Power — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Definition of average and instantaneous mechanical power, power as the rate of energy transfer, the power-force-velocity relation, calculation of power for constant and variable motion, and applications to forces acting on moving objects.

You should already know: Work done by constant and variable forces, The work-energy theorem for rigid bodies, Basic differentiation and integration for kinematics.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: Mechanics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Power?

Power is a fundamental quantity that describes how fast work is done, or equivalently how fast energy is transferred between forms or between objects in a mechanical system. Unlike work or energy, which are interval quantities that describe total change over a period, power is a rate quantity that can change at every instant of motion.

This topic makes up approximately 4-6% of the total exam weight for AP Physics C: Mechanics, aligned with the College Board CED for Unit 3, and it appears regularly in both multiple-choice (MCQ) questions (often standalone or paired with kinematics) and free-response (FRQ) questions (usually as a part of a larger energy or force problem).

Power is almost always denoted with a capital $P$, with average power written as $P_{\text{avg}}$ and instantaneous power written as $P$ when context is clear. The SI unit of power is the watt (W), where $1\text{ W}=1\text{ J/s}=1{\text{ kgcdotpm}}^2/{\text{s}}^3$. On the exam, you may also encounter the imperial unit horsepower, with $1\text{ hp}=746\text{ W}$, which you are expected to convert if required. Understanding power connects all prior concepts of work and energy to real-world systems where energy output rate matters.

2. Average Power

Average power is the total work done by a force (or total energy transferred) divided by the length of the time interval over which the work is done. It describes the overall constant rate that would result in the same total energy transfer as the actual varying process. From its definition, the core formula for average power is: $$P_{\text{avg}}=\frac{\Delta W}{\Delta t}=\frac{\Delta E_{\text{total}}}{\Delta t}$$ where $\Delta W$ is total work done over $\Delta t=t_2-t_1$, and $\Delta E_{\text{total}}$ is the total change in mechanical energy of the system. By the work-energy theorem, $\Delta W=\Delta K+\Delta U$ for systems with conservative forces, so we can always substitute total energy change for total work when calculating average power. This makes average power easy to calculate even when you do not know the exact force or displacement.

Worked Example

A 500 kg elevator accelerates upward from rest at $1.5{\text{ m/s}}^2$ 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: $T-mg=ma⟹T=m(g+a)=500(9.8+1.5)=5650\text{ N}$.
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 $P_{\text{avg}}=\Delta E/\Delta t$ first — this is often faster than calculating work from force and displacement, especially when acceleration changes.

3. Instantaneous Power

Instantaneous power is the power delivered by a force at a single moment in time, rather than averaged over an interval. We derive it by taking the limit of average power as the time interval approaches zero, which gives the derivative of work with respect to time: $$P=\underset{\Delta t\to 0}{\lim }\frac{\Delta W}{\Delta t}=\frac{dW}{dt}$$ For a force $\vec{F}$ acting on an object with instantaneous velocity $\vec{v}$, we substitute $dW=\vec{F}\cdot d\vec{r}$ into the derivative: $$P=\frac{d}{dt}\left(\vec{F}\cdot d\vec{r}\right)=\vec{F}\cdot \frac{d\vec{r}}{dt}=\vec{F}\cdot \vec{v}$$ This simplifies to $P=Fv\cos \theta$, where $\theta$ is the angle between the force and velocity vectors, and $F_{\parallel }=F\cos \theta$ is the component of force parallel to motion. 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 also directed downward, so $\theta =0^{\circ }$ and $\cos \theta =1$.
3. Force of gravity: $F_g=mg=0.5(9.8)=4.9\text{ N}$.
4. Calculate power: $P=F_{g}v\cos \theta =4.9\times 9.8\times 1\approx 48\text{ W}$.

Exam tip: A negative power result means the force is removing energy from the system (e.g., kinetic friction), not a calculation error — always pay attention to the sign of power, as it is often tested.

4. Power for Variable Motion and Constant Power Systems

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: since $P=dW/dt$, rearranging gives $dW=Pdt$, so integrating over a time interval gives: $$W={\int }_{t_1}^{t_2}P(t)dt$$ If power is constant, this simplifies to $W=P\Delta t$, which matches the average power formula (for constant power, $P=P_{\text{avg}}$). Using the work-energy theorem for a constant-power system starting from rest with no friction, we get $Pt=\frac{1}{2}m{v}^2$, so $v(t)=\sqrt{2Pt/m}$. This is a common result tested on both MCQ and FRQ.

