Power

AP Physics C: Mechanics· AP Physics C: Mechanics CED — Work, Energy, and Power· 14 min read

1. What is Power?★★☆☆☆⏱ 3 min

Power is a fundamental rate quantity that describes how fast work is done, or equivalently how fast energy is transferred between forms or objects in a mechanical system. Unlike work or energy (which describe total change over an interval), power can change at every instant of motion. This topic makes up ~4-6% of the AP Physics C: Mechanics total exam score, appearing in both multiple-choice and free-response questions.

📘 Definition

Power

$P$ (capital P), $P_{avg}$ for average power

Rate at which work is done or energy is transferred in a mechanical system

Example:

A 100 W light bulb transfers 100 J of energy every second.

The SI unit of power is the watt (W), where $1 W = 1 J/s = 1 kg \cdotp m^{2} / s^{3}$ . You may also encounter horsepower (hp) in real-world problems, with the conversion $1 hp = 746 W$ that you are expected to remember for the exam.

2. Average Power★★☆☆☆⏱ 4 min

Average power is the total work done by a force (or total energy transferred) divided by the length of the time interval over which work occurs. It describes the constant rate that would produce the same total energy transfer as the actual varying process.

P_{\text{avg}} = \frac{\Delta W}{\Delta t} = \frac{\Delta E_{\text{total}}}{\Delta t}

By the work-energy theorem, $Δ W = Δ K + Δ U$ for systems with conservative forces, so you can always substitute total energy change for total work when calculating average power. This lets you calculate average power even without knowing the exact force or displacement at every point.

📐 Worked Example

A 500 kg elevator accelerates upward from rest at $1.5 m/s^{2}$ for 4.0 s. What is the average power delivered by the elevator cable's tension force over this 4.0 s interval?

1
Use Newton's second law to find tension
$T - mg = ma \implies T = m(g+a) = 500(9.8 + 1.5) = 5650 \text{ N}$
2
Calculate total displacement over the interval, starting from rest
$d = \frac{1}{2} a t^2 = 0.5(1.5)(4.0)^2 = 12 \text{ m}$
3
Find total work done by tension
$W_T = Td = 5650 \times 12 = 67800 \text{ J}$
4
Divide by time to get average power
$P_{\text{avg}} = 67800 / 4.0 = 16950 \text{ W} \approx 17 \text{ kW}$

Exam tip:

When asked for average power, always try $P_{avg} = Δ E /Δ t$ first — this is often faster than calculating work from force and displacement, especially when acceleration changes.

3. Instantaneous Power and the Power-Force-Velocity Relation★★★☆☆⏱ 4 min

Instantaneous power is the power delivered by a force at a single moment in time, rather than averaged over an interval. It is found by taking the limit of average power as the time interval approaches zero, giving the derivative of work with respect to time.

P = \lim_{\Delta t \to 0} \frac{\Delta W}{\Delta t} = \frac{dW}{dt}

For a force $F$ acting on an object with instantaneous velocity $v$ , substitute $d W = F \cdot d r$ into the derivative to get the key relation:

P = \vec{F} \cdot \vec{v} = Fv\cos\theta

Here $θ$ is the angle between the force and velocity vectors, so only the component of force parallel to motion contributes to power. This is the most frequently tested power relation on the AP exam.

📐 Worked Example

A 0.5 kg ball is dropped from rest near Earth's surface. What is the instantaneous power delivered by gravity 1.0 s after release? Ignore air resistance.

1
Find velocity 1.0 s after release
$v = gt = 9.8(1.0) = 9.8 \text{ m/s}, directed downward$
2
Gravity is parallel to velocity, so $θ = 0^{\circ}$ and $cos θ = 1$
3
Calculate force of gravity
$F_g = mg = 0.5(9.8) = 4.9 \text{ N}$
4
Compute instantaneous power
$P = F_g v \cos\theta = 4.9 \times 9.8 \times 1 \approx 48 \text{ W}$

4. Power for Variable Motion and Constant Power Systems★★★★☆⏱ 5 min

Many AP problems involve systems where power is held constant (e.g., a car engine operating at maximum output) instead of force being constant. We can invert the definition of instantaneous power to find total work from power:

W = \int_{t_1}^{t_2} P(t) dt

If power is constant, this simplifies to $W = P Δ t$ , which matches the average power formula (for constant power, $P = P_{avg}$ ). For a constant-power system starting from rest with no friction or potential energy change, the work-energy theorem gives $P t = \frac{1}{2} m v^{2}$ , so $v (t) = 2 P t / m$ .

📐 Worked Example

A 1000 kg car accelerates from rest with a constant power output of 50 kW from its engine. Ignoring friction and air resistance, what is the car's speed after 10 s?

