text{kg}$ block from rest with a constant net force of $12\ \text{N}$ . Its acceleration is $4.0\ \text{m/s}^2$ , so at $t = 2.0\ \text{s}$ it moves at $8.0\ \text{m/s}$ and the force delivers $P = (12)(8.0) = 96\ \text{W}$ . Over those $2.0\ \text{s}$ the force did $W = \frac{1}{2}(3.0)(8.0)^2 = 96\ \text{J}$ , an average of $48\ \text{W}$ : half the final power, because $P = Fv$ grows linearly with time from zero.
Perpendicular forces deliver zero power. The tension in a string whirling a ball in a circle, or the normal force on a car on a level road, is perpendicular to the velocity at every instant, so it does no work (Topic 3.2) and delivers no power.

Power also depends on the frame, through $v$ , the same way kinetic energy does (Topic 3.1). Problems use the ground frame unless they say otherwise.

Constant power is not constant force

An engine or a motor tends to deliver roughly constant power, not constant force. If $P$ is fixed, then $F = \frac{P}{v}$ : the faster the object goes, the less force the same power provides. That is why a car accelerates hard from a stop and more gently at highway speed, and why it can't keep accelerating forever: at top speed, $\frac{P}{v}$ has fallen to equal the resistive forces.

It also gives speed a strong effect on power. Air drag at highway speeds is close to proportional to $v^2$ , so the power needed to hold a speed against it is

P = F_{drag}\,v \propto v^2 \cdot v = v^3.

Doubling the speed takes eight times the power. Worked example 2 puts numbers on this for a cyclist, and worked example 3 follows a car under constant power from rest.

Worked examples

Worked exampleRunning up the stairs

A $65\ \text{kg}$ student runs up a staircase that rises $4.0\ \text{m}$ in $3.2\ \text{s}$ , starting and ending at nearly the same speed. With $g = 9.8\ \text{m/s}^2$ , what average power does she deliver to lift herself, and where does that energy come from?

Show solution

Her kinetic energy doesn't change, so the energy she supplies all goes into the gravitational potential energy of the student-Earth system:

\Delta U = mg\Delta y = (65)(9.8)(4.0) = 2548\ \text{J}.

P_{avg} = \frac{\Delta U}{\Delta t} = \frac{2548\ \text{J}}{3.2\ \text{s}} = 8.0\times10^2\ \text{W}.

The energy is converted from chemical energy in her muscles, which is why her own body counts as the source, not the stairs. That $800\ \text{W}$ is only the rate of increase of potential energy: her body converts chemical energy considerably faster, with the rest becoming thermal energy.

Worked exampleA cyclist against air drag

On level ground, the air drag on a cyclist is $F = bv^2$ with $b = 0.20\ \text{kg/m}$ . Other resistance is negligible.

(a) What power must the cyclist deliver to ride at a constant $10\ \text{m/s}$ ?

(b) By what factor must the power increase to ride at $15\ \text{m/s}$ ?

Show solution

(a) At constant speed the net force is zero, so the cyclist's forward push equals the drag, $F = (0.20)(10)^2 = 20\ \text{N}$ . The push is parallel to the velocity:

P = Fv = (20\ \text{N})(10\ \text{m/s}) = 200\ \text{W}.

(b) $P = bv^2 \cdot v = bv^3$ , so the ratio is $\left(\frac{15}{10}\right)^3 = 3.375 \approx 3.4$ . The power rises to $675\ \text{W}$ for a $50\%$ increase in speed. Writing $P \propto v^3$ first and then taking the ratio is the cleanest route on a "by what factor" question, and it shows the functional dependence graders want.

Worked exampleA car at constant power

A car of mass $m$ starts from rest on a level road. Its engine delivers a constant power $P$ to the car's motion. Neglect friction losses and air resistance.

(a) Derive an expression for the car's speed $v$ as a function of time $t$ .

(b) Derive an expression for the distance traveled as a function of time.

(c) For $m = 1200\ \text{kg}$ and $P = 60\ \text{kW}$ , use the model to calculate the time to reach $25\ \text{m/s}$ , the distance covered in that time, and the forward force on the car at that speed.

(d) The model predicts an unlimited force as $t \to 0$ . Explain why that can't happen, and find the speed at which the model can start to apply if the coefficient of static friction between tires and road is $0.80$ and $g = 9.8\ \text{m/s}^2$ .

Show solution

(a) All the power goes into kinetic energy, so the energy delivered in time $t$ is $Pt$ , and by the work-energy theorem

\frac{1}{2}mv^2 = Pt \quad\Rightarrow\quad v = \sqrt{\frac{2Pt}{m}}.

