3.5 Power

Four little picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. Two friends, the same stairs

2. Same time, more weight

Mia and Jo each carry the same box up the same stairs. Mia takes 10 s and Jo takes 20 s. Who uses more power?

3. Two elevators

Read the story as text

An old building and a new building both have 30 m tall elevator shafts. Each lifts the same 800 kg of cab and people to the top floor. The old one takes 30 s. The new one takes 15 s.

Both raise the same mass the same height, so both give the same energy to the system. Yet the new motor is bigger, louder and costs more. What is the new motor doing "more" of, if not energy? It moves the same energy in half the time: it has twice the power.

A lamp is rated 60 W. That means it uses:

4. Check yourself

Think of your answer first, then tap to see it.

a) Two motors do the same work. Motor A takes 10 s and motor B takes 5 s. Which has more power, and how many times more?

Show answer

Motor B, twice as much. Same work in half the time means twice the power.

b) A 100 W lamp is on for 10 s. How much energy does it use?

Show answer

Energy = power × time = 100 J/s × 10 s = 1000 J.

c) You lift a 2.0 kg box 1.0 m straight up in 1.0 s, at a steady speed. About what power do you deliver?

Show answer

Work = m g h = 2.0 × 9.8 × 1.0 = 19.6 J. Power = 19.6 J ÷ 1.0 s = 19.6 W, about 20 W.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead.

The same work is done in half the time. The power is:

A motor delivers 500 W for 10 s. How much energy does it transfer?

You push a box with 100 N at a steady 2.0 m/s, along the motion. Your power is:

How fast energy is moved: P = W / t = ΔE / t, and P = F v. Average and instantaneous power, with an elevator motor to play with.

The idea

In pictures

  • Two friends climb the same stairs. They do the same work, but the one who gets there faster has more power.
  • Power is how fast energy moves: joules every second. Half the time means twice the power.
  • Lifting more weight in the same time also takes more power: more work, same time.
  • Pushing hard and moving fast both raise the power: power = push × speed.

Below is the same idea in words, and then with numbers.

Power is how fast energy is transferred, or how fast work is done.

Pavg = W / Δt = ΔE / Δt

The unit is the watt: 1 W = 1 J/s. A 60 W bulb uses 60 joules every second. (Car engines are sometimes rated in horsepower: 1 hp ≈ 746 W.)

If a force F pushes an object moving at speed v (force along the motion), then each second it does F × (distance per second) of work:

P = F v cos θ   (instantaneous power)

With numbers: the two elevators

m = 800 kg, height 30 m, both start and end at rest (so in the end all the work is ΔUg). Up is positive.

Show all steps as text
  1. Energy given to the cab + Earth system: ΔUg = m g h = 800 × 9.8 × 30 = 235 200 J. Same mass, same height: same energy for both.
  2. Old motor: Pavg = 235 200 / 30 = 7840 W ≈ 7.8 kW. Average power = energy / time.
  3. New motor: Pavg = 235 200 / 15 = 15 680 W ≈ 15.7 kW. Half the time, twice the power.
  4. Check with P = F v: the new elevator's average speed is 30 m / 15 s = 2.0 m/s. At constant speed F = m g = 7840 N, so P = 7840 × 2.0 = 15 680 W. Same answer from forces.
  5. A real elevator speeds up first. While accelerating at 1.0 m/s², the cable pulls F = m(g + a) = 800 × 10.8 = 8640 N; at the moment v = 2.0 m/s, P = 8640 × 2.0 = 17 280 W, more than the cruising power. Instantaneous power can be above the average for a moment.

Play: an elevator motor

The elevator speeds up, cruises, then slows to a stop at the top. The graph shows the motor's power, P = (cable force) × (speed), at every moment. The panel also shows the average power so far. Predict the shape of the power graph first.

Try: halve the cruise speed. The trip takes about twice as long and the cruising power halves, but the energy at the top (m g H) is the same.

Worked examples

Running up the stairs basic

A 60 kg student runs up stairs that rise 4.0 m in 5.0 s. What is her average power output (to the student + Earth system's Ug)?

Show solution
Show all steps as text
  1. ΔUg = m g h = 60 × 9.8 × 4.0 = 2352 J. Starts and ends at about the same speed.
  2. P = 2352 / 5.0 = 470 W. Energy per second.

Pushing a cart basic

You push a shopping cart with 200 N at a constant 3.0 m/s (force along the motion). What power do you deliver?

Show solution
Show all steps as text
  1. P = F v = 200 × 3.0 = 600 W. Force and velocity in the same direction, cos 0° = 1.

Average vs final power medium

A 1000 kg car speeds up from rest to 20 m/s in 8.0 s with a constant net force. Ignore friction. Find the average power and the power at the end.

