As Physics Topic 5: Work,Energy & Power

Complete Work, Energy and Power Notes - A-Level Physics

This comprehensive guide covers the A-Level Physics topic of Work, Energy and Power from foundational concepts to advanced applications, including all key concepts, derivations, formulas, and exam-style examples.

Foundation Concepts: These notes build from basic principles, making them accessible to all students regardless of their previous physics background.

Fundamental Energy Concepts

What is Energy?

Energy is defined as the stored ability to do work. It is a scalar quantity measured in joules (J). Energy can be transferred from one object to another or transformed from one form to another.

Basic Forms of Energy

Kinetic Energy (KE) - Energy due to motion
Examples: Moving car, flowing water, wind

Potential Energy (PE) - Stored energy due to position or configuration
Examples: Water in a raised reservoir, stretched spring, chemical bonds

Energy Transfer Processes

Mechanical Work: Force moving through distance
Heating: Energy transfer due to temperature difference
Electrical Work: Charges moving through potential difference
Radiation: Energy transfer by electromagnetic waves

5.1 Energy Conservation

Syllabus Coverage: Understand the concept of work, energy conservation, efficiency, and power.

1. Work Done

Work is defined as the product of a force and the distance moved in the direction of the force. It is a scalar quantity measured in joules (J), even though direction is mentioned in its definition. This is a case where two vectors (force and displacement) are multiplied to give a scalar product.

W = Fs cosθ

Where:

W = Work done (Joules, J)
F = Force applied (Newtons, N)
s = Displacement (metres, m)
θ = Angle between the force vector and the displacement vector

The Joule: The unit of work, the joule, is defined as the work done when a force of one newton moves its point of application a distance of one metre in the direction of the force.

Key Cases:

Force and displacement in same direction (θ = 0°): W = Fs (Maximum work)
Force perpendicular to displacement (θ = 90°): W = 0 (e.g., gravity does no work on horizontal motion)
Force opposes displacement (θ = 180°): W = -Fs (Negative work, e.g., friction)

Example 1: Work Done by a Crane

A crane lifts a boat of weight 5000 N from a height of 0.20 m above the water to a height of 3.70 m above the water at a constant speed. What work is done by the crane?

Solution:

Height difference = 3.70 - 0.20 = 3.50 m

Work done = Force × Distance = 5000 N × 3.50 m = 17,500 J

Example 2: Work on an Inclined Surface

An escalator lifts a family of total weight 2200 N a distance of 8.0 m at an angle of 37° to the horizontal. What work is done by the escalator on the family?

Solution:

Vertical height = 8.0 × sin 37° = 4.81 m

Work done = Force × Vertical Distance = 2200 N × 4.81 m = 10,600 J

Note: The step pushes vertically upwards, so only the vertical displacement matters for work against gravity.

2. Principle of Conservation of Energy

Conservation of Energy: Energy cannot be created or destroyed; it can only be transferred from one form to another. The total energy in a closed system remains constant.

This principle was established through the careful experiments of James Prescott Joule in the 19th century, who showed that different forms of energy could be converted into one another with no net loss of energy.

Real-World Application: Motorway Design

Motorway designers try to make slip roads upwards for getting off and downwards for getting on. This design uses gravitational potential energy efficiently:

Upward exit slip roads: Vehicles use kinetic energy to gain height
Downward entrance slip roads: Vehicles use height to gain kinetic energy

This natural slope design saves fuel and improves efficiency.

3. & 4. Efficiency

In real-world systems, energy transfers are never 100% efficient due to energy losses (typically as heat, sound, or other non-useful forms).

Efficiency = (Useful Energy Output / Total Energy Input) × 100%

Or in terms of power:

Efficiency = (Useful Power Output / Total Power Input) × 100%

Sankey Diagrams: These are flow diagrams where arrow width represents energy/power flow. They visually show energy inputs, useful outputs, and wasted energy. While not directly specified in the syllabus, they help understand energy losses in practical devices.

Typical Efficiencies:

Well-tuned diesel engines: ~55%
Well-tuned petrol engines: ~40%
Electric motors: 70-95%
These limitations come from thermodynamics - it's difficult to convert disorganized energy (random molecular motion) into organized energy (directed motion).

5. & 6. Power

Power is the rate at which work is done, or energy is transferred.

P = W / t

Where:

P = Power (Watts, W, where 1 W = 1 J/s)
W = Work done (J)
t = Time taken (s)

Note on Notation: Be careful with W (work) vs W (watt). Write "work" or "watt" to avoid confusion.

