DP Physics SL A.3 Work, Energy and Power

DP Physics SL A.3 Work, Energy and Power - Full notes

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DP Physics SL | A.3 Work, Energy and Power

DP PHYSICS · THEME A: SPACE, TIME AND MOTION

A.3

Work, Energy and Power

Full notes for SL · IB-style

Contents

1	Energy and work
2	Mechanical energy
3	Dissipation and the work–energy theorem
4	Power and efficiency
5	Energy resources
6	Exam technique and mixed problems
7	Investigating energy
8	Common misconceptions and links
9	Exam-style question with markscheme
10	Check your understanding
11	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
calculate work done by a force, including at an angle and from force–displacement graphs
calculate kinetic, gravitational and elastic potential energy
apply conservation of energy, including with resistive forces and the work–energy theorem
calculate power and efficiency and interpret Sankey diagrams
compare energy sources using specific energy and energy density
calculate power available from the wind
describe experiments on power and energy storage

There is no additional HL content in this subtopic. SL and HL students study the same material.

Guiding questions

How are concepts of work, energy and power used to predict changes within a system? How can a consideration of energetics be used as a method to solve problems in kinematics? How can transfer of energy be used to do work?

How this topic is examined

Energy questions appear on every paper, often combined with kinematics and momentum. Expect energy conservation, work against friction, power and efficiency calculations, Sankey diagrams and energy resources.

State the principle you use (conservation of energy, work–energy theorem) and show each energy term clearly.

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Energy and work

Energy is never lost, only transferred.

Think about it

When a bouncing ball rises a little lower each time, where has its energy gone?

Energy methods often solve mechanics problems faster than forces and suvat, because energy is a scalar and does not depend on the path taken. This section defines energy and work and introduces the most important idea in physics: conservation of energy.

1.1 Energy stores and transfers

Energy is a scalar measured in joules (J). It is stored in various forms and transferred by working (forces moving through a distance), heating, electrical work and radiation. The principle of conservation of energy states that energy cannot be created or destroyed; the total energy of a closed system is constant.

Energy store	Example	Formula (where used)
kinetic	moving car	½mv²
gravitational potential	raised book	mgΔh
elastic potential	stretched spring	½kΔx²
chemical	fuel, food, batteries	specific energy × mass
internal (thermal)	hot water	mcΔT
nuclear	uranium nuclei	Δmc²

Quick check

1. Describe the energy transfers for a pendulum released from rest.

Answers: 1. gravitational potential to kinetic and back; gradually to internal energy through air resistance and friction

1.2 Work done by a force

Work is done when a force moves its point of application. W = Fs cos θ, where θ is the angle between the force and the displacement. Work done equals energy transferred. Work is a scalar; 1 J = 1 N m.

Figure 1. Only the component of F along the displacement (F cos θ) does work.

WORKED EXAMPLE 1 Angled pull (standard)

A sledge is pulled 50 m by a rope with tension 80 N at 25° to the horizontal. Calculate the work done.

Solution

W = 80 × 50 × cos 25°M1

= 3.6 × 10³ JA1

WORKED EXAMPLE 2 Negative work (standard)

Friction of 30 N acts on a box sliding 4.0 m. Find the work done by friction.

Solution

θ = 180°: W = 30 × 4.0 × cos 180°M1

= −120 J (energy removed from the box)A1

1.3 Positive and negative work

Work is positive when the force has a component in the direction of motion (energy is transferred to the object) and negative when the component is opposite to the motion (energy is removed). Gravity does positive work on a falling ball and negative work on a rising one; friction always does negative work on a sliding object.

WORKED EXAMPLE 3 Lifting and lowering (standard)

A 2.0 kg box is lifted 1.5 m at constant speed and then lowered 1.5 m at constant speed. Find the work done by gravity in each stage and in total.

Solution

Lifting: W = −mgh = −29.4 J; lowering: +29.4 JA1

Total work by gravity = 0 (same start and end height)R1

Quick check

1. What is the sign of the work done by air resistance on a falling stone?

Answers: 1. negative

1.4 When is no work done?

No work is done when there is no displacement, or when the force is perpendicular to the displacement. A force can act without doing work.

