1. Kinetic Energy

Learning outcomes

• I can define kinetic energy as the energy associated with motion.
• I can explain how an object's mass and speed affect its kinetic energy.
• I can recall and use the equation\( E_k = \frac{1}{2}mv^2 \) .
• I can calculate the kinetic energy of moving objects in a variety of situations.
• I can analyze how changes in speed affect kinetic energy and explain why kinetic energy increases rapidly with velocity.

Kinetic Energy

Kinetic energy is the energy an object has because it is moving.

Any moving object has kinetic energy, including:

• a rolling ball
• a moving car
• a running person
• a flying aircraft
• flowing water
• a falling object
• a moving planet

An object that is stationary has zero kinetic energy relative to the chosen reference frame.

The amount of kinetic energy depends on two important quantities:

• the object's mass
• the object's speed

 

 

 

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The Kinetic Energy Equation

The equation for translational kinetic energy is:

Eₖ = 0.5mv²

or equivalently:

Eₖ = ½mv²

where:

• Eₖ = kinetic energy in joules (J)
• m = mass in kilograms (kg)
• v = speed in metres per second (m/s)

The equation shows that kinetic energy is directly proportional to mass but proportional to the square of speed.

That difference is extremely important.

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Units of Kinetic Energy

The SI unit of energy is the joule (J).

From the equation:

Eₖ = 0.5mv²

the units are:

kg × (m/s)²

which becomes:

kg·m²/s²

This combination of units is defined as one joule:

1 J = 1 kg·m²/s²

Larger quantities of energy may be expressed in:

• kilojoules (kJ)
• megajoules (MJ)

Remember:

1 kJ = 1000 J

1 MJ = 1,000,000 J

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Calculating Kinetic Energy

The basic procedure is:

Identify mass → identify speed → square the speed → multiply by mass → multiply by 0.5

For example, consider a 4 kg object moving at 3 m/s.

Eₖ = 0.5mv²

Substitute:

Eₖ = 0.5 × 4 × 3²

Square the speed:

Eₖ = 0.5 × 4 × 9

Therefore:

Eₖ = 18 J

The object has 18 J of kinetic energy.

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Worked Example: Moving Bicycle

A bicycle and rider have a combined mass of 80 kg and travel at 5 m/s.

Calculate their kinetic energy.

Eₖ = 0.5mv²

Eₖ = 0.5 × 80 × 5²

Eₖ = 0.5 × 80 × 25

Eₖ = 1000 J

The bicycle and rider have 1000 J, or 1.0 kJ, of kinetic energy.

 

 

 

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Worked Example: Moving Car

A car has a mass of 1200 kg and travels at 20 m/s.

Eₖ = 0.5mv²

Eₖ = 0.5 × 1200 × 20²

Eₖ = 0.5 × 1200 × 400

Eₖ = 240,000 J

Therefore:

Eₖ = 240 kJ

This large amount of energy must be transferred somewhere when the car stops.

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How Mass Affects Kinetic Energy

Consider two objects moving at the same speed.

Object A

Mass = 2 kg
Speed = 5 m/s

Eₖ = 0.5 × 2 × 5² = 25 J

Object B

Mass = 4 kg
Speed = 5 m/s

Eₖ = 0.5 × 4 × 5² = 50 J

Doubling the mass has doubled the kinetic energy.

Therefore:

mass × 2 → kinetic energy × 2

If speed remains constant, kinetic energy is directly proportional to mass.

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Comparing Mass and Kinetic Energy

If speed remains constant:

Mass	Kinetic Energy
m	Eₖ
2m	2Eₖ
3m	3Eₖ
4m	4Eₖ

This relationship is linear.

If mass increases by 50%, kinetic energy also increases by 50%, provided speed does not change.

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How Speed Affects Kinetic Energy

Speed has a much stronger effect because speed is squared.

Consider a 2 kg object.

At 2 m/s

Eₖ = 0.5 × 2 × 2² = 4 J

At 4 m/s

Eₖ = 0.5 × 2 × 4² = 16 J

The speed doubled:

2 m/s → 4 m/s

but the kinetic energy increased from:

4 J → 16 J

Therefore:

speed × 2 → kinetic energy × 4

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Why Speed Has Such a Large Effect

The reason is the v² term.

If speed doubles:

(2v)² = 4v²

Therefore:

kinetic energy × 4

If speed triples:

(3v)² = 9v²

Therefore:

kinetic energy × 9

If speed quadruples:

(4v)² = 16v²

Therefore:

kinetic energy × 16

This is one of the most important relationships to understand about kinetic energy.

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Speed and Kinetic Energy

For the same object:

Speed	Relative Kinetic Energy
v	Eₖ
2v	4Eₖ
3v	9Eₖ
4v	16Eₖ
5v	25Eₖ

Kinetic energy therefore does not increase linearly with speed.

It increases with the square of speed.

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Worked Example: Doubling Speed

A 1000 kg car travels at 10 m/s.

Initial kinetic energy:

Eₖ = 0.5 × 1000 × 10²

Eₖ = 50,000 J

Now the car travels at 20 m/s.

Eₖ = 0.5 × 1000 × 20²

Eₖ = 200,000 J

The speed doubled, but:

50,000 J → 200,000 J

The kinetic energy became four times greater.

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Worked Example: Tripling Speed

A ball has 20 J of kinetic energy while moving at speed v.

What happens if its speed becomes 3v?

Because:

Eₖ ∝ v²

then:

new Eₖ = 3² × 20

new Eₖ = 9 × 20

new Eₖ = 180 J

There is no need to know the ball's mass for this comparison.

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Why This Matters for Vehicles

The squared relationship between speed and kinetic energy has important consequences for transportation.

A faster vehicle has much more kinetic energy that must be transferred when it stops.

For the same vehicle:

50 km/h → Eₖ

100 km/h → 4Eₖ

The speed has doubled, so the kinetic energy has quadrupled.

This is one reason increasing speed can greatly increase the energy involved in a collision.

 

 

 

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Kinetic Energy and Braking

When a vehicle brakes, its kinetic energy must be transferred.

Much of the energy is ultimately transferred as thermal energy to:

• brakes
• tires
• road
• surrounding air

If a vehicle has more kinetic energy, more energy must be transferred before it can stop.