7. Power and Velocity

P = Fv

Where F is the force in the direction of motion and v is the velocity.

Derivation of P = Fv:
1. We know P = W/t
2. We know W = Fs
3. Substituting: P = Fs/t
4. Since s/t = v (velocity): P = Fv

Example 3: Power Output

A student of mass 60 kg climbs a vertical rope at a constant speed of 0.5 m/s. Calculate the student's useful power output.

Solution:

Force = weight = mg = 60 × 9.81 = 588.6 N

Power = Force × Velocity = 588.6 × 0.5 = 294.3 W ≈ 290 W (2 s.f.)

Note: A fit person would struggle to sustain 100 W continuously.

Example 4: Efficiency Calculation

If the student's muscles in Example 3 are 20% efficient, what is the total metabolic power they generate?

Solution:

Efficiency = (Useful Power Output / Total Power Input) × 100%

20% = (294.3 / Total Power Input) × 100%

Total Power Input = 294.3 / 0.20 = 1471.5 W ≈ 1500 W (2 s.f.)

5.2 Gravitational Potential Energy and Kinetic Energy

Syllabus Coverage: Derive and use formulas for gravitational potential energy and kinetic energy.

1. & 2. Gravitational Potential Energy (GPE or E_p)

Gravitational Potential Energy is the energy an object possesses due to its position in a gravitational field. It is the stored ability to do work.

ΔE_p = mgΔh

Where:

ΔE_p = Change in gravitational potential energy (J)
m = Mass (kg)
g = Gravitational field strength (N/kg)
Δh = Change in vertical height (m)

Derivation of ΔE_p = mgΔh:
1. To lift an object of mass m to a height Δh at constant speed, apply force equal to weight: F = mg
2. Work done = Force × Distance = mg × Δh
3. This work is stored as gravitational potential energy
4. Therefore, ΔE_p = mgΔh

3. & 4. Kinetic Energy (KE or E_k)

Kinetic Energy is the energy an object possesses due to its motion. It is the work an object can do against a resistive force due to its speed, and also the work needed to accelerate it to that speed.

E_k = ½mv²

Where:

E_k = Kinetic energy (J)
m = Mass (kg)
v = Speed (m/s)

Derivation of E_k = ½mv²:
Consider mass m accelerated from rest (u=0) to velocity v by constant force F over distance s:
1. Work done = Fs
2. From Newton's 2nd law: F = ma
3. From equations of motion: v² = u² + 2as = 2as (since u=0)
4. Therefore, as = v²/2
5. Work done = Fs = ma × s = m × (as) = m × (v²/2) = ½mv²
6. This work is transferred to kinetic energy: E_k = ½mv²

Example 5: Conservation of Energy - Free Fall

A 2.0 kg stone is dropped from rest from a cliff top 30 m above the water. Calculate:

a) The initial gravitational potential energy

b) The kinetic energy just before impact

c) The speed just before impact (ignore air resistance)

Solution:

a) ΔE_p = mgΔh = 2.0 × 9.81 × 30 = 588.6 J ≈ 590 J

b) By conservation of energy: Initial GPE = Final KE = 590 J

c) E_k = ½mv² → 588.6 = ½ × 2.0 × v² → v² = 588.6 → v = √588.6 = 24.26 m/s ≈ 24 m/s

Example 6: Work Against Air Resistance

A ball of mass 1.30 kg is thrown vertically upwards with a speed of 18.6 m/s and reaches a height of 15.0 m. Calculate the work done by the ball against air resistance.

Solution:

Initial KE = ½ × 1.30 × (18.6)² = 224.8 J

Final GPE = 1.30 × 9.81 × 15.0 = 191.3 J

By conservation of energy: Initial KE = Final GPE + Work against air resistance

224.8 = 191.3 + W_air

W_air = 224.8 - 191.3 = 33.5 J

Example 7: Kinetic Energy and Stopping Distance

A loaded truck of mass 40,000 kg travels at 20 m/s. Calculate:

a) Its kinetic energy

b) The stopping force needed to stop it in 100 m

Solution:

a) E_k = ½ × 40,000 × 20² = 8.0 × 10⁶ J

b) Work done by brakes = Force × Distance = E_k

F × 100 = 8.0 × 10⁶

F = 80,000 N

Note: For 10 m stopping distance, F = 800,000 N - this explains why severe damage occurs in short-distance collisions.

Example 8: Hydroelectric Power with Efficiency

Water falls through 120 m. At the water wheel, its kinetic energy is only 10% of its initial GPE.