Figure 2. Three situations in which a force does no work on the object.

Common mistake

Saying a person holding a heavy bag still does work on it. No mechanical work is done on the bag, although the person's muscles use chemical energy.

1.5 Work from force–displacement graphs

When the force varies, the work done is the area under the force–displacement graph. For a spring obeying Hooke's law the graph is a straight line through the origin, so the area is a triangle: W = ½Fx = ½kx².

Figure 3. Force–extension graph for a spring. The shaded triangle is the work done stretching it.

WORKED EXAMPLE 4 Spring (standard)

Using the graph, calculate the work done stretching the spring by 0.10 m.

Solution

Area = ½ × 0.10 × 20M1

= 1.0 JA1

WORKED EXAMPLE 5 Variable force (exam level)

A force rises uniformly from 0 to 40 N over 2.0 m, then stays at 40 N for a further 3.0 m. Find the total work done.

Solution

½(2.0)(40) + 3.0(40)M1

= 160 JA1

2 Mechanical energy

Kinetic, gravitational and elastic energy.

2.1 Kinetic energy

Ek = ½mv². Doubling the speed quadruples the kinetic energy, which is why stopping distances increase so rapidly with speed. Kinetic energy can also be written in terms of momentum: Ek = p²/2m.

Figure 4. Kinetic energy of a 1200 kg car against speed: a parabola.

WORKED EXAMPLE 6 Car (basic)

Calculate the kinetic energy of a 1200 kg car at 25 m s⁻¹.

Solution

½ × 1200 × 25²M1

= 3.75 × 10⁵ JA1

WORKED EXAMPLE 7 Momentum form (exam level)

Two objects have the same momentum. Object A has twice the mass of B. Compare their kinetic energies.

Solution

Ek = p²/2mM1

B has twice the kinetic energy of AA1

2.2 Gravitational potential energy

Near Earth's surface, the change in gravitational potential energy is ΔEp = mgΔh. Only changes in height matter, so any convenient zero level can be chosen. It is the work done against gravity raising the object.

WORKED EXAMPLE 8 Stairs (standard)

A 60 kg student climbs 3.5 m of stairs. Calculate the gain in gravitational potential energy.

Solution

60 × 9.81 × 3.5M1

= 2.06 × 10³ JA1

Quick check

1. Does the path taken up a hill affect the gain in gravitational potential energy?

Answers: 1. no, only the change in height matters

2.3 Elastic potential energy

For a spring obeying Hooke's law, the energy stored is EH = ½kΔx², equal to the area under the force–extension graph.

Figure 5. A compressed spring stores elastic potential energy that becomes kinetic energy of the block.

WORKED EXAMPLE 9 Launcher (exam level)

A spring of k = 400 N m⁻¹ is compressed 0.050 m and launches a 0.020 kg ball on a smooth surface. Find the launch speed.

Solution

EH = ½(400)(0.050)² = 0.50 JM1

½(0.020)v² = 0.50: v = 7.1 m s⁻¹A1

2.4 Conservation of mechanical energy

When only gravity and elastic forces do work (no friction or drag), mechanical energy Ek + Ep + EH is conserved. Write 'loss of one store = gain of another'. Mass often cancels, so the speed at the bottom of a frictionless slope is independent of mass.

Figure 6. A roller coaster track. The speed at each point depends only on the height dropped from A.

Figure 7. Energy bar chart for the coaster (no friction): the total stays the same.

WORKED EXAMPLE 10 Roller coaster (standard)

The coaster starts from rest at 32 m. Find its speed at B (height 0) and at C (height 12 m), ignoring friction.

Solution

B: v = √(2 × 9.81 × 32) = 25.1 m s⁻¹M1 A1

C: v = √(2 × 9.81 × 20) = 19.8 m s⁻¹A1

WORKED EXAMPLE 11 Pendulum (standard)

A pendulum bob is released 0.20 m above its lowest point. Find its maximum speed.

Solution

v = √(2 × 9.81 × 0.20)M1

= 1.98 m s⁻¹A1

2.5 Energy changes in a bungee jump

A bungee jumper first falls freely (gravitational potential energy to kinetic energy). Once the cord becomes taut, elastic potential energy builds up; the jumper slows and stops at the lowest point, where kinetic energy is zero and the cord stores the energy lost from the gravitational store.