Show solution
Show all steps as text
  1. W = ΔK = ½ × 1000 × 20² = 200 000 J. Pavg = 200 000 / 8.0 = 25 000 W. Work-energy theorem, then divide by time.
  2. a = 20 / 8.0 = 2.5 m/s², F = m a = 2500 N. Constant force.
  3. At the end: P = F v = 2500 × 20 = 50 000 W, twice the average. v grows steadily from 0, so the average speed (and average power) is half the final.

How fast can the motor lift? medium

A 1500 W motor lifts a 50 kg crate at constant speed. What is the fastest speed it can lift at?

Show solution
Show all steps as text
  1. Constant speed: F = m g = 50 × 9.8 = 490 N. Net force zero.
  2. v = P / F = 1500 / 490 = 3.1 m/s. P = F v rearranged.

A car climbing a hill AP

A 1200 kg car drives at a constant 25 m/s. Air drag and rolling friction together are 600 N. Find the engine power needed (a) on flat road and (b) up a 5.0° slope.

Show solution
Show all steps as text
  1. (a) Constant speed on flat road: driving force = 600 N. P = 600 × 25 = 15 000 W = 15 kW. The engine only replaces what drag takes out.
  2. (b) Uphill, gravity also pulls back along the road: m g sin 5.0° = 1200 × 9.8 × 0.0872 = 1025 N. Component of gravity along the slope.
  3. Driving force = 600 + 1025 = 1625 N, P = 1625 × 25 ≈ 40 600 W ≈ 41 kW. The extra 25.6 kW goes into Ug each second: m g (v sin 5°) = 1200 × 9.8 × 2.18 ≈ 25 600 W.

Practice

  1. Motor B does the same work as motor A in half the time. Compared with A, B's power is

    Worked answer

    P = W / t. Same W, half the t, so twice the P.

  2. An electricity bill lists 1 kilowatt-hour (kWh). How many joules is that?

    Worked answer

    1 kWh = 1000 J/s × 3600 s = 3 600 000 J. It is energy (power × time).

  3. A winch lifts a 10 kg bucket at a constant 0.50 m/s. Its power output is about

    Worked answer

    Constant speed: F = m g = 98 N. P = F v = 98 × 0.50 = 49 W.

  4. On a graph of power (vertical) against time (horizontal), what does the area under the line represent?

    Worked answer

    Area = P × t = (J/s) × s = J, which is energy.

  5. A cart starts from rest and is pushed by a constant net force on a frictionless track. Which describes the power delivered by the force?

    Worked answer

    P = F v. F is constant and v = a t grows steadily, so P grows as a straight line from 0.

  6. A truck drives at constant velocity on a level highway. Which is true about its engine?

    Worked answer

    The net power on the truck is zero (K constant), but the engine pushes forward against drag and friction with power F v, and that energy ends up as internal energy of the air and road.

  7. Short answer. Motor A lifts 200 kg up 12 m in 20 s. Motor B lifts 150 kg up 20 m in 25 s. Which has the larger average power? Show the numbers.

    Worked answer

    A: 200 × 9.8 × 12 / 20 = 23 520 / 20 = 1176 W. B: 150 × 9.8 × 20 / 25 = 29 400 / 25 = 1176 W. They are the same, about 1.2 kW, even though B does more work (B takes longer).

  8. Short answer (experimental design). Describe how you could measure your own power output running up a staircase, using only a tape measure, a stopwatch and a bathroom scale. Say which quantities you measure and how you calculate P. Name one source of error.

    Worked answer

    Measure your mass m with the scale (kg). Measure the height of one step and count the steps (or measure the total vertical rise h directly) with the tape. Time the run from bottom to top with the stopwatch, Δt; repeat 3 times and average.

    P = m g h / Δt. Use vertical height, not the slanted length of the stairs.

    Errors: reaction time of the stopwatch (small for a 5 s run); you may not start and end at the same speed; energy also goes into internal energy of your muscles, so this is the power going into Ug, less than the chemical power your body uses.

AP question types for this topic

Four short free response questions, one of each AP Physics 1 free response type, all about this topic. Write your answer first, then open the worked answer and score yourself with the points shown.

1. Mathematical Routines

Nadia, mass m, runs up a flight of stairs of vertical height h in time t at a steady pace.

(a) Derive an expression for her average power output against gravity in terms of m, g, h and t.

(b) Calculate it for m = 55 kg, h = 4.5 m and t = 6.0 s.

(c) Show that the same power comes from P = Fv, using her vertical speed.

Show worked answer and scoring
  1. (a) Work against gravity W = mgh; power = work ÷ time, so P = mgh/t. 1 point for W = mgh, 1 point for dividing by t.
  2. (b) mgh = 55 × 9.8 × 4.5 = 2425.5 J; P = 2425.5 ÷ 6.0 ≈ 404 W. 1 point.
  3. (c) Vertical speed v = 4.5 ÷ 6.0 = 0.75 m/s; force = weight = mg = 539 N; P = Fv = 539 × 0.75 ≈ 404 W, the same, because h/t is the speed. 1 point for the speed, 1 point for matching the power.