(i) Show that the water speed at the wheel is 15 m/s

(ii) The wheel produces 110 kW electrical power. Calculate the water mass flow rate (kg/s) if electricity production is 25% efficient.

Solution:

(i) Let m = water mass

Initial GPE = m × 9.81 × 120 ≈ 1177.2m J

KE at wheel = 10% of GPE = 0.10 × 1177.2m = 117.72m J

½mv² = 117.72m → ½v² = 117.72 → v² = 235.44 → v ≈ 15.3 m/s ≈ 15 m/s

(ii) Let ṁ = mass flow rate (kg/s)

KE per second at wheel = ½ṁv² = ½ṁ(15)² = 112.5ṁ W

Electrical power = 25% of KE power = 0.25 × 112.5ṁ = 28.125ṁ W

110,000 = 28.125ṁ

ṁ = 110,000 / 28.125 ≈ 3910 kg/s

Other Forms of Energy

Type of Energy	Explanation	Key Equation
Gravitational Potential	Energy from position in a gravitational field	ΔE_p = mgΔh
Kinetic	Energy from motion	E_k = ½mv²
Elastic Potential (Spring)	Energy stored in stretched/compressed spring (area under F-x graph)	E = ½Fx = ½kx² (for Hooke's law)
Elastic Potential (Gas)	Energy from compressed gas: work done = pΔV (area under p-V graph)	W = pΔV
Electrical Potential	Energy a charge has from position in electric field: W = Eqx	E = Eqx
Internal Energy	Sum of random molecular KE and PE: ~3kT/2 per molecule (k = Boltzmann constant)	U = 3NkT/2 (for monatomic gas)
Chemical Energy	Energy from atomic rearrangement in molecules	-
Nuclear Energy	Energy from nuclear structure of atoms	-
Thermal Energy	Energy due to temperature difference	Q = mcΔθ
Sound Energy	Organized molecular motion in sound waves	-
Radiation Energy	Electromagnetic wave energy (light, infrared, radio, etc.)	E = hf (for photons)

Practice Problems

Problem 1: Work and Power

A person of mass 60 kg goes upstairs, a vertical distance of 2.8 m, in 4.7 s. Determine the work done and the minimum power output of the person.

Problem 2: Stopping Distance

A vehicle has a kinetic energy of 48,000 J.

(a) Calculate the distance it will travel against a constant stopping force of 600 N.

(b) The mass of the vehicle is 800 kg. Calculate its initial speed.

Problem 3: Acceleration Work

A car of mass 1200 kg increases speed from 25 m/s to 30 m/s.

(a) Calculate the increase in kinetic energy.

(b) Calculate the force needed if acceleration occurs over 100 m.

Problem 4: Different Acceleration Work

Explain why the work needed to accelerate a train from 0 to 10 m/s is not the same as from 40 to 50 m/s, even though the speed increase is the same.

Problem 5: Braking Force

A person of mass 80 kg traveling in a car at 25 m/s is stopped in 43 m.

(a) Calculate the average force on the person during braking.

(b) Identify the forces that cause drag on the person during braking.

Exam Tip: Always show your working clearly in calculations. Pay attention to significant figures in final answers. Use the conservation of energy principle as your first approach for energy problems - it's often the most efficient method. Remember that work done = area under force-displacement graph, and power = work/time = force×velocity.

These comprehensive notes cover all essential concepts of Work, Energy and Power for A-Level Physics, from foundational principles to advanced applications. Practice applying these principles to various scenarios to build strong problem-solving skills for your exams.

Sankey Diagrams - Energy Flow Visualization

Sankey Diagrams: Visualizing Energy Flow

Definition: Sankey diagrams are flow diagrams where the width of arrows is proportional to the amount of energy or power being transferred. They provide a clear visual representation of energy inputs, useful outputs, and energy losses.