Because kinetic energy depends on v², speed has a particularly important effect.

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Kinetic Energy in Collisions

During a collision, kinetic energy may be transferred into:

• deformation
• thermal energy
• sound
• motion of other objects

For example, during a car crash, energy may be transferred into bending and crushing the vehicle's structure.

Modern vehicles use crumple zones partly to manage this energy transfer over a greater distance and time.

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Kinetic Energy and Falling Objects

A falling object gains speed as gravitational potential energy is transferred into kinetic energy.

Ignoring air resistance:

gravitational potential energy → kinetic energy

As the object falls:

• gravitational potential energy decreases
• speed increases
• kinetic energy increases

Just before reaching the ground, much of the original gravitational potential energy may have been transferred into kinetic energy.

 

 

 

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Example: Falling Object

Suppose a 2 kg object reaches a speed of 10 m/s while falling.

Eₖ = 0.5 × 2 × 10²

Eₖ = 100 J

The object has 100 J of kinetic energy.

If air resistance is negligible, this energy came from a decrease in gravitational potential energy.

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Kinetic Energy and Roller Coasters

Roller coasters demonstrate the transfer between gravitational potential energy and kinetic energy.

At the top of a hill:

• height is large
• gravitational potential energy is large
• speed may be relatively low

As the coaster travels downhill:

gravitational potential energy → kinetic energy

Near the bottom:

• height is lower
• speed is greater
• kinetic energy is greater

Friction and air resistance transfer some energy to the surroundings.

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Kinetic Energy Is a Scalar

Kinetic energy is a scalar quantity.

This means it has:

• magnitude
• no direction

Velocity has direction, but kinetic energy does not.

Consider two identical cars:

Car A: 20 m/s east

Car B: 20 m/s west

Both have the same mass and the same speed.

Therefore, they have the same kinetic energy.

The direction of motion does not change kinetic energy.

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Why Squaring Velocity Removes Direction

Suppose:

v = +5 m/s

Then:

v² = (+5)² = 25

If:

v = −5 m/s

Then:

v² = (−5)² = 25

Both produce the same kinetic energy.

Therefore, kinetic energy cannot be negative.

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Can Kinetic Energy Be Zero?

Yes.

If:

v = 0

then:

Eₖ = 0.5m(0)²

so:

Eₖ = 0 J

A stationary object has zero translational kinetic energy relative to the chosen reference frame.

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Reference Frames

Whether an object is moving depends on the reference frame.

Imagine sitting in a train moving at constant speed.

Relative to the train:

you are stationary

Relative to the ground:

you are moving

Therefore, your calculated kinetic energy depends on the reference frame used to measure your speed.

In most introductory problems, the reference frame is Earth's surface.

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Rearranging the Kinetic Energy Equation

Sometimes kinetic energy is known and mass or speed must be calculated.

Starting with:

Eₖ = 0.5mv²

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Finding Mass

Multiply both sides by 2:

2Eₖ = mv²

Divide by v²:

m = 2Eₖ/v²

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Worked Example: Finding Mass

An object has:

Kinetic energy = 400 J

Speed = 10 m/s

Find its mass.

m = 2Eₖ/v²

m = (2 × 400)/10²

m = 800/100

m = 8 kg

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Finding Speed

Starting with:

Eₖ = 0.5mv²

Multiply by 2:

2Eₖ = mv²

Divide by m:

v² = 2Eₖ/m

Take the square root:

v = √(2Eₖ/m)

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Worked Example: Finding Speed

A 5 kg object has 360 J of kinetic energy.

v = √(2Eₖ/m)

v = √((2 × 360)/5)

v = √144

v = 12 m/s

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Unit Conversions

Mass must normally be expressed in kilograms.

If mass is given in grams:

1000 g = 1 kg

Therefore:

mass in kg = mass in g ÷ 1000

For example:

500 g = 0.500 kg

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Worked Example: Mass in Grams

A 200 g ball travels at 15 m/s.

First convert mass:

200 g = 0.200 kg

Then:

Eₖ = 0.5 × 0.200 × 15²

Eₖ = 0.1 × 225

Eₖ = 22.5 J

The ball has 22.5 J of kinetic energy.

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Converting km/h to m/s

Vehicle speeds are often given in km/h, while kinetic-energy calculations normally use m/s.

To convert:

m/s = km/h ÷ 3.6

For example:

72 km/h ÷ 3.6 = 20 m/s

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Worked Example: Car Speed in km/h

A 1500 kg car travels at 72 km/h.

First:

72 ÷ 3.6 = 20 m/s

Then:

Eₖ = 0.5 × 1500 × 20²

Eₖ = 300,000 J

or:

Eₖ = 300 kJ

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Comparing Two Objects

Consider:

Object A

Mass = 4 kg
Speed = 6 m/s

Eₖ = 0.5 × 4 × 6² = 72 J

Object B

Mass = 2 kg
Speed = 10 m/s

Eₖ = 0.5 × 2 × 10² = 100 J

Object A has more mass, but Object B has more kinetic energy.

This demonstrates that speed can have a very strong influence because it is squared.

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Percentage Changes

Suppose speed increases by 20%.

The new speed is:

1.2v

Because kinetic energy depends on speed squared:

new Eₖ = (1.2)²Eₖ

new Eₖ = 1.44Eₖ

Therefore, a 20% increase in speed produces a 44% increase in kinetic energy, assuming mass remains constant.

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Another Percentage Example

Suppose speed decreases by 50%.

New speed:

0.5v

New kinetic energy:

(0.5)²Eₖ = 0.25Eₖ

The object now has only 25% of its original kinetic energy.

Therefore, reducing speed by half reduces kinetic energy by 75%.

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Graph of Kinetic Energy and Mass

If speed is constant:

Eₖ ∝ m

A graph of kinetic energy against mass is therefore a straight line through the origin.

Double the mass:

double the kinetic energy.

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Graph of Kinetic Energy and Speed

If mass is constant:

Eₖ ∝ v²

A graph of kinetic energy against speed is therefore curved, not a straight line.

At higher speeds, relatively small increases in speed can produce large increases in kinetic energy.

This graphical relationship is important when analyzing motion and transportation.

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Energy Transfers

Kinetic energy is one part of a larger energy system.

