Power is how fast energy is transferred or converted—energy is the “what,” power is the “how quickly.” Per the CED: power is the rate of change of energy (3.5.A.1). Average power: P_avg = ΔE / Δt (3.5.A.2); since work changes energy, P_avg = W / Δt (3.5.A.3). Instantaneous power: P_inst = dW/dt (3.5.A.4), and for a force with component parallel to velocity: P_inst = F∥ v = F v cosθ (3.5.A.5). Units: watt (W) = J/s. Quick example: lifting a 10 kg mass by 2.0 m does W = mgh ≈ 196 J. If you do that in 2.0 s, P_avg = 98 W; do it in 1.0 s, P_avg = 196 W—same energy, different power. For AP exam work, know the definitions, the equations above, and how to apply P = F·v in problems. For a focused review, see the Topic 3.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM) and try practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Work and energy tell you how much energy is transferred or changed; power tells you how fast that transfer happens. The CED defines power as the rate of energy change (P = ΔE/Δt) and gives average and instantaneous forms (Pavg = W/Δt, Pinst = dW/dt, and for a constant force Pinst = F·v = Fv cosθ). That matters because two processes can do the same work but take very different times—a light bulb and a laser might move the same energy but at totally different rates. On the exam, Topic 3.5 asks you to describe energy transfer in terms of power (3.5.A), use the formulas above, and convert units (1 W = 1 J/s). If you want targeted practice and a quick recap of these ideas, check the Topic 3.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM) and the unit overview (https://library.fiveable.me/ap-physics-c-mechanics/unit-3). For extra problems, Fiveable’s practice bank is useful (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Average power is just how fast energy changes: P_avg = ΔE / Δt. Pick a system, decide what energy changed (work in/out, ΔK, ΔU, heat, etc.), find the net energy transfer ΔE (in joules), and divide by the time interval Δt (in seconds). Units: 1 watt = 1 J/s. Quick steps: 1. Identify the energy change of the system (ΔE = E_final − E_initial). If the problem gives work done, use W for ΔE. 2. Measure the time interval Δt during which that change occurred. 3. Compute P_avg = ΔE/Δt and include sign if direction matters (positive = into system). Example: If 200 J of work is done on an object in 5.0 s, P_avg = 200 J / 5.0 s = 40 W. Remember AP C: Mechanics expects you to connect power to work and energy (P_avg = W/Δt) and to know units and when to use instantaneous power (P_inst = dW/dt or F·v). For a short AP review, check the Topic 3.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Average power tells you how much energy was transferred or converted over a finite time interval: P_avg = ΔE/Δt (or P_avg = W/Δt). Instantaneous power is the rate of energy transfer at a single moment—mathematically the time derivative: P_inst = dW/dt. For a constant force with a component parallel to velocity, the AP CED gives P_inst = F‖ v = F v cosθ. Same units (W = J/s). Use P_avg when you only know total work or energy change over an interval. Use P_inst when you need the power at a specific time (take the limit Δt → 0 or compute dW/dt; often you’ll express W in terms of v(t) or F(t) then differentiate). On the exam you may be asked to derive or apply both forms (multiple-choice or free-response), so be comfortable converting total work/time to P_avg and taking derivatives or using P = F·v for instantaneous power. Review Topic 3.5 on Fiveable (study guide: https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Use P = W/Δt when you want average power—the rate energy (or work) is transferred/converted over a finite time interval. That comes straight from the CED: P_avg = ΔE/Δt or P_avg = W/Δt (Topic 3.5.A.2–3). Use it for problems that give total work done over a time interval and ask for the average power. Use P = F v cosθ for instantaneous mechanical power delivered by a force: P_inst = dW/dt = F·v. That form (Topic 3.5.A.4–5) is best when you know a force and the object's instantaneous velocity (or the component of force parallel to velocity). If F is constant and velocity is changing, F·v(t) gives the instantaneous power at each moment; integrating P_inst over time gives total work. Quick checklist: - Given W and Δt → P_avg = W/Δt. - Given F and v (and angle θ between them) → P_inst = F v cosθ. - To go between them: integrate P_inst(t) to get W, or average P_inst over Δt to get P_avg. For more AP-style examples and practice, check the Topic 3.