Figure 8. Energy of a 70 kg bungee jumper (cord natural length 20 m, k = 120 N m⁻¹) against distance fallen.

WORKED EXAMPLE 12 Lowest point (exam level)

Using the data in the caption, find the distance fallen when the jumper momentarily stops.

Solution

mg(20 + x) = ½kx²: 686.7(20 + x) = 60x²M1

x = 21.9 m; distance fallen = 41.9 mA1

WORKED EXAMPLE 13 Maximum speed (exam level)

Explain where the jumper's speed is greatest.

Solution

Speed increases while weight > tensionR1

Maximum where kx = mg: x = 5.7 m beyond 20 m, i.e. about 25.7 m fallenA1

2.6 Energy–position graphs for a pendulum

For a pendulum, gravitational potential energy is greatest at the ends of the swing and least at the centre, while kinetic energy does the opposite. Without friction, their sum is constant; with damping, the total slowly falls.

Figure 9. Energy against displacement for an undamped pendulum.

Quick check

1. Where is the speed of the pendulum greatest?

Answers: 1. at the centre (lowest point)

2.7 Loop-the-loop

Combining energy conservation with circular motion: at the top of a vertical loop of radius r, the minimum speed to stay on the track is √(gr), when the normal reaction is zero. Energy conservation then gives the minimum release height h = 2.5r (ignoring friction and rolling).

Figure 10. A marble released from height h travels around a vertical loop.

WORKED EXAMPLE 14 Minimum height (exam level)

Show that the minimum release height for a loop of radius r is 2.5r.

Solution

At top: mg = mv²/r, so v² = grM1

mgh = mg(2r) + ½m(gr)M1

h = 2.5rAG

Quick check

1. Find the minimum release height for a loop of radius 0.40 m.

Answers: 1. 1.0 m

2.8 Energy and projectiles

Energy methods give the speed of a projectile at any height without needing its direction: ½mv² = ½mu² − mgΔh.

WORKED EXAMPLE 15 Thrown ball (exam level)

A ball is thrown at 15 m s⁻¹ from a 20 m cliff at any angle. Find its speed when it hits the sea (no air resistance).

Solution

v² = 15² + 2(9.81)(20)M1

v = 24.8 m s⁻¹ (independent of the angle)A1

Quick check

1. Why is the landing speed independent of the throwing angle?

Answers: 1. energy is a scalar; only the height change matters

3 Dissipation and the work–energy theorem

Real systems lose useful energy.

3.1 Work done against friction and drag

When resistive forces act, work done against them transfers energy into internal energy of the surroundings (dissipation). Energy equation: initial energy = final energy + work done against resistance (F × distance along the path).

WORKED EXAMPLE 16 Skier (exam level)

A 50 kg skier starts from rest and descends 40 m vertically along a 120 m slope against an average friction of 60 N. Find her speed at the bottom.

Solution

Loss of Ep = 50 × 9.81 × 40 = 19 620 JA1

Work against friction = 60 × 120 = 7200 JA1

½(50)v² = 12 420: v = 22.3 m s⁻¹A1

Common mistake

Multiplying the friction force by the vertical height. Friction acts along the path, so use the path length.

3.2 The work–energy theorem

The net work done on an object equals its change in kinetic energy: Wnet = ΔEk. This links forces and energy without needing time, and is especially useful for braking problems.

WORKED EXAMPLE 17 Braking (exam level)

A 1000 kg car slows from 20 m s⁻¹ to rest in 40 m. Find the average braking force.

Solution

ΔEk = ½(1000)(20)² = 2.0 × 10⁵ JM1

F = 2.0 × 10⁵/40 = 5.0 × 10³ NA1

WORKED EXAMPLE 18 Doubling speed (exam level)

Show that doubling the speed of a car quadruples its braking distance for the same braking force.

Solution

Fs = ½mv², so s ∝ v²R1

Doubling v multiplies s by 4A1

3.3 Bouncing balls

A dropped ball converts gravitational potential energy to kinetic energy, then briefly to elastic potential energy on impact, and back. Each bounce dissipates some energy, so each rebound is lower; the ratio of rebound height to drop height is the fraction of energy kept.

