IB Physics HL · guided topic map
Work, energy and power for IB Physics HL
Work, energy and power for IB Physics HL, organized into 1 syllabus topic and 5 mapped concept guides.
- Syllabus topics
- 1
- Mapped concept guides
- 5
- Educational level
- IB Diploma Physics Higher Level
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A.3Work, energy and power
Space, time and motion
5 guides+
Work, energy and power
Space, time and motion
- 01Work done and signed energy transferSL + HL
- 02Energy conversion and conservationSL + HL
- 03Kinetic energySL + HL
- 04Gravitational potential energySL + HL
- 05Power and efficiencySL + HL
Diagrams
Work, energy and power as IB Physics HL draws it
The figures from the IB Physics HL practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.
01Figure 4Work, energy and powerIB
Figure comment
Figure 4Side view, drawn not to scale. A road climbs to the right from level ground, and the angle between the road and a dashed horizontal line drawn from the foot of the slope is marked 4.0°. The cyclist and bicycle are drawn as one wheeled body on the road, labelled total mass 78 kg. An arrow from the front of the body points up the slope and is labelled 5.5 m s⁻¹; a second arrow from the rear points down the slope and is labelled 25 N.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Guided questions 5 parts
Reading cue. The 25 N arrow points down the slope, so it is the resistance and not the drive; the 4.0° is measured from the dashed horizontal, so the weight component along the road uses sin, not cos.
aState State the resultant force acting on the cyclist and bicycle while they travel up the slope at the speed shown, and give a reason.
Check answer 2 marks
- zero
- the velocity is constant, so by Newton's first law the forward force, the weight component and the 25 N resistance balance
bDetermine Determine the gravitational potential energy gained by the cyclist and bicycle during one minute of riding at the speed shown.
Check answer 3 marks
- distance along the road = 5.5 × 60 = 330 m
- vertical rise = 330 × sin 4.0° = 23.0 m
- ΔE_p = 78 × 9.81 × 23.0 = 1.76 × 10⁴ J
cDetermine Determine the work done against the resistive force over the same minute, and hence determine the percentage of the cyclist's useful work that goes into raising her and the bicycle.
Check answer 3 marks
- work against resistance = 25 × 330 = 8.25 × 10³ J
- total useful work = 1.76 × 10⁴ + 0.825 × 10⁴ = 2.59 × 10⁴ J
- fraction raising the cyclist = 1.76 × 10⁴ / 2.59 × 10⁴ = 68%
dDiscuss Further on, the road steepens to 8.0° and the cyclist holds the same useful power output of 431 W. Discuss how her steady speed changes, supporting your answer with a calculation, and whether the resistive force would still be 25 N.
Check answer 4 marks
- component of weight down an 8.0° slope = 78 × 9.81 × sin 8.0° = 106 N
- forward force needed = 106 + 25 = 131 N, so v = 431 / 131 = 3.3 m s⁻¹
- the force needed has risen from 78.4 N only to 131 N, a factor of 1.7, so the speed falls by 1.7 and not by the factor of 2 the doubled angle might suggest — the 25 N resistance is unchanged while only the weight component doubles
- air resistance is part of the 25 N and falls as the speed falls, so the true steady speed is a little above 3.3 m s⁻¹ and this figure is a lower bound
Transfer challenge
A lift of total mass 850 kg is raised vertically at a steady 1.2 m s⁻¹. Frictional forces on the lift total 400 N. Determine the useful power output of the motor.
Check answer 4 marks
- the motion is now vertical, so the whole weight opposes it: 850 × 9.81 = 8.34 × 10³ N
- total upward force required = 8.34 × 10³ + 400 = 8.74 × 10³ N
- P = Fv = 8.74 × 10³ × 1.2 = 1.0 × 10⁴ W
- notes that the sine factor of the cyclist's slope has become 1, which is why so much more power is needed at a lower speed