Examples include:

Falling Object

gravitational potential → kinetic

Car Accelerating

chemical energy → kinetic + thermal

Electric Vehicle Accelerating

electrical energy → kinetic + thermal

Braking

kinetic → thermal

Collision

kinetic → deformation + thermal + sound + other kinetic energy

Energy is transferred between stores rather than simply disappearing.

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Real-World Connection: Sports

Kinetic energy is important in many sports.

Consider a moving ball.

A faster ball has much more kinetic energy because:

Eₖ ∝ v²

Increasing the speed of a:

• football
• tennis ball
• baseball
• hockey puck

can greatly increase its kinetic energy.

Mass also matters, which is why objects of different masses moving at the same speed can carry different amounts of kinetic energy.

 

 

 

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Real-World Connection: Spacecraft

Spacecraft can travel at extremely high speeds.

Even relatively small objects can therefore possess enormous kinetic energies.

This is particularly important for objects in orbit.

Because kinetic energy depends on speed squared, even small pieces of space debris moving at orbital speeds can carry enough kinetic energy to damage spacecraft.

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Real-World Connection: Wind Energy

Moving air has kinetic energy.

Wind turbines capture some of this energy and convert it into electrical energy.

Faster-moving air carries much more energy than slower-moving air.

This is one reason wind speed is extremely important when determining suitable locations for wind turbines.

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Common Mistakes

Forgetting to Square the Speed

Incorrect:

Eₖ = 0.5mv

Correct:

Eₖ = 0.5mv²

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Squaring the Mass

Incorrect:

Eₖ = 0.5m²v

Only the speed is squared.

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Assuming Doubling Speed Doubles Kinetic Energy

Incorrect:

speed × 2 → Eₖ × 2

Correct:

speed × 2 → Eₖ × 4

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Forgetting to Convert Grams to Kilograms

Use kilograms in standard SI calculations.

750 g = 0.750 kg

not:

750 kg

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Forgetting to Convert km/h to m/s

For standard calculations:

m/s = km/h ÷ 3.6

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Giving the Answer in Newtons

Kinetic energy is measured in joules, not newtons.

Newtons measure force.

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Thinking Direction Changes Kinetic Energy

Kinetic energy depends on speed, not direction.

Two identical objects moving at equal speeds in opposite directions have equal kinetic energies.

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Thinking a Negative Velocity Produces Negative Kinetic Energy

Velocity may be negative depending on direction, but:

v² is always positive

so kinetic energy cannot be negative.

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Check Your Understanding

1. Define kinetic energy.
2. State the equation for kinetic energy.
3. What unit is used for kinetic energy?
4. Calculate the kinetic energy of a 5 kg object moving at 4 m/s.
5. Calculate the kinetic energy of a 1000 kg car moving at 15 m/s.
6. A 500 g ball travels at 8 m/s. Calculate its kinetic energy.
7. Explain how doubling an object's mass affects its kinetic energy if speed remains constant.
8. Explain how doubling speed affects kinetic energy.
9. What happens to kinetic energy if speed triples?
10. An object has 50 J of kinetic energy. Its speed doubles without changing its mass. What is its new kinetic energy?
11. An object has 80 J of kinetic energy. Its speed is halved. What is its new kinetic energy?
12. Why does speed have a greater effect on kinetic energy than mass?
13. A 1500 kg car travels at 25 m/s. Calculate its kinetic energy.
14. Convert 90 km/h to m/s.
15. A 1200 kg vehicle travels at 90 km/h. Calculate its kinetic energy.
16. An object has 900 J of kinetic energy and moves at 15 m/s. Calculate its mass.
17. A 10 kg object has 500 J of kinetic energy. Calculate its speed.
18. Two identical balls move at 10 m/s east and 10 m/s west. Compare their kinetic energies.
19. Explain why higher vehicle speeds can make collisions more severe.
20. Explain the energy transfers that occur when a moving car brakes to a stop.

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Key Terms

Kinetic energy — Energy associated with the motion of an object.

Energy — Capacity of a system to cause change or transfer energy.

Joule (J) — SI unit of energy.

Mass — Measure of the amount of matter in an object, measured in kilograms.

Speed — Distance traveled per unit time.

Velocity — Speed in a particular direction.

Scalar quantity — Quantity with magnitude but no direction.

Reference frame — Coordinate system or viewpoint relative to which motion is measured.

Energy transfer — Movement of energy from one store or system to another.

Translational kinetic energy — Kinetic energy associated with an object's overall motion from one position to another.

Direct proportionality — Relationship in which multiplying one quantity by a factor multiplies another by the same factor.

Squared relationship — Relationship in which one quantity depends on the square of another.

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Key Takeaways

• Kinetic energy is the energy associated with motion.
• A stationary object has zero translational kinetic energy relative to the chosen reference frame.
• Kinetic energy depends on both mass and speed.
• The kinetic energy equation is Eₖ = 0.5mv².
• Kinetic energy is measured in joules (J).
• Doubling mass doubles kinetic energy if speed remains constant.
• Tripling mass triples kinetic energy if speed remains constant.
• Kinetic energy is therefore directly proportional to mass.
• Speed has a much stronger effect because it is squared.
• Doubling speed produces four times the kinetic energy.
• Tripling speed produces nine times the kinetic energy.
• Quadrupling speed produces sixteen times the kinetic energy.
• A 20% increase in speed produces a 44% increase in kinetic energy.
• Halving speed leaves an object with only one-quarter of its original kinetic energy.
• Kinetic energy is a scalar and therefore has no direction.
• Objects traveling in opposite directions can have identical kinetic energies.
• Vehicle speed strongly affects the amount of energy that must be transferred during braking or a collision.
• Kinetic energy can be transferred into thermal energy, sound, deformation, gravitational potential energy, or the motion of other objects.
• Mass should normally be converted to kilograms and speed to metres per second before using the equation.
• When solving problems, always check that the final answer includes the correct unit.

The central relationship to remember is:

Eₖ = 0.5mv²

and especially:

mass × 2 → kinetic energy × 2

but

speed × 2 → kinetic energy × 4

That squared relationship is why speed has such a powerful effect on kinetic energy.

2. Gravitational Potential Energy

Gravitational Potential Energy

Gravitational potential energy is energy associated with an object's position in a gravitational field.