Key Features of Sankey Diagrams

📏 Width Proportional to Quantity: Thicker arrows = more energy/power

🎯 Direction Shows Flow: Arrows show energy from source to destination

📊 Conservation of Energy: Total input width = Total output width

🔍 Efficiency Visualization: Clearly shows useful vs wasted energy

Car Engine Sankey Diagram

Typical Petrol Engine Energy Flow

ENERGY INPUT

Chemical Energy from Fuel 100%

USEFUL OUTPUTS

Kinetic Energy (Motion) 25%

Electrical Energy 5%

Total Useful: 30%

ENERGY LOSSES

Heat to Cooling 30%

Exhaust Heat 25%

Friction 10%

Sound 5%

Total Wasted: 70%

Overall Efficiency: 30%
Only 30% of fuel energy becomes useful work

📝 How to Read This Diagram:

Input (Blue): 100% chemical energy from fuel
Useful Output (Green): 25% kinetic + 5% electrical = 30% total useful
Wasted Energy (Red): 30% cooling + 25% exhaust + 10% friction + 5% sound = 70% total wasted
Efficiency: (30% ÷ 100%) × 100% = 30% efficiency

LED Light Bulb Example

Modern LED Light Bulb

ELECTRICAL INPUT

100J Electrical Energy 100%

USEFUL OUTPUT

Light Energy 40J (40%)

ENERGY LOSSES

Heat Energy 60J (60%)

Efficiency: 40%
Much more efficient than old incandescent bulbs (5%)

Why Sankey Diagrams Matter

✅ Visual Impact: Immediately see where most energy is lost

✅ Efficiency Analysis: Easy to calculate and compare efficiencies

✅ Design Improvement: Identifies areas for energy-saving improvements

✅ Educational Value: Makes abstract energy concepts tangible

Exam Tips

🎯 Remember: While Sankey diagrams aren't directly on the syllabus, understanding them helps with efficiency calculations and energy transfer concepts.

📐 Drawing Tips: When sketching Sankey diagrams, ensure arrow widths are proportional to the energy quantities they represent.

🧮 Calculation Ready: Always be prepared to extract efficiency calculations from Sankey diagram data.

💡 Key Takeaway: Sankey diagrams turn abstract energy numbers into visual stories, making it easy to see efficiency at a glance and identify where energy is being wasted.

Work, Force Components & Vertical Work — Interactive

Work & Force Components — Interactive

Change the force, angle, and displacement to see the horizontal component \(F\cos\theta\) and the work \(W=F\cos\theta\cdot s\). Toggle vertical mode to view lifting/lowering (with gravity).

Force magnitude, F (N) Displacement, s (m) Angle θ (degrees)

θ = 30°

cosθ = 0.866

Mode

Resulting component along motion:

F·cosθ = 25.98 N

Work done:

W = 103.9 J

Reset

Formulas

F_parallel = F cos θ
W = (F cos θ)·s

Vertical mode (lifting): W = mgh (if F = mg). When lowering with gravity the work done by gravity = +mgh; by you = −mgh.

Current values

F = 30 N

θ = 30°

s = 4 m

Worked example

If F = 30 N, θ = 30°, s = 4 m → F cos θ = 25.98 N, W = 103.9 J

Work When Force and Displacement Are in the Same Direction

If a force F acts in the same direction as the displacement s, then the work done is:

W = Fs

Work When Force is at an Angle θ to Displacement

Only the component of the force in the direction of movement does work.

The useful (horizontal) component of the force is:

F_parallel = F cos θ

So, the work done becomes:

W = F s cos θ

Example:
A 50 N force pulls a box through 6 m at 40° to the horizontal.
F cos θ = 50 cos 40° = 38.3 N
W = 38.3 × 6 = 229.8 J

Work Done Vertically (Against Gravity)

When lifting a body vertically, the force must overcome gravity (mg).

W = mgh

Example:
Lifting a 12 kg object through 2 m:
W = mgh = 12 × 9.8 × 2 = 235.2 J

Summary Table

Situation	Work Done Formula
Force in same direction as displacement	W = Fs
Force at angle θ to displacement	W = F s cos θ
Vertical lifting against gravity	W = mgh

Work Done by an Electric Current

Definition: Work done by an electric current is the electrical energy transferred (or converted into other forms such as heat, light or mechanical energy) when charge flows through a potential difference.

Key formulas

W = V I t

Where: W = work (J), V = potential difference (V), I = current (A), t = time (s).

Since P = V I, then W = P t

Using Ohm's law (V = I R): W = I² R t = (V² / R) t

Power relations (for quick reference)

P = V I
P = I² R
P = V² / R

Worked example
A 12 V lamp draws 0.5 A for 4 minutes.
t = 4 × 60 = 240 s
W = V I t = 12 × 0.5 × 240 = 1440 J

Convert to kilowatt-hours (useful for electricity bills):

1 kWh = 3.6 × 10⁶ J, so 1440 J = 1440 / 3.6×10⁶ = 0.0004 kWh

Quick summary table

Quantity	Formula
Work (general)	W = V I t = P t
Using resistance	W = I² R t = (V² / R) t
Power	P = V I = I² R = V² / R
Energy units	1 J (joule); 1 kWh = 3.6 × 10⁶ J

