Short on time? This page is the recap for the calculus-based course (separate from Physics 1, which uses no calculus). For explanations, diagrams and worked examples, use the full study guide.
Recap
- Power is the rate at which energy changes. Energy can be transferred into or out of a system by work from outside forces, or converted from one form to another inside the system.
- Choose the system first. Friction on a sliding block transfers energy out of a “block only” system but converts kinetic energy to internal energy in a “block + floor” system.
- Average power uses the total energy change (or total work) and the time taken.
- Instantaneous power is a derivative. With dW = F·dr, it becomes the dot product P = F·v.
- The net power on an object is the rate of change of its kinetic energy.
Key relationships
| Relationship | Units | Meaning |
|---|---|---|
| P_avg = ΔE/Δt = W/Δt | W = J/s | total energy change ÷ time |
| P = dW/dt | W | instantaneous power |
| P = F·v = Fv cos θ = F∥v | W | only the component of F along v counts |
| P = F_x v_x + F_y v_y + F_z v_z | W | component form of the dot product |
| P_net = F_net·v = dK/dt | W | positive: speeding up; negative: slowing down |
| W = ∫P dt | J | area under the P–t graph |
| constant P from rest: v = √(2Pt/m) | m/s | from K = Pt, only that force doing work |
Assumptions behind the numbers
- Inertial reference frame; air resistance ignored unless stated; g = 9.8 m/s².
- Point object: forces act at one point, so F·v uses the object’s velocity.
- Constant-power models fail near v = 0, where F = P/v would be unlimited.
Mistakes to avoid
- Watts for energy. A kilowatt-hour (3.6 × 10⁶ J) is energy, not power.
- Using the whole force in P = Fv. Take the component along v, or use the dot product.
- Dropping the sign. Forces opposing the motion deliver negative power.
- Constant power = constant force. At constant P, the force falls as v rises.
- P_avg = (average F) × (average v). Wrong when both change. Use W/Δt.
- Forgetting perpendicular forces. Normal force on level ground, or tension in uniform circular motion, deliver zero power.
- Mixing up the system. Say whether energy is transferred out or converted inside.
Quick self-check
- A 40 N force acts at 60° to a velocity of 3.0 m/s. What power does it deliver? (40 × 3.0 × cos 60° = 60 W)
- P(t) = 2.0t² W. How much work is done from 0 to 3.0 s? (∫₀³ 2.0t² dt = 18 J)
- A 250 kg cart starts from rest and is pushed with a constant 500 W, no other work done. What is its speed at 4.0 s? (v = √(2 × 500 × 4.0 ÷ 250) = 4.0 m/s)
Next: practice questions.