When an object is raised, energy is transferred to its gravitational potential energy store. When the object falls, that energy can be transferred to other energy stores, particularly kinetic energy.

Examples include:

• a book on a shelf
• water stored behind a dam
• a roller coaster at the top of a hill
• a raised weight
• a climber on a mountain
• a ball held above the ground

The higher an object is raised, the greater the change in its gravitational potential energy.

 

 

 

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What Determines Gravitational Potential Energy?

Near Earth's surface, the change in gravitational potential energy depends on three quantities:

• mass
• gravitational field strength
• change in vertical height

The relationship is:

ΔEₚ = mgΔh

where:

• ΔEₚ = change in gravitational potential energy in joules (J)
• m = mass in kilograms (kg)
• g = gravitational field strength in newtons per kilogram (N/kg)
• Δh = change in vertical height in metres (m)

For many school calculations on Earth:

g = 9.8 N/kg

Sometimes questions use the approximation:

g = 10 N/kg

Always use the value provided in the question when one is given.

Understanding Δ

The Greek letter Δ (delta) means change in.

Therefore:

Δh = final height − initial height

and:

ΔEₚ = final gravitational potential energy − initial gravitational potential energy

If an object moves upward:

Δh is positive

so:

ΔEₚ is positive

If an object moves downward:

Δh is negative

so:

ΔEₚ is negative

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Raising an Object

Suppose an object moves from:

2 m → 7 m

Then:

Δh = 7 − 2

Δh = +5 m

Its gravitational potential energy increases.

If the object instead moves from:

7 m → 2 m

then:

Δh = 2 − 7

Δh = −5 m

Its gravitational potential energy decreases.

The important quantity is the change in height, not simply the final height.

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Worked Example: Raising a Box

A 5 kg box is lifted vertically by 3 m.

Use:

g = 9.8 N/kg

ΔEₚ = mgΔh

ΔEₚ = 5 × 9.8 × 3

ΔEₚ = 147 J

The gravitational potential energy increases by:

147 J

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Worked Example: Using g = 10 N/kg

A 12 kg object is raised 4 m.

Use:

g = 10 N/kg

ΔEₚ = mgΔh

ΔEₚ = 12 × 10 × 4

ΔEₚ = 480 J

The object's gravitational potential energy increases by 480 J.

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How Mass Affects Gravitational Potential Energy

Consider two objects raised through the same height in the same gravitational field.

Object A

Mass = 2 kg
Height change = 5 m

Using g = 10 N/kg:

ΔEₚ = 2 × 10 × 5 = 100 J

Object B

Mass = 4 kg
Height change = 5 m

ΔEₚ = 4 × 10 × 5 = 200 J

Doubling the mass doubles the change in gravitational potential energy.

Therefore:

mass × 2 → ΔEₚ × 2

Gravitational potential energy is directly proportional to mass.

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How Height Affects Gravitational Potential Energy

Now keep mass constant.

A 3 kg object is raised by 2 m:

ΔEₚ = 3 × 10 × 2 = 60 J

Raise the same object by 4 m:

ΔEₚ = 3 × 10 × 4 = 120 J

Doubling the height change doubles the change in gravitational potential energy.

Therefore:

height × 2 → ΔEₚ × 2

This is different from kinetic energy, where doubling speed produces four times the kinetic energy.

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Comparing Kinetic and Gravitational Potential Energy

For kinetic energy:

Eₖ = 0.5mv²

Speed is squared.

Therefore:

speed × 2 → Eₖ × 4

For gravitational potential energy near Earth's surface:

ΔEₚ = mgΔh

Height is not squared.

Therefore:

height × 2 → ΔEₚ × 2

This distinction is important.

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Gravitational Field Strength

The symbol g represents gravitational field strength.

Near Earth's surface:

g ≈ 9.8 N/kg

This means that every kilogram of mass experiences approximately 9.8 N of gravitational force.

Recall the equation for weight:

W = mg

Therefore:

ΔEₚ = mgΔh

can also be written as:

ΔEₚ = WΔh

So the change in gravitational potential energy equals:

weight × vertical height change

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Worked Example Using Weight

A crate has a weight of 300 N and is lifted vertically 2.5 m.

ΔEₚ = WΔh

ΔEₚ = 300 × 2.5

ΔEₚ = 750 J

The crate gains 750 J of gravitational potential energy.

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Why g Matters

Gravitational field strength is not the same everywhere.

Different planets and moons have different values of g.

For the same object raised through the same height:

stronger gravitational field → greater change in gravitational potential energy

weaker gravitational field → smaller change in gravitational potential energy

This is because more work must be done against a stronger gravitational force.

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Mass and Weight Are Different

Mass describes the amount of matter in an object.

Unit:

kilogram (kg)

Weight is the gravitational force acting on an object.

Unit:

newton (N)

The relationship is:

W = mg

Mass remains the same when an object travels to another planet, but its weight can change because gravitational field strength changes.

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Worked Example: Different Gravitational Fields

A 10 kg object is raised 5 m.

On Earth

Use:

g = 9.8 N/kg

ΔEₚ = 10 × 9.8 × 5

ΔEₚ = 490 J

In a gravitational field of 2 N/kg

ΔEₚ = 10 × 2 × 5

ΔEₚ = 100 J

The same object gains less gravitational potential energy because the gravitational field is weaker.

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Lowering an Object

When an object is lowered, its gravitational potential energy decreases.

Suppose a 6 kg object moves downward by 4 m.

Using:

g = 10 N/kg

The height change is:

Δh = −4 m

Therefore:

ΔEₚ = 6 × 10 × (−4)

ΔEₚ = −240 J

The negative sign means the gravitational potential energy has decreased by 240 J.

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What Happens to the Lost Energy?

When gravitational potential energy decreases, the energy does not disappear.

It is transferred elsewhere.

For a freely falling object:

gravitational potential energy → kinetic energy

If air resistance is significant:

gravitational potential energy → kinetic energy + thermal energy

Energy is conserved.

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Falling Objects

Imagine a ball held above the ground.