Figure 11. Height of a ball bouncing on a hard floor: each bounce keeps 70% of the energy.

WORKED EXAMPLE 19 Rebound (exam level)

A 0.058 kg tennis ball dropped from 2.0 m rebounds to 1.4 m. Calculate the energy lost and the percentage lost.

Solution

ΔE = 0.058 × 9.81 × 0.60 = 0.34 JM1 A1

Percentage = 0.60/2.0 = 30%A1

3.4 Stopping distances

Stopping distance = thinking distance + braking distance. Thinking distance (speed × reaction time) is proportional to speed; braking distance (from Fs = ½mv²) is proportional to speed squared. Tiredness, alcohol and phone use increase reaction time; wet or icy roads reduce the braking force.

Figure 12. Thinking, braking and total stopping distances against speed.

WORKED EXAMPLE 20 Stopping distance (exam level)

A car at 30 m s⁻¹ has a driver reaction time of 0.70 s and decelerates at 6.0 m s⁻². Calculate the stopping distance.

Solution

Thinking: 30 × 0.70 = 21 m; braking: 30²/(2 × 6.0) = 75 mM1 A1

Total = 96 mA1

3.5 Energy in collisions

Use momentum conservation to find velocities after a collision, then energy calculations for kinetic energy changes. In elastic collisions both are conserved; in inelastic collisions kinetic energy becomes internal energy and sound; in explosions kinetic energy increases.

WORKED EXAMPLE 21 Lost energy (standard)

A 2.0 kg mass at 6.0 m s⁻¹ sticks to a 1.0 kg mass at rest. Find the kinetic energy lost.

Solution

v = 12/3.0 = 4.0 m s⁻¹M1

36 − 24 = 12 J lostA1

3.6 Energy transfers on slopes

The change in gravitational potential energy depends only on vertical height. A longer, gentler ramp needs a smaller force, but on a smooth ramp the work done is the same: force × length = mgh.

WORKED EXAMPLE 22 Ramp (standard)

A 50 kg crate is pushed up a smooth 4.0 m ramp onto a 1.0 m high platform. Find the minimum force along the ramp and the work done.

Solution

Work = mgh = 50 × 9.81 × 1.0 = 491 JA1

F = 491/4.0 = 123 NA1

4 Power and efficiency

How fast energy is transferred, and how much is useful.

Think about it

Two cranes lift the same load to the same height, one in half the time. Which has done more work? Which is more powerful?

4.1 Power

Power is the rate of transfer of energy (or rate of doing work): P = ΔE/Δt = W/t, measured in watts (1 W = 1 J s⁻¹). For a constant force moving an object at constant velocity, P = Fv.

WORKED EXAMPLE 23 Crane (standard)

A crane lifts 500 kg through 20 m in 25 s. Calculate the useful power.

Solution

P = mgh/t = 500 × 9.81 × 20/25M1

= 3.9 kWA1

WORKED EXAMPLE 24 Car at constant speed (standard)

A car travels at a constant 30 m s⁻¹ against total resistance of 900 N. Calculate the output power.

Solution

P = Fv = 900 × 30M1

= 27 kWA1

Quick check

1. Convert 1 kWh into joules.

Answers: 1. 3.6 × 10⁶ J

4.2 Maximum speed of vehicles

A vehicle reaches top speed when its driving force equals total resistance. Since drag increases roughly with v², the power needed rises roughly with v³.

Figure 13. Power needed to overcome resistance rises steeply with speed; top speed is where it meets the maximum power.

WORKED EXAMPLE 25 Top speed (exam level)

A car's engine has a maximum useful power of 60 kW. At top speed the total resistance is 1500 N. Calculate the top speed.

Solution

v = P/F = 60 000/1500M1

= 40 m s⁻¹A1

4.3 Efficiency

Efficiency = useful energy output/total energy input = useful power output/total power input. Real machines are always less than 100% efficient because some energy is dissipated.

WORKED EXAMPLE 26 Motor (standard)

An electric motor takes 600 W and lifts a 20 kg load at 2.0 m s⁻¹. Calculate its efficiency.

