Power measures how quickly a force does work
You will be able to: Calculate instantaneous power with a dot product and distinguish it from work.
Why can the same force deliver different power at different speeds?
Push a cart with 10 N along its motion. At 2 m/s you deliver 20 joules each second, or 20 watts. At 4 m/s the same force delivers 40 watts because its point of application travels twice as far each second.
A useful starting point: Measure spring stiffness with an energy graph →
Words and symbols before equations
- Power P
- Rate of energy transfer or conversion, measured in watts (W).
- Watt
- 1 W = 1 J/s; this unit W is not the work variable W.
- Instantaneous
- Evaluated at a particular moment rather than averaged over an interval.
- Application-point velocity
- Velocity of the location where a force acts; for our particle it is the object’s velocity.
What this picture assumes
Instantaneous force acts on a translating point object in one inertial frame. Force angle is measured from +x; negative velocity reverses motion. Power of this force is shown; other forces may also transfer energy.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Force component along +x = 10 N; vₓ = 2 m/s; P = 20 W. This is instantaneous power of this force, not necessarily net power.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
A small work contribution is dW = F · dr. Divide by dt: P = dW/dt = F · v. Thus P = Fv cos θ at an instant, using the angle between force and velocity. This identity also applies when the force changes, provided its instantaneous value is used.
Positive power transfers energy in through that force, negative power removes it, and perpendicular force has zero power. The sum of force powers on a particle is dK/dt. A motor’s power alone need not equal dK/dt if resistance simultaneously removes energy.
For an ideal constant-power drive with no losses, F = P/v along the motion. It supplies less force as speed rises. The formula becomes singular at v = 0, so an actual motor needs a finite-force startup regime; do not interpret an ideal constant-power model as unlimited real force at rest.
A worked example, step by step
A 2 kg particle has instantaneous velocity (3, 4) m/s and total force (6, −2) N. Find net power and the current rate of change of K.
- Use the velocity and force components in the same inertial frame.
- P_net = Fₓvₓ + Fᵧvᵧ = (6)(3) + (−2)(4).
- P_net = 18 − 8 = 10 W = 10 J/s.
- Therefore dK/dt = 10 J/s at that instant. This does not mean power remains 10 W at later times.
Power is not force and not energy. A large force at zero application-point speed has zero instantaneous mechanical power.
Can a force accelerate an object at an instant when its power is zero?
Compare with an explanation
Yes. At rest F can be nonzero while F · v = 0. Just after that instant the speed and power may become nonzero.
Predict. Change one thing. Explain.
Hold force fixed and change velocity through zero. Predict the sign of power. Then set the force perpendicular and explain why it can change direction without increasing K.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Force component along +x = 10 N; vₓ = 2 m/s; P = 20 W. This is instantaneous power of this force, not necessarily net power.
Instantaneous force acts on a translating point object in one inertial frame. Force angle is measured from +x; negative velocity reverses motion. Power of this force is shown; other forces may also transfer energy.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant work, system boundary, energy or calculus relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA constant horizontal force of 4 N accelerates a 2 kg cart from rest with no losses. Find v at 3 s, instantaneous power then, work over the first 3 s, and average power over that interval.
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Compare with the answer and four-point rubric
- 1 point: Acceleration is 4/2 = 2 m/s², so v(3 s) = 6 m/s.
- 1 point: Instantaneous P = Fv = 24 W at 3 s.
- 1 point: Distance = ½at² = 9 m, so work = 4×9 = 36 J.
- 1 point: Average P = 36 J / 3 s = 12 W, smaller than the final instantaneous value.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1How does force connect to power?
Through P = F · v at the force’s application point.
RECALL 2Can power be negative?
Yes. It means energy leaves the chosen object through that force.
RECALL 3Why is constant power problematic exactly at rest?
F = P/v would require infinite force for nonzero P, beyond the idealization’s useful domain.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Power measures how quickly a force does work
- P = dW/dt = F · v = Fv cos θ.
- For a fixed-mass particle, ΣP = dK/dt.
- Constant positive drive power along motion: F = P/v for v > 0.
Remember: Power is not force and not energy. A large force at zero application-point speed has zero instantaneous mechanical power.
Conditions: Instantaneous force acts on a translating point object in one inertial frame. Force angle is measured from +x; negative velocity reverses motion. Power of this force is shown; other forces may also transfer energy.
Refresh Kid · AP Physics C: Mechanics Unit 3 (official Unit 3) · Objectives 3.5.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.5, objectives 3.5.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 3: Work, Energy, and Power. The unit covers Topics 3.1–3.5. Calculus connects work to force integrals, force to potential-energy derivatives, and power to the rate of energy transfer. Models distinguish object-only and multi-object systems; translational models exclude rotational energy unless explicitly noted. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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