Before release:

• gravitational potential energy is relatively high
• kinetic energy is zero if the ball is stationary

As it falls:

• gravitational potential energy decreases
• kinetic energy increases

Immediately before impact:

• gravitational potential energy relative to the ground is lower
• kinetic energy is greater

 

 

 

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Energy Conservation During a Fall

Ignoring air resistance:

decrease in gravitational potential energy = increase in kinetic energy

Therefore:

−ΔEₚ = ΔEₖ

For example, if an object loses:

500 J of gravitational potential energy

it gains:

500 J of kinetic energy

assuming no energy is transferred to the surroundings.

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Worked Example: Falling Object

A 2 kg object falls through 10 m.

Use:

g = 9.8 N/kg

Change in gravitational potential energy:

ΔEₚ = 2 × 9.8 × (−10)

ΔEₚ = −196 J

The object loses 196 J of gravitational potential energy.

Ignoring air resistance, it gains 196 J of kinetic energy.

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Connecting GPE and Kinetic Energy

Recall:

Eₖ = 0.5mv²

and:

ΔEₚ = mgΔh

For an object falling from rest with negligible air resistance:

lost gravitational potential energy = gained kinetic energy

Therefore:

mgΔh = 0.5mv²

This relationship can be used to calculate the speed of a falling object without directly using acceleration equations.

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Worked Example: Finding Speed from Energy

A ball falls 5 m from rest.

Ignore air resistance and use:

g = 9.8 N/kg

Energy conservation gives:

mgΔh = 0.5mv²

Mass appears on both sides, so it cancels:

gΔh = 0.5v²

Substitute:

9.8 × 5 = 0.5v²

49 = 0.5v²

v² = 98

v ≈ 9.9 m/s

Notice that the mass was not required.

Ignoring air resistance, objects falling from the same height reach the same speed regardless of mass.

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Choosing a Reference Height

Gravitational potential energy is usually measured relative to a chosen reference level.

For example, we might choose:

• the floor
• ground level
• the bottom of a hill
• the lowest point of a roller coaster

as:

h = 0

What usually matters physically is the change in gravitational potential energy, not the absolute value.

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Example: Shelf Heights

A 4 kg book moves from a shelf 1 m above the floor to a shelf 3 m above the floor.

The change in height is:

Δh = 3 − 1

Δh = 2 m

Using g = 9.8 N/kg:

ΔEₚ = 4 × 9.8 × 2

ΔEₚ = 78.4 J

A common mistake would be to use 3 m rather than the 2 m change in height.

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Vertical Height Matters

The equation uses vertical height change, not the total distance traveled.

Imagine carrying a box up a long ramp.

Suppose:

Ramp distance = 10 m

Vertical height gained = 3 m

For gravitational potential energy, use:

Δh = 3 m

not:

10 m

 

 

 

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Different Paths to the Same Height

Suppose identical objects are moved from the same starting height to the same final height.

Object A is lifted vertically.

Object B moves up a ramp.

Object C follows a curved path.

If all three have the same:

• mass
• starting height
• final height

then they have the same change in gravitational potential energy.

The path does not matter for ΔEₚ.

Only the vertical height difference matters.

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Rearranging the Equation

Starting with:

ΔEₚ = mgΔh

we can rearrange the equation to find other quantities.

Finding Mass

m = ΔEₚ/(gΔh)

Finding Height Change

Δh = ΔEₚ/(mg)

Finding Gravitational Field Strength

g = ΔEₚ/(mΔh)

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Worked Example: Finding Height

A 20 kg object gains 1960 J of gravitational potential energy.

How high was it raised?

Use:

Δh = ΔEₚ/(mg)

Δh = 1960/(20 × 9.8)

Δh = 1960/196

Δh = 10 m

The object was raised 10 m.

---

Worked Example: Finding Mass

An object gains 1500 J of gravitational potential energy when raised 5 m.

Use:

g = 10 N/kg

m = ΔEₚ/(gΔh)

m = 1500/(10 × 5)

m = 1500/50

m = 30 kg

---

Roller Coasters

Roller coasters provide an excellent example of energy transfer.

At the top of a hill:

• gravitational potential energy is high
• kinetic energy may be relatively low

As the coaster moves downhill:

gravitational potential energy → kinetic energy

The coaster speeds up.

At the bottom:

• gravitational potential energy is lower
• kinetic energy is greater

 

 

 

6

Why the First Roller-Coaster Hill Is Often High

The first major hill gives the coaster a large gravitational potential energy store.

As the coaster descends, this energy can be transferred into kinetic energy.

However, some energy is continuously transferred to the surroundings through:

• friction
• air resistance
• sound

Therefore, without additional energy input, later hills usually cannot be higher than the initial hill.

---

Hydroelectric Dams

Water stored high behind a dam has gravitational potential energy.

When the water is released:

gravitational potential energy → kinetic energy of water

The moving water turns turbines:

kinetic energy → rotational kinetic energy

The generator then transfers energy electrically.

A simplified pathway is:

gravitational potential → kinetic → mechanical → electrical

 

 

 

5

Why Dam Height Matters

From:

ΔEₚ = mgΔh

greater height means more gravitational potential energy for the same mass of water.

Therefore, a larger vertical drop can provide more energy per kilogram of water.

This height difference is often called the head in hydroelectric systems.

---

Pumped Hydroelectric Storage

Gravitational potential energy can also be used to store energy.

When excess electricity is available:

electricity → pumps → water moved uphill

The water gains gravitational potential energy.

Later:

water moves downhill → turbines → electricity

A simplified sequence is:

electrical energy → gravitational potential energy → electrical energy

This is called pumped hydroelectric storage.

---

Cranes and Lifting

A crane transfers energy when it raises a load.

Suppose a crane lifts a 2000 kg load through 15 m.

Using:

g = 9.8 N/kg

ΔEₚ = 2000 × 9.8 × 15

ΔEₚ = 294,000 J

or:

294 kJ

At least this much energy must be transferred to the gravitational potential energy store of the load.

Real cranes require more energy because they are not 100% efficient.

---

Elevators

An elevator also demonstrates gravitational potential energy.

When an elevator rises:

electrical energy → gravitational potential energy + other energy transfers

When it descends:

gravitational potential energy decreases

Some modern elevator systems can recover part of this energy using regenerative systems.

---

Human Body and GPE

When you climb stairs, your body gains gravitational potential energy.