Total: 5 points.

2. Translation Between Representations

Tariq's toy motor pulls a 1.5 kg cart from rest along a level, frictionless track. The motor delivers a constant power of 6.0 W.

yt
A
yt
B
yt
C
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D

Graphs against time. A: a straight line rising from zero. B: a curve that rises steeply, then flattens. C: a flat line. D: a curve bending upward.

(a) Which graph shows the cart's kinetic energy against time? Justify using the meaning of power.

(b) Which graph shows its speed against time? Justify with an equation.

(c) Calculate the kinetic energy and the speed at t = 3.0 s, and describe how the kinetic energy graph changes if the power is doubled.

Show worked answer and scoring
  1. (a) A. Power is the rate of energy transfer, so with constant P the kinetic energy grows by the same amount each second: K = Pt, a straight line from zero. 1 point for A, 1 point for the reasoning.
  2. (b) B. ½mv2 = Pt, so v = √(2Pt/m); a square root rises steeply then flattens. 1 point.
  3. (c) K = 6.0 × 3.0 = 18 J; v = √(2 × 18 ÷ 1.5) = √24 ≈ 4.9 m/s. Doubling the power doubles the slope of the straight line. 1 point for 18 J and 4.9 m/s, 1 point for the doubled slope.

Total: 5 points.

3. Experimental Design and Analysis

Elif wants to find the output power of a small electric winch motor. She has the motor, a string, slotted masses, a metre stick and a stopwatch. The motor lifts each load at a steady speed through a height of 1.20 m.

(a) Describe how she should collect her data.

(b) Assuming the power is the same for every load, derive what she should plot to get a straight line, and what the slope means.

(c) Use her results to find the power.

m (kg)0.100.200.300.40
t (s)0.400.771.191.56

Show worked answer and scoring
  1. (a) Mark 1.20 m with the metre stick. Hang a load, run the motor and time the lift between the marks once the speed is steady. Repeat three times for each of several loads and average. 1 point for the marked height and timing, 1 point for repeats over several loads.
  2. (b) Pt = mgh, so t = (gh/P) m. Plot t against m: slope = gh/P, so P = gh ÷ slope. 1 point for the equation, 1 point for the graph and slope meaning.
  3. (c) Best-fit slope ≈ 3.9 s/kg (line passes almost through the origin). P = 9.8 × 1.20 ÷ 3.9 ≈ 3.0 W. 1 point for the slope, 1 point for 3.0 W.

Total: 6 points.

4. Qualitative/Quantitative Translation

Two elevators each lift an 800 kg load 12 m up at a steady speed. The old one takes 20 s; the new one takes 10 s. Kofi claims: “The new elevator does twice as much work.”

(a) Explain in words whether Kofi is right, and what really is twice as big.

(b) Derive expressions for the work and the power of each elevator, and calculate them.

(c) Connect your expressions in (b) to your reasoning in (a).

Show worked answer and scoring
  1. (a) No. Both raise the same load through the same height, so both transfer the same energy: the same work. The new one does it in half the time, so its power (rate of doing work) is twice as big. 1 point for same work, 1 point for twice the power.
  2. (b) W = mgh = 800 × 9.8 × 12 = 94 080 J ≈ 94 kJ for both. P = W/t: old 94 080 ÷ 20 ≈ 4.7 kW, new 94 080 ÷ 10 ≈ 9.4 kW. 1 point for the work, 1 point for both powers.
  3. (c) Time does not appear in W = mgh, so work cannot depend on how fast the lift is; t is only in the denominator of P = W/t, so halving t doubles P, exactly the words of (a). 1 point.

Total: 5 points.

Common mistakes

The mistake: mixing up power and energy ("a 100 W bulb uses 100 W of energy").

Why it is wrong: watts measure a rate, joules per second. Energy is power × time.

How to spot it: check the units of your answer: J for energy, W for power.

The mistake: treating kWh as a unit of power.

Why it is wrong: kW × h is power × time, which is energy: 1 kWh = 3.6 × 10⁶ J.

How to spot it: "per hour" would be a rate; "times hours" is an amount.

The mistake: "Constant speed means the motor's power is zero."

Why it is wrong: the net work is zero, but the motor still pushes along the motion: P = F v ≠ 0. Other forces (gravity, drag) take the energy away as fast as the motor supplies it.

How to spot it: ask "where does the motor's energy go?" (into Ug, or thermal energy).

The mistake: using the final speed in P = F v to get the average power.

Why it is wrong: P = F v is the power at one instant. For the average use W / Δt (or F times the average speed when F is constant).

How to spot it: if the speed changes, your "average" from the final speed is too big.