# Power

> AP Physics C: Mechanics · Unit 3: Work, Energy, and Power
> Source: https://www.owlsprep.com/study/ap-physics-c-mech-u3-power/

This module covers average and instantaneous mechanical power, the power-force-velocity relation, and calculations for constant and variable motion. It includes worked examples, common exam pitfalls, and a quick reference cheatsheet for AP Physics C Mechanics exam prep.

**Prerequisites:** [Work done by constant and variable forces](https://www.owlsprep.com/study/ap-physics-c-mech-u3-work-done-by-variable-forces/); [Work-energy theorem for rigid bodies](https://www.owlsprep.com/study/ap-physics-c-mech-u3-work-energy-theorem/); Basic differentiation and integration for kinematics

## Learning objectives

- Define average and instantaneous mechanical power
- Apply the power-force-velocity relation to solve problems
- Calculate power for constant and variable motion systems
- Solve problems involving constant power output

## What is Power?

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.

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

*Notation:* $P$ (capital P), $P_{\text{avg}}$ for average power

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

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

## 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 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, $\Delta W = \Delta K + \Delta U$ 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 $1.5 \text{ m/s}^2$ 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

   $$T - mg = ma \implies T = m(g+a) = 500(9.8 + 1.5) = 5650 \text{ N}$$
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 $P_{\text{avg}} = \Delta E / \Delta t$ first — this is often faster than calculating work from force and displacement, especially when acceleration changes.

## Instantaneous Power and the Power-Force-Velocity Relation

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 $\vec{F}$ acting on an object with instantaneous velocity $\vec{v}$, substitute $dW = \vec{F} \cdot d\vec{r}$ into the derivative to get the key relation:

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

Here $\theta$ 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 $\theta = 0^\circ$ and $\cos\theta = 1$
3. Calculate force of gravity

   $$F_g = mg = 0.5(9.8) = 4.9 \text{ N}$$
4. Compute instantaneous power

   $$P = F_g v \cos\theta = 4.9 \times 9.8 \times 1 \approx 48 \text{ W}$$

> **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.

## 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:

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

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

**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 $v$

   $$v = \sqrt{\frac{2W}{m}} = \sqrt{\frac{2 \times 5 \times 10^5}{1000}} = \sqrt{1000} \approx 32 \text{ m/s}$$

> **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.

**Worked example:** 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. 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 $P=Fv$ for instantaneous power

   $$F = 10(2) = 20 \text{ N}, \quad P = Fv = 20(1.5) = 30 \text{ W}$$

**Check your understanding**

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

1. 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 $t = 3.0 \text{ s}$?

   - 3.0 W
   - 6.75 W
   - 13.5 W
   - 27 W

   *Why:* Correct. Acceleration $a(t) = 0.5t$, integrating gives $v(t) = 0.25t^2$, so $P = Fv = (6.0)(2.25) = 13.5$ W.

## Common pitfalls

- **Wrong:** Calculating average power by averaging initial and final instantaneous power ($P_{\text{avg}} = (P_1 + P_2)/2$) for arbitrary motion
  - Why it fails: Students confuse average power with average velocity for constant acceleration, where linear averaging works. This only holds if power changes linearly with time.
  - Correct: Always use $P_{\text{avg}} = \Delta W/\Delta t$ regardless of how power changes.
- **Wrong:** Dropping the $\cos\theta$ term and using full force magnitude when force is perpendicular to velocity
  - Why it fails: Students memorize $P=Fv$ and forget the dot product. For example, centripetal force is always perpendicular to velocity, so it delivers zero power.
  - Correct: Always calculate the component of force parallel to velocity before computing power.
- **Wrong:** 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 it fails: Students only account for potential energy change and forget that work done during acceleration also increases kinetic energy.
  - Correct: Always add all energy changes (kinetic + potential) when calculating average power for accelerating systems.
- **Wrong:** Using constant-acceleration kinematics ($v=at$) for constant-power acceleration problems
  - Why it fails: Students associate constant output with constant acceleration, and do not check the relation between force and velocity.
  - Correct: Always use work-energy for constant-power problems, starting from $W = Pt = \Delta K$.
- **Wrong:** Using the conversion $1 \text{ hp} = 550 \text{ W}$ instead of $746 \text{ W}$
  - Why it fails: Students mix up 1 hp = 550 ft-lb per second with the watt conversion.
  - Correct: Memorize $1 \text{ hp} = 746 \text{ W}$ for the AP exam, and confirm unit conversions before finalizing your answer.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Average Power | $P_{\text{avg}} = \frac{\Delta W}{\Delta t} = \frac{\Delta E_{\text{total}}}{\Delta t}$ | Applies to any force, any interval; works for constant or variable power |
| Instantaneous Power | $P = \frac{dW}{dt} = \vec{F} \cdot \vec{v} = F v \cos\theta$ | $\theta$ is angle between force and velocity; negative power means force removes energy from the system |
| Instantaneous Power (Parallel Force) | $P = F_{\parallel} v$ | Simplified form when only the parallel component of force contributes |
| Total Work from Time-Varying Power | $W = \int_{t_1}^{t_2} P(t) dt$ | Integrates rate of energy transfer to get total work done over an interval |
| Total Work (Constant Power) | $W = P \Delta t$ | Special case of the integral when power is constant |
| Velocity (Constant Power, Starting from Rest) | $v(t) = \sqrt{\frac{2 P t}{m}}$ | No friction, no potential energy change; acceleration is not constant here |
| Horsepower-Watt Conversion | $1 \text{ hp} = 746 \text{ W}$ | Required for all real-world engine problems on the AP exam |

## 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 and energy-based approaches to rotational motion, where power relates torque and angular velocity.

- [Systems of Particles and Linear Momentum Overview](https://www.owlsprep.com/study/ap-physics-c-mech-u4-overview/)
- [Center of Mass](https://www.owlsprep.com/study/ap-physics-c-mech-u4-center-of-mass/)
- [Impulse and Momentum](https://www.owlsprep.com/study/ap-physics-c-mech-u4-impulse-and-momentum/)

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