Suppose a 60 kg person climbs vertically 4 m.

Using:

g = 9.8 N/kg

ΔEₚ = 60 × 9.8 × 4

ΔEₚ = 2352 J

or approximately:

2.35 kJ

The person's body must transfer at least this much energy mechanically to increase gravitational potential energy.

Because the body is not perfectly efficient, the chemical energy used is greater.

---

GPE and Mountains

Climbing a mountain increases gravitational potential energy.

For a 70 kg climber gaining 1000 m of vertical height:

ΔEₚ = 70 × 9.8 × 1000

ΔEₚ = 686,000 J

ΔEₚ = 686 kJ

This represents the increase in gravitational potential energy of the climber alone.

---

Gravitational Potential Energy Beyond Earth's Surface

The equation:

ΔEₚ = mgΔh

works well near Earth's surface when gravitational field strength is approximately constant.

Over very large distances, however, g changes significantly.

For satellites, planets, and spacecraft, a more general gravitational potential energy equation is required.

For ordinary near-Earth calculations, however:

ΔEₚ = mgΔh

is an excellent approximation.

---

Graph of GPE Against Height

For constant mass and gravitational field strength:

ΔEₚ ∝ Δh

Therefore, a graph of gravitational potential energy against height is a straight line.

If height doubles:

GPE doubles

If height triples:

GPE triples

This differs from kinetic energy against speed, which produces a curved relationship because speed is squared.

---

Graph of GPE Against Mass

For constant height and gravitational field strength:

ΔEₚ ∝ m

The graph is again a straight line.

Doubling the mass doubles the gravitational potential energy change.

---

Real-World Comparison

Imagine two roller-coaster cars at the same height.

Car A

Mass = 500 kg

Car B

Mass = 1000 kg

Car B has twice the mass.

Therefore, at the same height:

Car B has twice the gravitational potential energy relative to the same reference level.

If both descend the same vertical distance, Car B loses twice as much gravitational potential energy.

---

Common Mistakes

Using the Total Height Instead of the Change in Height

If an object moves from 4 m to 10 m:

Incorrect:

Δh = 10 m

Correct:

Δh = 10 − 4 = 6 m

---

Using Distance Instead of Vertical Height

If an object moves 20 m along a ramp but rises only 5 m:

Use:

Δh = 5 m

---

Forgetting Mass Must Be in Kilograms

Convert grams to kilograms first.

500 g = 0.500 kg

---

Confusing Mass and Weight

Mass is measured in:

kg

Weight is measured in:

N

Remember:

W = mg

---

Confusing g with Mass

In:

ΔEₚ = mgΔh

g represents gravitational field strength, not grams.

---

Forgetting the Sign of Δh

Moving upward:

Δh > 0

Moving downward:

Δh < 0

---

Assuming Gravitational Potential Energy Disappears During a Fall

Energy is transferred.

Typically:

gravitational potential → kinetic

with some energy possibly transferred thermally because of air resistance.

---

Using Horizontal Distance

Moving horizontally at the same height does not change gravitational potential energy.

If:

Δh = 0

then:

ΔEₚ = 0

---

Check Your Understanding

1. Define gravitational potential energy.
2. State the equation for the change in gravitational potential energy.
3. What does Δ mean?
4. State the units for mass, gravitational field strength, height, and energy.
5. Calculate the GPE gained by a 5 kg object raised 6 m. Use g = 9.8 N/kg.
6. Calculate the GPE gained by a 20 kg object raised 3 m. Use g = 10 N/kg.
7. A 500 g ball is raised 4 m. Calculate its increase in GPE using g = 9.8 N/kg.
8. Explain what happens to GPE when mass doubles.
9. Explain what happens to GPE when height doubles.
10. An object moves from 3 m above the ground to 11 m above the ground. What is Δh?
11. A 10 kg object moves from 12 m to 5 m. Calculate ΔEₚ using g = 9.8 N/kg.
12. What does a negative value of ΔEₚ mean?
13. An object gains 980 J of GPE when raised 5 m. Calculate its mass.
14. A 25 kg object gains 4900 J of GPE. Calculate the height through which it was raised.
15. A ramp is 12 m long but raises an object vertically by 3 m. Which value should be used for Δh?
16. Explain why different paths to the same final height produce the same change in GPE.
17. Describe the energy transfers as a ball falls.
18. Explain how a roller coaster demonstrates transfers between GPE and kinetic energy.
19. Explain how a hydroelectric dam uses gravitational potential energy.
20. A 1000 kg roller-coaster car descends 20 m. Calculate the GPE lost using g = 9.8 N/kg.
21. Ignoring friction, how much kinetic energy would the coaster gain?
22. Explain why later roller-coaster hills are usually lower than the first major hill.
23. Explain why water stored higher behind a dam has greater potential to generate energy.
24. Describe how pumped hydroelectric storage stores energy.
25. Compare the effects of doubling height on GPE with doubling speed on kinetic energy.

---

Key Terms

Gravitational potential energy (GPE) — Energy associated with an object's position in a gravitational field.

Energy store — A way of describing where energy is held within a physical system.

Joule (J) — SI unit of energy.

Mass — Measure of the amount of matter in an object, measured in kilograms.

Weight — Gravitational force acting on an object.

Gravitational field — Region in which an object with mass experiences a gravitational force.

Gravitational field strength (g) — Gravitational force per unit mass, measured in N/kg.

Height — Vertical position relative to a chosen reference level.

Change in height (Δh) — Difference between final and initial vertical position.

Delta (Δ) — Symbol meaning "change in."

Reference level — Chosen position from which height or gravitational potential energy is measured.

Energy transfer — Movement of energy between energy stores or systems.

Kinetic energy — Energy associated with motion.

Conservation of energy — Principle that energy cannot be created or destroyed, although it can be transferred between stores.

Head — Vertical height difference available to water in a hydroelectric system.

Pumped hydroelectric storage — Energy-storage system that uses electricity to pump water uphill and later recovers energy as the water flows downhill.