Solution

Useful power = 20 × 9.81 × 2.0 = 392 WM1

Efficiency = 392/600 = 65%A1

WORKED EXAMPLE 27 Pump (exam level)

A pump raises 300 kg of water per minute through 12 m with 60% efficiency. Calculate the input power.

Solution

Useful power = 300 × 9.81 × 12/60 = 589 WM1 A1

Input = 589/0.60 = 980 WA1

In the real world

Large electric motors can exceed 95% efficiency, while a typical petrol car engine converts only about 25–30% of the fuel energy into useful work.

4.4 Power and energy graphs

The gradient of an energy–time graph is power; the area under a power–time graph is energy transferred.

Figure 14. Power delivered by a motor during a lift journey. The area is the energy used.

WORKED EXAMPLE 28 Area (standard)

Find the total energy delivered by the motor in the figure.

Solution

½(10)(20) + 20(20) + ½(10)(20) kJM1

= 600 kJA1

4.5 Power in sport

Athletes' power outputs can be estimated from energy changes. A sprinter accelerating from rest to 10 m s⁻¹ in 2.0 s gains kinetic energy rapidly; a cyclist at steady speed works against air resistance, which dominates at high speeds.

WORKED EXAMPLE 29 Sprinter (standard)

A 75 kg sprinter reaches 10 m s⁻¹ in 2.0 s from rest. Estimate the average useful power.

Solution

Ek = ½(75)(10)² = 3750 JM1

P = 3750/2.0 = 1.9 kWA1

WORKED EXAMPLE 30 Cyclist (exam level)

A cyclist at a steady 12 m s⁻¹ experiences drag of 25 N and rolling resistance of 5 N. Calculate the power output.

Solution

P = (25 + 5) × 12M1

= 360 WA1

4.6 Efficiency of energy chains

When energy passes through several devices in turn, the overall efficiency is the product of the individual efficiencies. Losses at each stage add up quickly.

WORKED EXAMPLE 31 Power chain (exam level)

A power station (40%), transmission (92%) and an electric motor (85%) are used in series. Find the overall efficiency.

Solution

0.40 × 0.92 × 0.85M1

= 31%A1

Quick check

1. Why is overall efficiency always less than the smallest single efficiency?

Answers: 1. multiplying by numbers less than 1 makes the product smaller

4.7 Sankey diagrams

A Sankey diagram shows energy transfers with arrows whose widths are proportional to the energy. The input arrow splits into useful and wasted outputs.

Figure 15. Sankey diagram for a power station with 38% efficiency.

WORKED EXAMPLE 32 Reading (standard)

Use the diagram to find the efficiency of the power station.

Solution

380/1000M1

= 38%; 620 MJ wastedA1

5 Energy resources

Where our energy comes from.

5.1 Specific energy and energy density

Specific energy is the energy released per unit mass (J kg⁻¹); energy density is energy per unit volume (J m⁻³). High specific energy matters where mass is critical (aircraft); high energy density where volume is limited (cars).

Fuel	Specific energy / MJ kg⁻¹	Energy density / MJ dm⁻³
wood	16	about 10
coal	24	about 30
petrol	46	34
natural gas (at 1 atm)	55	0.036
lithium-ion battery	0.5–0.9	1–2.5
uranium-235 (fission)	8 × 10⁷	1.5 × 10⁹

Figure 16. Specific energy and energy density of some fuels.

WORKED EXAMPLE 33 Fuel mass (standard)

How much petrol must burn to release 230 MJ?

Solution

230/46M1

= 5.0 kgA1

5.2 Renewable and non-renewable sources

Non-renewable sources (coal, oil, gas, nuclear fuel) are finite; renewable sources (solar, wind, hydro, tidal, geothermal, biomass) are replenished naturally. Choices balance cost, reliability, emissions, land use and environmental impact.

Source	Advantages	Disadvantages
fossil fuels	reliable, high energy density	CO₂ emissions, finite
nuclear	low CO₂, reliable	radioactive waste, high cost
wind	no emissions in use	intermittent, visual impact
solar	no emissions, falling cost	intermittent, needs large area
hydro	reliable, storage possible	floods land, ecosystem damage

5.3 Power from the wind

The kinetic energy per second of wind passing through area A at speed v is P = ½ρAv³. Doubling the wind speed increases the available power eightfold. Real turbines extract at most about 59% (the Betz limit).