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. 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: $W=\Delta K=\frac{1}{2}m{v}^2-0$.
3. Rearrange to solve for $v$: $v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2\times 5\times 10^5}{1000}}=\sqrt{1000}\approx 32\text{ m/s}$.

Exam tip: Constant power does NOT mean constant acceleration. Since $P=Fv$, $F=P/v$, so acceleration $a=P/(mv)$ decreases as speed increases — never use constant-acceleration kinematics for constant-power problems.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Calculating average power by averaging initial and final instantaneous power ($P_{\text{avg}}=(P_1+P_2)/2$) 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 $P_{\text{avg}}=\Delta W/\Delta t$ regardless of how power changes.
• Wrong move: Dropping the $\cos \theta$ term and using full force magnitude when force is perpendicular to velocity. Why: Students memorize $P=Fv$ 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 $P_{\text{avg}}=mg\Delta y/\Delta t$, 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 ($v=at$) 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 $W=Pt=\Delta K$.
• Wrong move: Using the conversion $1\text{ hp}=550\text{ W}$ instead of $746\text{ W}$. Why: Students mix up 1 hp = 550 ft-lb per second with the watt conversion. Correct move: Memorize $1\text{ hp}=746\text{ W}$ for the AP exam, and confirm unit conversions before finalizing your answer.

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

A force $F(t)=(2.0\text{ N/s})t$ 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 $t=3.0\text{ s}$? A) 3.0 W B) 6.75 W C) 13.5 W D) 27 W

Worked Solution: First, find acceleration as a function of time: $a(t)=F(t)/m=(2.0t)/4.0=0.5t{\text{ m/s}}^3$. Velocity is the integral of acceleration from 0 to $t$: $v(t)={\int }_0^{t}0.5\tau d\tau =0.25t^2$. At $t=3.0\text{ s}$, $F=2.0(3)=6.0\text{ N}$, and $v=0.25(3^2)=2.25\text{ m/s}$. Power is $P=Fv=6.0\times 2.25=13.5\text{ W}$. The correct answer is C.

Question 2 (Free Response)

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

Worked Solution: (a) Work done by a variable force is the integral of force over displacement: $$W={\int }_0^{2}10xdx=5x^2{|}_0^2=5(4)=20\text{ J}$$ (b) For constant speed, total time is $\Delta t=\Delta x/v=2.0/1.5=4/3\text{ s}$. Average power is: $$P_{\text{avg}}=W/\Delta t=20/(4/3)=15\text{ W}$$ (c) At $x=2.0\text{ m}$, $F=10(2)=20\text{ N}$. Velocity is constant at $1.5\text{ m/s}$, so: $$P=Fv=20(1.5)=30\text{ W}$$

Question 3 (Application / Real-World Style)

A 1500 kg car with a 120 horsepower engine accelerates from rest up a hill with a constant incline of 5 degrees above the horizontal. Ignoring friction and air resistance, what is the maximum constant speed the car can maintain moving up this incline? Give your answer in m/s and convert to km/h for context. (Use $g=9.8{\text{ m/s}}^2$, $1\text{ hp}=746\text{ W}$)

Worked Solution: First convert engine power to watts: $P=120\times 746=89520\text{ W}$. For constant speed, the engine force equals the component of gravity parallel to the incline: $$F=mg\sin \theta =1500(9.8)(\sin 5^{\circ })\approx 12810\text{ N}$$ Use $P=Fv$ to solve for speed: $$v=P/F=89520/12810\approx 7.0\text{ m/s}$$ Convert to km/h: $v=7.0\times 3.6=25.2\text{ km/h}$. In context, this means a 120 hp car can maintain a maximum speed of roughly 25 km/h up this steep incline, ignoring drag and rolling friction.

7. Quick Reference Cheatsheet

8. 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 $P=Fv$ relation, is also required to analyze simple harmonic motion, where you calculate the average power delivered by the restoring force over a full oscillation. Without a solid understanding of how power connects work, force, and velocity, you will struggle with energy-based approaches to rotational motion, where power is used to relate torque and angular velocity.

• Work done by variable forces
• Work-energy theorem
• Systems of particles and linear momentum
• Rotation of rigid bodies