---

Key Takeaways

• Gravitational potential energy is associated with position in a gravitational field.
• Raising an object increases its gravitational potential energy.
• Lowering an object decreases its gravitational potential energy.
• Near Earth's surface, the change in gravitational potential energy is calculated using ΔEₚ = mgΔh.
• Mass must normally be measured in kilograms.
• Height must be the vertical change in height, not the total distance traveled.
• Gravitational field strength near Earth's surface is approximately 9.8 N/kg.
• Some school problems approximate g as 10 N/kg.
• Doubling mass doubles the change in gravitational potential energy.
• Doubling height doubles the change in gravitational potential energy.
• A stronger gravitational field produces a greater change in gravitational potential energy for the same mass and height.
• Moving upward produces a positive ΔEₚ.
• Moving downward produces a negative ΔEₚ.
• When an object falls, gravitational potential energy can be transferred into kinetic energy.
• Air resistance can transfer some energy to thermal stores in the object and surroundings.
• Energy is conserved during these transfers.
• Roller coasters continually transfer energy between gravitational potential and kinetic energy stores.
• Water stored behind a dam possesses gravitational potential energy that can ultimately be transferred into electrical energy.
• Pumped hydroelectric systems use gravitational potential energy to store energy.
• The equation ΔEₚ = mgΔh is most appropriate near Earth's surface, where g can be treated as approximately constant.

The central relationship is:

ΔEₚ = mgΔh

So:

mass × 2 → ΔEₚ × 2

height × 2 → ΔEₚ × 2

gravitational field strength × 2 → ΔEₚ × 2

Unlike kinetic energy, none of these quantities is squared. Gravitational potential energy near Earth's surface therefore changes linearly with mass, gravitational field strength, and vertical height.

 

 

 

 

3. Elastic Potential Energy

Elastic Potential Energy

Elastic potential energy is energy stored when an elastic object is stretched or compressed.

When a force changes the shape of an elastic object, work is done on the object and energy is transferred into its elastic potential energy store.

When the object is released, this stored energy can be transferred into other forms, especially kinetic energy.

Examples include:

• a stretched spring
• a compressed spring
• a stretched elastic band
• a drawn bow
• a trampoline
• a bungee cord
• a spring-loaded toy
• a pole used in pole vaulting

 

 

 

5

Elastic Deformation

An object undergoes deformation when its shape or size changes.

A spring can be:

stretched → extension

or:

compressed → compression

If the object returns to its original shape when the force is removed, the deformation is elastic.

For example:

normal spring → stretch → release → normal spring

Energy can be stored while the spring is deformed and released when it returns toward its original shape.

Elastic and Inelastic Deformation

Not all deformation is elastic.

Elastic deformation

The object returns to its original shape when the force is removed.

Examples:

• slightly stretched spring
• elastic band
• trampoline spring

Inelastic deformation

The object does not completely return to its original shape.

Examples:

• permanently bent wire
• crushed metal can
• overstretched spring

The elastic potential energy equation applies most simply when the object behaves elastically and follows Hooke's law.

Natural Length

The natural length of a spring is its length when it is neither stretched nor compressed.

The deformation, x, is measured from this natural position.

For a stretched spring:

extension = stretched length − natural length

For a compressed spring:

compression = natural length − compressed length

For energy calculations, x represents the magnitude of the deformation.

The Elastic Potential Energy Equation

For an ideal spring obeying Hooke's law:

Eₑ = ½kx²

where:

• Eₑ = elastic potential energy in joules (J)
• k = spring constant in newtons per metre (N/m)
• x = extension or compression in metres (m)

This equation tells us that elastic potential energy depends on:

• the stiffness of the spring
• the amount of deformation

Spring Constant

The spring constant, k, measures the stiffness of a spring.

A large value of k means:

stiffer spring

A small value of k means:

less stiff spring

For example:

Spring A:

k = 50 N/m

Spring B:

k = 500 N/m

Spring B is much stiffer.

It requires a greater force to produce the same extension.

Hooke's Law

For an ideal spring within its elastic region:

F = kx

where:

• F = force in newtons (N)
• k = spring constant in N/m
• x = extension or compression in metres

The restoring force itself acts in the direction opposite to the deformation, which can be represented more fully as:

F = −kx

The negative sign indicates direction.

For many introductory magnitude calculations, we simply use:

F = kx

Connecting Hooke's Law and Elastic Energy

As a spring stretches, the force required to stretch it increases.

At:

x = 0

the stretching force is:

F = 0

At maximum extension x:

F = kx

The average force during the stretch is therefore:

F_average = ½kx

Work done is:

work = average force × distance

Therefore:

Eₑ = ½kx × x

which gives:

Eₑ = ½kx²

This is why the factor ½ appears in the elastic potential energy equation.

Force-Extension Graph

For a spring obeying Hooke's law, force is directly proportional to extension.

A force-extension graph is therefore a straight line through the origin.

The gradient is:

gradient = F/x = k

So the gradient represents the spring constant.

The area under the force-extension graph represents the work done stretching the spring, which equals the elastic potential energy stored.

Because the area is triangular:

area = ½ × base × height

Therefore:

Eₑ = ½ × x × F

Since:

F = kx

we obtain:

Eₑ = ½kx²

Worked Example: Stretched Spring

A spring has:

k = 200 N/m

and is stretched:

x = 0.10 m

Calculate the elastic potential energy.

Eₑ = ½kx²

Eₑ = 0.5 × 200 × 0.10²

Square the extension:

0.10² = 0.01

Therefore:

Eₑ = 0.5 × 200 × 0.01

Eₑ = 1 J

The spring stores 1 J of elastic potential energy.

Worked Example: Compressed Spring

A spring has a spring constant of:

k = 500 N/m

It is compressed by:

x = 0.20 m

Calculate the stored elastic potential energy.

Eₑ = ½kx²

Eₑ = 0.5 × 500 × 0.20²

Eₑ = 0.5 × 500 × 0.04

Eₑ = 10 J

The compressed spring stores 10 J.

A compressed spring can store elastic potential energy just as a stretched spring can.

 

 

 

6

How Spring Stiffness Affects Energy

Consider two springs stretched by the same amount.

Spring A

k = 100 N/m

x = 0.20 m

Eₑ = 0.5 × 100 × 0.20²

Eₑ = 2 J

Spring B

k = 200 N/m

x = 0.20 m

Eₑ = 0.5 × 200 × 0.20²

Eₑ = 4 J

Doubling the spring constant doubles the stored energy when extension remains constant.