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Why sin θ is Better on an Incline

For an object moving a distance s up an incline of angle θ, the vertical height gained is:

h = s sin θ

Therefore, the gain in gravitational potential energy is:

ΔE_p = m g h = m g (s sin θ)

Choosing the Correct Formula

Situation	Best Formula
Work done along slope	W = F s cos θ
Work done vertically	W = F s sin θ
Work done against gravity	W = m g h = m g (s sin θ)

Example:
A 6 kg box is pulled 4 m up a 30° incline.
Height gained: h = 4 sin 30° = 2 m.
Work against gravity = m g h = 6 × 9.8 × 2 = 117.6 J.

Proof (components only): Why W = F s cos θ on an incline

Set-up: Let a force of magnitude F act at angle θ to the direction of motion along the incline. The object moves distance s along the incline.

1. Resolve the force into components

Draw the right triangle formed by F and its components parallel and perpendicular to the incline. The component parallel to the incline (the part that causes motion) is adjacent to θ, so by trigonometry:

F_parallel = F cos θ

2. Work = (force along motion) × (distance)

Only the parallel component does work for displacement along the incline, so

W = F_parallel · s = (F cos θ) s = F s cos θ

Example:
F = 60 N, θ = 20°, s = 5 m → F_parallel = 60 cos 20° ≈ 56.38 N.
W ≈ 56.38 × 5 = 281.9 J.

Note: θ must be measured between the force direction and the direction of motion along the incline. If θ is instead given relative to horizontal, adjust the geometry so the angle used is the angle between F and the slope direction.

Work Done On an Object vs Work Done By an Object

Work done on an object

Definition: Energy transferred to an object by external forces; the object gains energy (kinetic or potential).

Example: You push a stationary box and it speeds up — you do work on the box (it gains kinetic energy).

Work done by an object

Definition: Energy the object transfers to its surroundings; the object loses energy (kinetic or stored energy decreases).

Example: A moving car pushes air and overcomes friction — the car does work by itself on the surroundings (it loses kinetic energy).

Simple relations

Work done on object = increase in the object's energy (ΔE > 0).
Work done by object = decrease in the object's energy (ΔE < 0).

Comparison table

Aspect	Work done on object	Work done by object
Who supplies energy?	External agent (person, motor, battery)	The object itself (stored or kinetic energy)
Object's energy change	Energy increases	Energy decreases
Sign convention (work on object)	Positive when energy added	Negative if object loses energy (or positive if describing work done by it)
Typical examples	Lifting a mass (you do work on mass); battery powering a motor	Falling mass does work on ground/air; heated object does work on surroundings

Exam tip

State clearly whether energy is being transferred to the object (work on) or from the object (work by). Use sign convention consistently: positive work adds energy to the system you're tracking; negative work removes energy from it.

Problem 1: Work and Power

A person of mass 60 kg goes upstairs, a vertical distance of 2.8 m, in 4.7 s. Determine the work done and the minimum power output of the person.

Solution:

Step 1: Calculate the work done.

The work done (W) against gravity is equal to the gain in gravitational potential energy.

W = m × g × h

W = (60 kg) × (9.8 m/s²) × (2.8 m)

W = 1646.4 J

The work done is 1650 J (rounded to 3 significant figures).

Step 2: Calculate the minimum power output.

Power (P) is the rate of doing work.

P = W / t

P = 1646.4 J / 4.7 s

P ≈ 350.3 W

The minimum power output is 350 W (rounded to 2 significant figures).

Problem 2: Stopping Distance

A vehicle has a kinetic energy of 48,000 J.

(a) Calculate the distance it will travel against a constant stopping force of 600 N.

(b) The mass of the vehicle is 800 kg. Calculate its initial speed.

Solution:

(a) Calculate the stopping distance.

The work done by the stopping force must equal the initial kinetic energy.

Work = Force × Distance

48,000 J = (600 N) × d

d = 48,000 J / 600 N = 80 m

The stopping distance is 80.0 m (3 significant figures).

(b) Calculate the initial speed.

Using the kinetic energy formula: KE = ½mv²

48,000 J = ½ × (800 kg) × v²

48,000 = 400 × v²

v² = 48,000 / 400 = 120

v = √120 ≈ 10.954 m/s

The initial speed is 11.0 m/s (rounded to 3 significant figures).

Problem 3: Acceleration Work

A car of mass 1200 kg increases speed from 25 m/s to 30 m/s.