Figure 17. Air passing through the swept area of a turbine carries kinetic energy.

WORKED EXAMPLE 34 Turbine (exam level)

A turbine has blades of length 40 m. Air density is 1.2 kg m⁻³ and wind speed 10 m s⁻¹. Calculate the power in the wind and the electrical output if the turbine is 35% efficient.

Solution

A = π(40)² = 5030 m²; P = ½(1.2)(5030)(1000) = 3.0 MWM1 A1

Output = 0.35 × 3.0 = 1.1 MWA1

5.4 Hydroelectric and pumped storage

Water stored behind a dam at height h has gravitational potential energy. As it falls through turbines, power P = (Δm/Δt)gh × efficiency. Pumped storage schemes pump water uphill when demand is low and release it when demand is high, acting as giant batteries.

Figure 18. A hydroelectric scheme: water falls through height h to drive a turbine.

WORKED EXAMPLE 35 Hydro power (exam level)

Water flows at 50 m³ s⁻¹ through turbines 120 m below the reservoir surface; efficiency is 85%. Calculate the electrical power output.

Solution

Δm/Δt = 5.0 × 10⁴ kg s⁻¹; P = 5.0 × 10⁴ × 9.81 × 120M1

× 0.85 = 50 MWA1

5.5 Solar power

Solar panels convert a fraction of the incident solar intensity into electrical power: P = IAη. Typical panel efficiencies are 18–22%, and the average intensity at the surface depends on latitude, season and weather.

WORKED EXAMPLE 36 Rooftop panels (exam level)

Panels of total area 20 m² receive an average 200 W m⁻² and are 20% efficient. Calculate the energy produced per day in kWh.

Solution

P = 200 × 20 × 0.20 = 800 WM1

E = 0.80 kW × 24 h = 19 kWhA1

In the real world

Solar photovoltaic electricity has become one of the cheapest sources of new electricity generation in many countries, with costs falling about 90% between 2010 and 2020.

5.6 Energy and society

Improving efficiency (insulation, LED lighting, efficient motors) reduces energy demand and emissions. Energy calculations are central to evaluating transport, buildings and power generation.

WORKED EXAMPLE 37 LED saving (standard)

A 60 W bulb is replaced by an 8 W LED used 4 hours a day. Calculate the energy saved in a year.

Solution

52 W × 4 h × 365M1

= 76 kWhA1

6 Exam technique and mixed problems

Choosing the right method.

6.1 Energy or forces?

Use energy methods when the question involves heights, speeds and distances but not time or direction: they are quicker and work for curved paths. Use forces and suvat when time, acceleration or direction are needed. For collisions, use momentum first, then energy.

Question asks for	Best method
speed after falling or sliding	energy conservation
time taken	forces + suvat (or kinematics)
average resistive force over a distance	work–energy theorem
velocity after a collision	momentum conservation
power of an engine at steady speed	P = Fv

Key point

Energy is a scalar: no components, no directions. That is why energy methods are so often the quickest route.

6.2 Multi-step problem: ski jump

Long questions often chain several ideas. Work in stages and label each energy store.

WORKED EXAMPLE 38 Ski jump (exam level)

A 60 kg skier starts from rest at the top of a 45 m high ramp. Friction does 6.0 kJ of work against her on the ramp. She leaves the end of the ramp horizontally, 5.0 m above the landing slope. Find (a) her speed at the end of the ramp; (b) her speed on landing, ignoring air resistance.

Solution

(a) mgh = 60 × 9.81 × 40 = 23.5 kJ (ramp end 5.0 m up)M1

½(60)v² = 23.5 − 6.0 = 17.5 kJ: v = 24.2 m s⁻¹A1

(b) v² = 24.2² + 2(9.81)(5.0): v = 26.1 m s⁻¹M1 A1

6.3 Multi-step problem: lift motor

Power, efficiency and energy questions often combine.