(b) Integrate the speed:

x = \int_0^t \sqrt{\frac{2P}{m}}\,t'^{1/2}\,dt' = \frac{2}{3}\sqrt{\frac{2P}{m}}\,t^{3/2}.

(c) In the model, from (a), $t = \frac{mv^2}{2P} = \frac{(1200)(25)^2}{2(60\,000)} = 6.25\ \text{s}$ . Here $\sqrt{\frac{2P}{m}} = 10\ \text{m/s}^{3/2}$ , so

x = \frac{2}{3}(10)(6.25)^{3/2} = 104\ \text{m}.

From $P = Fv$ , the forward force at $25\ \text{m/s}$ is $F = \frac{60\,000}{25} = 2.4\times10^3\ \text{N}$ .

(d) $F = \frac{P}{v}$ grows without limit as $v \to 0$ , but the forward force on the car is static friction from the road, which can't exceed $\mu_s mg = (0.80)(1200)(9.8) = 9.4\times10^3\ \text{N}$ . Below the speed where $\frac{P}{v}$ equals that maximum, the tires would spin, so the model only applies above

v = \frac{P}{\mu_s mg} = \frac{60\,000}{9408} = 6.4\ \text{m/s}.

A real car is limited by traction at low speed and by power at high speed, so the numbers in (c) belong to the model, not the car. To correct them, split the run at $6.4\ \text{m/s}$ . Below it the force is at most $\mu_s mg$ , so the acceleration is at most $\mu_s g = 7.84\ \text{m/s}^2$ , and the car needs $\frac{6.38}{7.84} = 0.81\ \text{s}$ and $\frac{(6.38)^2}{2(7.84)} = 2.6\ \text{m}$ to get there (the model claimed $0.41\ \text{s}$ and $1.7\ \text{m}$ ). Above it, the engine's constant power takes over, and the work-energy theorem gives the time from $6.38$ to $25\ \text{m/s}$ as $\frac{m(25^2 - 6.38^2)}{2P} = 5.84\ \text{s}$ . The real car reaches $25\ \text{m/s}$ at about $6.7\ \text{s}$ , after about $105\ \text{m}$ : the model is $0.4\ \text{s}$ fast, but its distance is close, because the traction-limited stretch is short.

Where students lose points

Check yourself

Question 1
A boat moves at constant speed, and the water's drag on it is proportional to the square of its speed. To double the boat's constant speed, the engine's power delivered to the motion must be multiplied by
1. A
  $2$
2. B
  $4$
3. C
  $6$
4. D
  $8$
Show answer
D
At constant speed the thrust equals the drag, $bv^2$ , so $P = bv^2 \cdot v = bv^3$ and doubling $v$ multiplies $P$ by $2^3 = 8$ . Choice B is the factor for the force alone; it leaves out that the force is also applied at twice the speed.
Question 2
The power delivered to a motor's load increases steadily from $0$ to $400\ \text{W}$ over $5.0\ \text{s}$ . How much energy is delivered in that time?
Show answer
$1.0\times10^3\ \text{J}$
Energy is the area under the power-time graph, here a triangle: $\frac{1}{2}(5.0\ \text{s})(400\ \text{W}) = 1000\ \text{J}$ . Multiplying the final power by the time gives twice that.
Question 3
A box slides across a floor at $3.0\ \text{m/s}$ while kinetic friction of $6.0\ \text{N}$ acts on it. At that instant, what is the power delivered to the box by friction?
1. A
  $+18\ \text{W}$
2. B
  $-18\ \text{W}$
3. C
  $0\ \text{W}$
4. D
  $-2.0\ \text{W}$
Show answer
B
$P = \vec{F}\cdot\vec{v} = Fv\cos 180^\circ = -(6.0)(3.0) = -18\ \text{W}$ : friction removes $18\ \text{J}$ of the box's kinetic energy each second. Choice A has the right size but misses that friction opposes the velocity.
Question 4
At one instant, a force $\vec{F} = (3.0\ \text{N})\hat{i} + (4.0\ \text{N})\hat{j}$ acts on an object moving with velocity $\vec{v} = (2.0\ \text{m/s})\hat{i} - (1.0\ \text{m/s})\hat{j}$ . At what rate is the force doing work on the object?
Show answer
$2.0\ \text{W}$
$P = \vec{F}\cdot\vec{v} = (3.0)(2.0) + (4.0)(-1.0) = 2.0\ \text{W}$ . The $\hat{j}$ part of the force points against the $\hat{j}$ part of the velocity, so it removes energy, partly canceling what the $\hat{i}$ part delivers.

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OpenStax, University Physics Volume 1, 7.4 Power