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM) and lots of practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Instantaneous power is P = F · v = F v cos θ because power measures how fast force does work = rate of energy transfer. Work done in a tiny time dt is dW = F · ds, and ds/dt = v, so P = F · v (CED 3.5.A.4–A.5). The cosine appears because only the component of the force parallel to the displacement (or velocity) does work. θ is the angle between the force vector and the velocity (or instantaneous displacement) vector. If θ = 0°, F is fully parallel to v and P = Fv (maximum); if θ = 90°, F is perpendicular to v and P = 0 (no power delivered). Negative power (cos θ < 0) means the force removes energy from the object (braking). This is exactly what the CED expects you to describe for instantaneous power (3.5.A.5). For more review and examples, see the Topic 3.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM) and plenty of practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Pick the form that matches what the problem gives and what it asks for: - If the question gives total energy change or total work over a time interval and asks for average power, use P_avg = ΔE/Δt or P_avg = W/Δt (CED 3.5.A.2–3). Good for FRs that give energy numbers and a time interval. - If it asks for instantaneous power or gives instantaneous quantities (force and velocity at an instant, or v(t)), use P_inst = dW/dt (CED 3.5.A.4). - If a constant force and the object’s velocity (or velocity component) are known, use the handy form P_inst = F‖ v = F v cosθ (CED 3.5.A.5). That’s the most common FR shortcut: dot the force with velocity. - If force varies with time/position but they give W(t) or ask for dW/dt, take the derivative or compute W then divide by Δt. On the exam, state the fundamental relation you’re using (energy rate or dW/dt) before plugging numbers—that earns method points. For a quick review, check the Topic 3.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM) and more practice at the Unit 3 page (https://library.fiveable.me/ap-physics-c-mechanics/unit-3) or practice bank (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Instantaneous power is P = F · v = F v cosθ (CED 3.5.A.5). If the force is perpendicular to the velocity, θ = 90°, cos90° = 0, so P = 0. That means the force does no work at that instant and doesn’t change the object’s kinetic energy—it can only change the direction of velocity, not its magnitude. Common example: centripetal force on an object in uniform circular motion is always perpendicular to v, so instantaneous power from that force is zero even though the object’s direction is changing. This is exactly what the AP asks you to connect in Topic 3.5 (rate of energy transfer). For a quick refresher, see the Topic 3.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM) and try practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Think of power as how fast energy (work) is being transferred. Average power = W/Δt (CED 3.5.A.2–3.5.A.3). To get instantaneous power use calculus. Work done by a force along a path: W = ∫ F · ds. Differentiate that with respect to time: dW/dt = d/dt ∫ F · ds = F · (ds/dt) = F · v, because ds/dt is the instantaneous velocity vector. So Pinst = dW/dt = F · v (CED 3.5.A.4–3.5.A.5). If F is at angle θ to v, that gives P = F v cosθ, which shows only the force component parallel to motion does power. Why it makes sense: dW is the tiny amount of energy added in a tiny time dt; dividing gives the rate of energy transfer. This is exactly what the AP asks you to know for Topic 3.5—see the study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics) for worked examples.

If you have energy vs time data, power is just the rate of change of energy. Use the CED formulas: average power Pav = ΔE/Δt and instantaneous power Pinst = dE/dt. Practically: - For a graph of average power over an interval: compute Pav = (E(t2) − E(t1)) / (t2 − t1) and plot that value at the midpoint time (t1+t2)/2. Example: ΔE = 20 J over Δt = 4 s → Pav = 5 W at tmid. - For a smooth instantaneous-power curve from discrete data, use numerical derivatives. Best choice: central difference for interior points: P_i ≈ (E_{i+1} − E_{i−1}) / (t_{i+1} − t_{i−1}). Use forward/backward difference at the ends. - Watch units (J/s = W) and noisy data—apply a small moving-average or fit a low-order polynomial before differentiating to reduce amplification of noise. - If you have force & velocity instead, you can also use P = F·v (CED 3.5.A.5). Practice plotting skills are tested on the exam (create quantitative graphs), so try problems at Fiveable (practice link: https://library.fiveable.me/practice/ap-physics-c-mechanics) and review Topic 3.5 (https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM).