1
Calculate total work done by the engine over 10 s
$W = P\Delta t = 50000 \text{ W} \times 10 \text{ s} = 5 \times 10^5 \text{ J}$
2
By work-energy, all work becomes kinetic energy (starts from rest)
$W = \Delta K = \frac{1}{2}mv^2 - 0$
3
Rearrange to solve for $v$
$v = \sqrt{\frac{2W}{m}} = \sqrt{\frac{2 \times 5 \times 10^5}{1000}} = \sqrt{1000} \approx 32 \text{ m/s}$

📐 Worked Example

A 2.0 kg block is dragged along a rough horizontal surface by a horizontal force $F (x) = 10 x$ , where $F$ is in newtons and $x$ is in meters, from $x = 0$ to $x = 2.0 m$ . The block moves at a constant speed of $1.5 m/s$ during this motion. Find (a) total work done by $F (x)$ , (b) average power, (c) instantaneous power at $x = 2.0$ m.

1
(a) Work done by a variable force is the integral of force over displacement
$W = \int_0^2 10x dx = 5x^2 \bigg|_0^2 = 20 \text{ J}$
2
(b) Calculate total time for the displacement, then find average power
$\Delta t = \frac{\Delta x}{v} = \frac{2.0}{1.5} = \frac{4}{3} \text{ s}, \quad P_{\text{avg}} = \frac{W}{\Delta t} = \frac{20}{4/3} = 15 \text{ W}$
3
(c) Find force at $x = 2.0$ m, then use $P = F v$ for instantaneous power
$F = 10(2) = 20 \text{ N}, \quad P = Fv = 20(1.5) = 30 \text{ W}$

✓ Quick check

Test your understanding of instantaneous power with this AP-style multiple choice question:

A force $F (t) = (2.0 N/s) t$ acts parallel to the motion of a 4.0 kg object that starts from rest at $t = 0$ . What is the instantaneous power delivered by the force at $t = 3.0 s$ ?
- 3.0 W
- 6.75 W
- 13.5 W
- 27 W
Reveal answer
13.5 W —
Correct. Acceleration $a (t) = 0.5 t$ , integrating gives $v (t) = 0.25 t^{2}$ , so $P = F v = (6.0) (2.25) = 13.5$ W.

5. Common Pitfalls

Wrong move:

Calculating average power by averaging initial and final instantaneous power ( $P_{avg} = (P_{1} + P_{2}) /2$ ) for arbitrary motion

Why:

Students confuse average power with average velocity for constant acceleration, where linear averaging works. This only holds if power changes linearly with time.

Correct move:

Always use $P_{avg} = Δ W /Δ t$ regardless of how power changes.

Wrong move:

Dropping the $cos θ$ term and using full force magnitude when force is perpendicular to velocity

Why:

Students memorize $P = F v$ and forget the dot product. For example, centripetal force is always perpendicular to velocity, so it delivers zero power.

Correct move:

Always calculate the component of force parallel to velocity before computing power.

Wrong move:

Assuming all work done lifting an accelerating elevator goes into gravitational potential energy, so $P_{avg} = m g Δ y /Δ t$ , ignoring kinetic energy

Why:

Students only account for potential energy change and forget that work done during acceleration also increases kinetic energy.

Correct move:

Always add all energy changes (kinetic + potential) when calculating average power for accelerating systems.

Wrong move:

Using constant-acceleration kinematics ( $v = a t$ ) for constant-power acceleration problems

Why:

Students associate constant output with constant acceleration, and do not check the relation between force and velocity.

Correct move:

Always use work-energy for constant-power problems, starting from $W = P t = Δ K$ .

Wrong move:

Using the conversion $1 hp = 550 W$ instead of $746 W$

Why:

Students mix up 1 hp = 550 ft-lb per second with the watt conversion.

Correct move:

Memorize $1 hp = 746 W$ for the AP exam, and confirm unit conversions before finalizing your answer.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Average Power	$P_{avg} = \frac{Δ W}{Δ t} = \frac{Δ E _{total}}{Δ t}$	Applies to any force, any interval; works for constant or variable power
Instantaneous Power	$P = \frac{d W}{d t} = F \cdot v = F v cos θ$	$θ$ is angle between force and velocity; negative power means force removes energy from the system
Instantaneous Power (Parallel Force)	$P = F_{∥} v$	Simplified form when only the parallel component of force contributes
Total Work from Time-Varying Power	$W = \int_{t_{1}}^{t_{2}} P (t) d t$	Integrates rate of energy transfer to get total work done over an interval
Total Work (Constant Power)	$W = P Δ t$	Special case of the integral when power is constant
Velocity (Constant Power, Starting from Rest)	$v (t) = \frac{2 P t}{m}$	No friction, no potential energy change; acceleration is not constant here
Horsepower-Watt Conversion	$1 hp = 746 W$	Required for all real-world engine problems on the AP exam

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Constant power car acceleration
2022 · FRQ
Power on inclined hill problem

What's Next

Power is the capstone of Unit 3: Work, Energy, and Power, and it is a prerequisite for almost all advanced topics in mechanics that rely on energy analysis. Next, you will apply power and energy concepts to systems of particles and center of mass motion, where you will calculate the power delivered to the entire system by external forces. Mastery of power, particularly the $P = F v$ relation, is also required to analyze simple harmonic motion and energy-based approaches to rotational motion, where power relates torque and angular velocity.