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3.5Work, Energy, and Power

Power — AP PHYSICS 1 Revision Notes & Topical Questions

5 min read 9 parts Lesson 12 of 14 1 quiz

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Lesson summary

3.5 Power — AP PHYSICS 1 notes: Describe energy transfer into, out of or within a system in terms of power.Power = the rate at which energy is transferred into, out of or within a system, or the rate at which work is done.Unit — watt, 1 W = 1 J/s; 1 kW = 1000 W.Scalar — power can…

  • Comparison
  • Pavg = ΔE/Δt
  • Pavg = W/Δt
  • Power–time graph
  • Work–time graph
  • P = F∥v = Fv cosθ

Section 1 of 9Power

Power
Power the rate at which energy is transferred into, out of or within a system, or the rate at which work is done.
Unit
  • Unit watt, 1 W = 1 J/s; 1 kW = 1000 W.
Scalar
  • Scalar power can be negative when a force removes energy.
    • (friction, braking)
Meaning
  • Meaning it tells how quickly energy changes, not how much energy there is.
Comparison
  • Comparison of two machines doing the same work, the one taking less time has more power.

Section 2 of 9Average power

Pavg
ΔE/Δt
  • energy transferred or converted divided by the time taken.
Pavg
W/Δt
  • since work is the change in energy due to a force.
Energy
  • E = Pt.
Power–time graph
  • Power–time graph area under the curve equals the energy transferred.
Work–time graph
  • Work–time graph slope is the power.

Section 3 of 9Instantaneous power

Instantaneous power
P
F∥v = Fv cosθ
  • power delivered by a constant force, with θ the angle between force and velocity.
Constant velocity
  • Constant velocity the driving force equals the total resistive force, so P = Fresistivev.
Lifting a load at constant speed
  • P = mgv.

Section 4 of 9Energy units

Kilowatt-hour
  • Kilowatt-hour is a unit of energy, not power: 1 kWh = 3.6 × 106 J.
Appliance rating
  • Appliance rating watts give the rate only; the energy depends on how long it runs.

Section 5 of 9Efficiency

Efficiency
Efficiency useful output power (or energy) divided by total input, with no unit and never above 100%.
Example
  • 400 W delivered from a 500 W input gives 0.80.

Section 6 of 9Calculating power from a situation

1
Identify the energy transferred or the force doing work.
2
Decide whether average (W/Δt) or instantaneous (Fv) power is needed.
3
At constant velocity use Newton's first law to find the force from the resistive forces.
4
Substitute into P = W/Δt or P = Fv cosθ.
5
State the answer in watts.

Section 7 of 9More from the specification

More from the specification
Component
  • Part of a vector along one axis.
Parallel
  • Components on separate branches, each with the full e.m.f.

Section 8 of 9Key terms

TermMeaning
PowerRate of energy transfer.
Watt1 joule per second.
Kilowatt-hourEnergy unit equal to 3.6 × 106 J.
Average powerEnergy transferred divided by time taken.
EfficiencyUseful output divided by total input.

Section 9 of 9Exam technique

1
Confusing power (J/s) with energy.
2
Not converting minutes or hours to seconds.
3
Treating a kilowatt-hour as a unit of power.
4
Believing larger power always means more work — time matters.

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