Power is the rate energy changes with time—so its SI unit is joule per second (J/s), called the watt (W). From the CED: Pavg = ΔE/Δt and Pinst = dW/dt (or P = F·v cosθ for a force-velocity case). Useful unit conversions: - 1 W = 1 J/s (exact) - 1 kW = 10^3 W - 1 MW = 10^6 W - 1 horsepower (hp) ≈ 745.7 W (mechanical) - 1 ft·lb/s ≈ 1.3558 W (because 1 ft·lb = 1.3558 J) - 1 erg/s = 1e-7 W (since 1 erg = 1e-7 J) So if you compute work in joules and divide by seconds, you get watts. On the AP exam you’ll most often use W = J/s or convert to kW/h for real-world numbers; remember P = Fv cosθ when given forces and velocities (CED 3.5.A.5). For a quick topic review and practice problems, see the Topic 3.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM) and more practice at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Power is just the rate energy changes with time—so on an energy vs. time graph the slope equals power. Instantaneous power = dE/dt, and average power = ΔE/Δt (so the slope of a tangent = instantaneous power; the slope of a secant between two times = average power). Positive slope → net energy flowing into/being added to the system; negative slope → energy leaving/being removed. Units: watts (J/s). For forces doing work, instantaneous power can also be written P = F·v = F v cosθ (CED 3.5.A.1–5). This idea shows up on the AP in both conceptual and lab-style questions (you may be asked to interpret slopes or use ΔE/Δt in Experimental Design/Analysis). Review Topic 3.5 on Fiveable (study guide: https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics) if you want some quick examples and graph practice.

Power matters in real machines because it tells you how quickly energy is transferred or converted—not just how much energy you have. In engines and motors you care about both energy (work) and the rate it’s delivered: higher power means the device can do the same amount of work in less time (Pavg = ΔE/Δt) or produce larger instantaneous effects (Pinst = F·v = F v cosθ). Practically, that controls acceleration, top speed, how fast a pump moves fluid, or how quickly an elevator climbs. Power also links to efficiency and heat: more power output often means more fuel or electrical input and more waste heat you must manage. On the AP C exam, expect to use P = W/Δt or P = Fv in problems and compare power input vs output. For a quick review see the Topic 3.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM) and try practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Power is how fast energy is transferred or converted: P = ΔE/Δt (or Pinst = F·v). For machines, efficiency compares useful power out to power in: efficiency = Pout / Pin (often written as η). So if a motor delivers 100 W of mechanical power and η = 0.80, the electrical power it draws is Pin = 100 W / 0.80 = 125 W (25 W lost as heat/friction). Efficiency is about how much of the input energy appears as the desired output; power is about the rate. They connect because for a required Pout a less-efficient machine needs a larger Pin and therefore wastes more power per second. On the AP exam you should be able to use Pavg = W/Δt and Pinst = Fv to compute input/output powers and then form η = Pout/Pin (CED 3.5.A.2–5). For more practice and concise review on Topic 3.5, see the Fiveable study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM) and lots of practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Power is just how fast energy is transferred or converted—that’s the key link to work and energy for the exam. Memorize these forms from the CED: average power P_avg = ΔE/Δt (energy rate) and because work changes energy, P_avg = W/Δt. Instantaneous power is P_inst = dW/dt, and for a constant force the component parallel to velocity gives P_inst = F·v = F v cosθ. Units: watt = J/s. Use P = Fv when you know force and speed (remember the cosine). On the AP exam this shows up in both MCQ and FR problems in Unit 3 (Work, Energy & Power)—you may be asked to derive, compute, or explain power as a rate of energy change. For a quick refresher see the Topic 3.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-3/5-power/study-guide/aXNJ6Af1NIzpF1IM) and grab practice problems from the Unit 3 page (https://library.fiveable.me/ap-physics-c-mechanics/unit-3) or the practice bank (https://library.fiveable.me/practice/ap-physics-c-mechanics).