AP Physics 1 · Lesson 3

Power and energy lost to friction

Know Know it

Power is the rate of energy transfer: P=ΔE/ΔtP = \Delta E/\Delta t, or P=W/ΔtP = W/\Delta t for work. Its unit is the watt, 1 W = 1 J/s. The same work done in half the time takes twice the power.

For a force F pushing along an object’s motion at speed v, the power it delivers is P=FvP = Fv. At constant speed against friction or drag, the driving force equals the resisting force, so the power needed is the resisting force times the speed.

When an object slides with kinetic friction fkf_k through a distance d, the mechanical energy lost becomes thermal energy, ΔEth=fkd\Delta E_{\text{th}} = f_kd.

Two ways to find power
When you knowUseExample
The energy transferred and the timeP=ΔE/ΔtP = \Delta E/\Delta tA pump that gives water 6,000 J in 20 s delivers 300 W
A force along the motion and the speedP=FvP = FvA 200 N pull at a steady 1.5 m/s delivers 300 W

Worked example Easy

A 52 kg student climbs 4.5 m up a rope at a steady speed in 15 s. What average power does she deliver to raise herself against gravity?

  1. 16 W
  2. 150 WAnswer
  3. 510 W
  4. 2,300 W

How to solve it

  1. At steady speed, all her work goes into gravitational energy: mgh=(52)(9.8)(4.5)=2,293mgh = (52)(9.8)(4.5) = 2{,}293 J.
  2. P=ΔE/Δt=2293/15P = \Delta E/\Delta t = 2293/15.
  3. P ≈ 150 W.

Why each choice is right or wrong

  • A. Incorrect. 16 W leaves out g, so it uses mass times height instead of the gain in gravitational energy.
  • B. Correct. The energy gained is mgh=(52)(9.8)(4.5)=2,293mgh = (52)(9.8)(4.5) = 2{,}293 J, so P=2293/15≈150P = 2293/15 \approx 150 W.
  • C. Incorrect. 510 W is her weight in newtons, 510 N, not a power.
  • D. Incorrect. 2,300 W is the energy, 2,293 J, without dividing by the 15 s.

Trap Faster is not more work

Running up a flight of stairs and walking up it take the same work against gravity, mghmgh. Running takes more power, because the same energy is transferred in less time. Keep the joule (energy) and the watt (energy per second) apart.

Explain Explain it

Task verbs: calculate, compare

To compare two machines or two people, compare the work and the time separately. Equal work in less time means more power; more work in the same time also means more power.

With friction in a system, add the thermal energy to the energy budget: work done by outside forces equals the change in kinetic energy plus the change in potential energy plus the thermal energy produced.

Worked example Medium

A worker drags a 2.0 kg box across a level floor at a steady 0.60 m/s for 5.0 s. The coefficient of kinetic friction between the box and the floor is 0.35. How much thermal energy does friction produce in that time?

  1. 4.1 J
  2. 6.9 J
  3. 21 JAnswer
  4. 34 J

How to solve it

  1. Kinetic friction: fk=μkmg=(0.35)(2.0)(9.8)=6.86f_k = \mu_kmg = (0.35)(2.0)(9.8) = 6.86 N.
  2. Distance at a steady speed: d = (0.60)(5.0) = 3.0 m.
  3. ΔEth=fkd=(6.86)(3.0)≈21\Delta E_{\text{th}} = f_kd = (6.86)(3.0) \approx 21 J.

Why each choice is right or wrong

  • A. Incorrect. 4.1 J is the power of the pull, (6.86 N)(0.60 m/s) = 4.1 W, not the energy over 5.0 s.
  • B. Incorrect. 6.9 J is the friction force, 6.86 N, which still has to be multiplied by the 3.0 m distance.
  • C. Correct. Friction is μkmg=(0.35)(2.0)(9.8)=6.86\mu_kmg = (0.35)(2.0)(9.8) = 6.86 N, and the box slides (0.60)(5.0) = 3.0 m, so the thermal energy is (6.86)(3.0) ≈ 21 J.
  • D. Incorrect. 34 J multiplies the friction force by the time, 5.0 s, instead of the distance, 3.0 m.

Compare Your turn

Motor X raises a 20 kg crate 6.0 m in 12 s, and motor Y raises a 30 kg crate 4.0 m in 6.0 s, both at steady speeds. Use g = 9.8 m/s². Compare the work the two motors do on the crates and the power each one delivers.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Says the two motors do equal work on the crates
  • Gives a value from 95 to 100 W for motor X’s power
  • Gives a value from 190 to 200 W for motor Y’s power
  • Says motor Y delivers more power, about twice as much, because it does the same work in half the time

Model answer

Motor X does W=mgh=(20)(9.8)(6.0)=1,176W = mgh = (20)(9.8)(6.0) = 1{,}176 J of work and motor Y does (30)(9.8)(4.0)=1,176(30)(9.8)(4.0) = 1{,}176 J, so the two motors do equal work on the crates, about 1,200 J each. Motor X’s power is 1176/12 = 98 W, and motor Y’s is 1176/6.0 = 196 W. Motor Y delivers more power, twice as much, because it does the same work in half the time.

Apply Apply it

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