WORKED EXAMPLE 39 Lift (exam level)

A 1200 kg lift (with passengers) rises 30 m at constant speed in 20 s. The motor is 70% efficient and the supply is 400 V. Find (a) the useful power; (b) the current drawn.

Solution

(a) P = mgh/t = 1200 × 9.81 × 30/20 = 17.7 kWM1 A1

(b) input = 17.7/0.70 = 25.2 kW; I = 25 200/400 = 63 AM1 A1

6.4 Multi-step problem: spring launcher

Springs, gravity and friction can all appear in one question.

WORKED EXAMPLE 40 Launcher on a slope (exam level)

A spring (k = 800 N m⁻¹) compressed by 0.10 m launches a 0.25 kg block up a slope. The block rises 1.2 m vertically before stopping. Find the energy dissipated by friction.

Solution

EH = ½(800)(0.10)² = 4.0 JM1

Gain in Ep = 0.25 × 9.81 × 1.2 = 2.94 JM1

Dissipated = 1.06 JA1

6.5 Common exam errors

Examiners repeatedly see the same slips in energy questions.

•Using height instead of path length for work against friction.
•Forgetting to square the speed in ½mv², or squaring the mass.
•Mixing kW and W, or kWh and J (1 kWh = 3.6 MJ).
•Quoting efficiency greater than 100% after dividing input by output.
•Using P = Fv when the speed is changing (it gives instantaneous power only).

7 Investigating energy

Practical work and data analysis.

7.1 Measuring the power of a person

Time a student running up stairs; measure the vertical height and the student's mass. Power = mgh/t. Repeat and average; the main uncertainty is the timing.

WORKED EXAMPLE 41 Stair climb (standard)

A 55 kg student climbs 4.2 m in 3.5 s. Calculate the power and its percentage uncertainty if the time is ±0.2 s.

Solution

P = 55 × 9.81 × 4.2/3.5 = 650 WA1

% uncertainty ≈ 0.2/3.5 = 6%A1

Try this

Measure your own power output running up a flight of stairs. Compare it with a 100 W light bulb and a 1.5 kW kettle.

7.2 Measuring the efficiency of a motor

A small electric motor lifts a mass through a measured height. The electrical energy input is IVt (from an ammeter, voltmeter and timer); the useful output is mgh. Efficiency = mgh/IVt. Repeat for different masses to see how efficiency varies with load.

WORKED EXAMPLE 42 Motor experiment (exam level)

A motor running at 6.0 V and 0.50 A lifts 0.20 kg through 0.80 m in 4.0 s. Find its efficiency.

Solution

Input = 6.0 × 0.50 × 4.0 = 12 J; output = 0.20 × 9.81 × 0.80 = 1.57 JM1

Efficiency = 13%A1

Quick check

1. Suggest two reasons for the low efficiency of a small motor.

Answers: 1. friction in bearings; heating of coils (I²R)

7.3 Investigating stopping distance and speed

Roll a trolley down a ramp from different heights onto a rough surface and measure the stopping distance. Since ½mv² = Fs and v² ∝ release height, the stopping distance should be proportional to the release height.

Figure 19. Stopping distance on carpet against release height: a straight line through the origin.

WORKED EXAMPLE 43 Friction from the gradient (exam level)

For a 0.50 kg trolley, the gradient of the graph is 2.4. Find the average friction force on the carpet.

Solution

mgh = Fs, so s/h = mg/FM1

F = 0.50 × 9.81/2.4 = 2.0 NA1

7.4 Energy stored in a stretched rubber band

Rubber bands do not obey Hooke's law, so the energy stored must be found from the area under the force–extension graph, for example by counting squares. The unloading curve lies below the loading curve; the area between them is energy dissipated (hysteresis).

Figure 20. Loading and unloading curves for a rubber band. The area between them is energy dissipated.