(a) Calculate the increase in kinetic energy.

(b) Calculate the force needed if acceleration occurs over 100 m.

Solution:

(a) Calculate the increase in kinetic energy.

ΔKE = ½m(v_f² - v_i²)

ΔKE = ½ × 1200 kg × ((30 m/s)² - (25 m/s)²)

ΔKE = 600 × (900 - 625)

ΔKE = 600 × 275 = 165,000 J

The increase in kinetic energy is 165 kJ (3 significant figures).

(b) Calculate the force needed over 100 m.

Work done by force = Increase in kinetic energy

F × d = ΔKE

F × 100 m = 165,000 J

F = 165,000 J / 100 m = 1,650 N

The force needed is 1.65 kN (3 significant figures).

Problem 4: Different Acceleration Work

Explain why the work needed to accelerate a train from 0 to 10 m/s is not the same as from 40 to 50 m/s, even though the speed increase is the same.

Solution:

The work needed to accelerate an object is equal to its change in kinetic energy.

Work from 0 m/s to 10 m/s:

W₁ = ½m(10² - 0²) = ½m(100) = 50m

Work from 40 m/s to 50 m/s:

W₂ = ½m(50² - 40²) = ½m(2500 - 1600) = ½m(900) = 450m

Explanation: Even though the speed increase is the same (10 m/s), the change in kinetic energy is not the same. Kinetic energy depends on the square of the speed (v²). The difference between the squares of the final and initial speeds is much larger for the higher speed interval (900 vs. 100). Therefore, significantly more work (450m J vs. 50m J) is required to accelerate the train from 40 m/s to 50 m/s.

Problem 5: Braking Force

A person of mass 80 kg traveling in a car at 25 m/s is stopped in 43 m.

(a) Calculate the average force on the person during braking.

(b) Identify the forces that cause drag on the person during braking.

Solution:

(a) Calculate the average force on the person.

First, find the initial kinetic energy:

KE = ½mv² = ½ × 80 kg × (25 m/s)² = 25,000 J

Work done by braking force = Kinetic energy

F × d = KE

F × 43 m = 25,000 J

F = 25,000 J / 43 m ≈ 581.4 N

The average force on the person is 580 N (rounded to 2 significant figures).

(b) Identify the forces that cause drag on the person.

During braking, the car decelerates. According to Newton's First Law (inertia), the person's body wants to continue moving forward. The forces that drag the person backward are:

The seat belt: It applies a backward force on the person's torso.
Friction between the person and the seat: This acts to pull the person backward.
Contact forces with the car's interior: Any surface the person is leaning against pushes them backward.

As Physics Topic 5: Work,Energy & Power

Complete Work, Energy and Power Notes - A-Level Physics

Fundamental Energy Concepts

What is Energy?

Basic Forms of Energy

Energy Transfer Processes

5.1 Energy Conservation

1. Work Done

2. Principle of Conservation of Energy

3. & 4. Efficiency

5. & 6. Power

7. Power and Velocity

5.2 Gravitational Potential Energy and Kinetic Energy

1. & 2. Gravitational Potential Energy (GPE or Ep)

3. & 4. Kinetic Energy (KE or Ek)

Other Forms of Energy

Practice Problems

Sankey Diagrams: Visualizing Energy Flow

Key Features of Sankey Diagrams

Car Engine Sankey Diagram

Typical Petrol Engine Energy Flow

📝 How to Read This Diagram:

LED Light Bulb Example

Modern LED Light Bulb

Why Sankey Diagrams Matter

Exam Tips

Work & Force Components — Interactive

Work When Force and Displacement Are in the Same Direction

Work When Force is at an Angle θ to Displacement

Work Done Vertically (Against Gravity)

Summary Table

Work Done by an Electric Current

Key formulas

Power relations (for quick reference)

Quick summary table

Why sin θ is Better on an Incline

Choosing the Correct Formula

Proof (components only): Why W = F s cos θ on an incline

1. Resolve the force into components

2. Work = (force along motion) × (distance)

Work Done On an Object vs Work Done By an Object

Work done on an object

Work done by an object

Simple relations

Comparison table

Exam tip

Problem 1: Work and Power

Solution:

Problem 2: Stopping Distance

Solution:

Problem 3: Acceleration Work

Solution:

Problem 4: Different Acceleration Work

Solution:

Problem 5: Braking Force

Solution:

1. & 2. Gravitational Potential Energy (GPE or E_p)

3. & 4. Kinetic Energy (KE or E_k)