Quick check

1. Why does a rubber band warm up when stretched and released repeatedly?

Answers: 1. energy dissipated by hysteresis becomes internal energy

8 Common misconceptions and links

Students often think…	Actually…
work is done whenever a force acts	only when the force has a component along the displacement
energy is used up	energy is transferred to less useful stores (dissipated)
power is energy	power is the rate of energy transfer
efficiency can exceed 100%	it is always less than 100% in real machines
heavier objects reach the bottom of a frictionless slope faster	speed is independent of mass
friction work uses the height	it uses the distance along the path

How A.3 connects to the rest of the course

Topic	Connection
A.1 Kinematics	energy methods replace suvat
A.2 Forces and momentum	collisions and impulse
B.1 Thermal energy transfers	internal energy
B.5 Current and circuits	electrical power P = IV
E.4 Fission	nuclear energy resources

9 Exam-style question with markscheme

Try this under exam conditions (about 20 minutes) before reading the markscheme.

A 0.60 kg trolley is released from rest at the top of a ramp 0.80 m high. It travels 2.4 m along the ramp to the bottom, then compresses a spring of spring constant 500 N m⁻¹ on a horizontal surface.

(a) Calculate the speed at the bottom of the ramp, ignoring friction. [2]

(b) The measured speed is 3.5 m s⁻¹. Calculate the average frictional force on the ramp. [3]

(d) Calculate the average power dissipated by friction if the descent takes 1.3 s. [2]

(e) Suggest one reason why the spring's compression might be less than calculated. [1]

Markscheme

Part	Answer	Marks
(a)	v = √(2 × 9.81 × 0.80)	M1
	3.96 m s⁻¹	A1
(b)	energy lost = 4.71 − ½(0.60)(3.5)² = 4.71 − 3.68	M1
	= 1.03 J	A1
	F = 1.03/2.4 = 0.43 N	A1
(c)	½(500)x² = 3.68	M1
	x = 0.12 m	A1
(d)	1.03/1.3	M1
	0.79 W	A1
(e)	friction on the horizontal surface / energy dissipated in the spring	R1

Examiner's comments. In (b), many candidates forgot that friction does work along the slope length (2.4 m), not the vertical height.

10 Check your understanding

Section A: multiple choice

1. The unit of power is

A. J s⁻¹ B. J s C. N s D. N m s

2. Work done is zero when the force is

A. parallel to the displacement B. large C. perpendicular to the displacement D. constant

3. Doubling the speed of a car multiplies its kinetic energy by

A. 2 B. 4 C. 8 D. √2

4. The area under a force–extension graph gives

A. power B. momentum C. spring constant D. work done

5. A motor's input is 500 W and useful output 350 W. Its efficiency is

A. 30% B. 70% C. 143% D. 35%

6. Wind power available is proportional to

A. v³ B. v C. v² D. 1/v

7. A 1 kg mass falls 10 m. Its gain in kinetic energy (no drag) is about

A. 10 J B. 1000 J C. 50 J D. 100 J

8. Power for a constant force at constant velocity is

A. F/v B. Fv² C. Fv D. ½Fv

Section B: structured

9. A 45 kg child slides down a 3.0 m high slide and reaches 5.5 m s⁻¹. Calculate the energy dissipated. [3]

10. A lift motor raises a 800 kg lift at 2.5 m s⁻¹. The motor is 75% efficient. Calculate the electrical input power. [3]

11. Explain why the speed of a frictionless roller coaster at the bottom is independent of its mass. [2]

12. Compare petrol and batteries as energy stores for vehicles using specific energy. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. 1324 − 681 = 643 J.

10. Useful 800 × 9.81 × 2.5 = 19.6 kW; input 26.2 kW.

11. mgh = ½mv²; m cancels.

12. petrol 46 MJ kg⁻¹, batteries about 0.5–0.9 MJ kg⁻¹; batteries much heavier per unit energy, though electric motors are more efficient.

11 Summary and topic map

Idea	Remember
Work	W = Fs cos θ; area under F–s
Energy stores	½mv², mgΔh, ½kΔx²
Conservation	energy in = energy out + dissipated
Work–energy	Wnet = ΔEk
Power	P = E/t = Fv
Efficiency	useful/total; Sankey diagrams
Resources	specific energy, energy density, ½ρAv³

Before the exam, make sure you…

•write every energy term with its formula
•use path length for work against friction
•convert units (kW, kWh, MJ)
•use P = Fv for constant velocity problems

Work, Energy and Power on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 21. Topic map for Work, Energy and Power.

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