Therefore:

k × 2 → Eₑ × 2

Elastic potential energy is directly proportional to the spring constant.

How Deformation Affects Energy

The effect of extension or compression is much stronger because x is squared.

Suppose:

Eₑ = ½kx²

If the deformation doubles:

x → 2x

then:

Eₑ = ½k(2x)²

Since:

(2x)² = 4x²

the stored energy becomes four times greater.

Therefore:

deformation × 2 → elastic energy × 4

Comparing Different Deformations

For the same spring:

This is a squared relationship.

It is similar to the relationship between speed and kinetic energy.

Worked Example: Doubling Extension

A spring stores 5 J when stretched by a certain distance.

What happens if the extension is doubled?

Since:

Eₑ ∝ x²

then:

new Eₑ = 2² × 5

new Eₑ = 4 × 5

new Eₑ = 20 J

Doubling the extension increases the stored energy from 5 J to 20 J.

Worked Example: Tripling Compression

A compressed spring stores 8 J.

The compression is increased to three times its original value.

new Eₑ = 3² × 8

new Eₑ = 9 × 8

new Eₑ = 72 J

Tripling the deformation produces nine times the elastic potential energy.

Why Deformation Has Such a Large Effect

The equation contains:

x²

This means that elastic potential energy does not increase linearly with extension.

For example:

x = 0.10 m → x² = 0.0100 m²

x = 0.20 m → x² = 0.0400 m²

Although extension doubled, x² became four times larger.

This explains why highly stretched or compressed elastic systems can store much more energy.

Converting Centimetres to Metres

Extension is often given in centimetres.

Before using:

Eₑ = ½kx²

convert x to metres.

Remember:

100 cm = 1 m

Therefore:

x in metres = x in centimetres ÷ 100

For example:

5 cm = 0.05 m

20 cm = 0.20 m

35 cm = 0.35 m

Worked Example: Extension in Centimetres

A spring has:

k = 400 N/m

and is stretched by:

5 cm

First convert:

5 cm = 0.05 m

Then:

Eₑ = 0.5 × 400 × 0.05²

Eₑ = 0.5 × 400 × 0.0025

Eₑ = 0.5 J

The spring stores 0.5 J.

Worked Example: Larger Spring

A spring with:

k = 800 N/m

is compressed by:

15 cm

Convert:

15 cm = 0.15 m

Calculate:

Eₑ = 0.5 × 800 × 0.15²

Eₑ = 400 × 0.0225

Eₑ = 9 J

The spring stores 9 J.

Rearranging the Equation

Starting with:

Eₑ = ½kx²

we can rearrange it to calculate other quantities.

Finding the Spring Constant

Multiply by 2:

2Eₑ = kx²

Therefore:

k = 2Eₑ/x²

Worked Example: Finding k

A spring stores 16 J when stretched by 0.40 m.

Calculate its spring constant.

k = 2Eₑ/x²

k = (2 × 16)/(0.40²)

k = 32/0.16

k = 200 N/m

Finding Extension

Starting with:

Eₑ = ½kx²

Multiply by 2:

2Eₑ = kx²

Divide by k:

x² = 2Eₑ/k

Take the square root:

x = √(2Eₑ/k)

Worked Example: Finding Extension

A spring has:

k = 250 N/m

and stores:

Eₑ = 5 J

Calculate its extension.

x = √(2Eₑ/k)

x = √((2 × 5)/250)

x = √0.04

x = 0.20 m

The spring is stretched by 0.20 m, or 20 cm.

Energy Transfers in a Spring

When you stretch a spring:

chemical energy in muscles → elastic potential energy

When the spring is released:

elastic potential energy → kinetic energy

In a real system, some energy may also be transferred into:

• thermal energy
• sound

Therefore, not all stored elastic energy necessarily becomes useful kinetic energy.

Spring-Loaded Toy

Imagine compressing a spring inside a toy launcher.

While compressing the spring:

work is done on spring → elastic potential energy increases

When released:

elastic potential energy → kinetic energy of projectile

If the spring stores 20 J, the projectile could theoretically gain up to 20 J of kinetic energy if energy losses were negligible.

Elastic Energy and Kinetic Energy

If energy losses are negligible:

elastic potential energy lost = kinetic energy gained

Therefore:

½kx² = ½mv²

This can be used to connect spring deformation with the speed of a launched object.

Worked Example: Spring Launcher

A spring stores 10 J of elastic potential energy.

It launches a 0.20 kg object.

Ignoring energy losses:

Eₖ = 10 J

Using:

Eₖ = 0.5mv²

10 = 0.5 × 0.20 × v²

10 = 0.10v²

v² = 100

v = 10 m/s

The object leaves with a speed of 10 m/s.

Elastic Energy and Gravitational Potential Energy

Elastic potential energy can also be transferred into gravitational potential energy.

For example, a spring can launch an object upward.

The energy transfer might be:

elastic potential → kinetic → gravitational potential

Ignoring losses:

½kx² = mgΔh

at the maximum height, if the object momentarily stops.

Worked Example: Spring Launching Upward

A spring stores 20 J.

It launches a 0.50 kg object vertically upward.

Ignoring energy losses:

elastic energy = gained gravitational potential energy

20 = mgΔh

Using:

g = 10 N/kg

20 = 0.50 × 10 × Δh

20 = 5Δh

Δh = 4 m

The stored elastic energy could raise the object approximately 4 m above its starting point under these simplified conditions.

Bows and Arrows

When an archer pulls back a bowstring, the bow bends.

Energy is transferred from the archer into the elastic potential energy store of the bow.

When released:

elastic potential energy → kinetic energy of arrow

 

 

 

4

A greater deformation can store more energy, but only while the bow remains within its safe operating range.

Trampolines

When a person lands on a trampoline:

kinetic energy → elastic potential energy

The trampoline stretches.

As it returns toward its original shape:

elastic potential energy → kinetic energy

The person accelerates upward.

Some energy is transferred to:

• thermal energy
• sound

so the system is not perfectly efficient.

 

 

 

6

Bungee Jumping

A bungee jump involves several energy transfers.

Initially:

gravitational potential energy

As the jumper falls:

gravitational potential → kinetic

As the cord stretches:

kinetic + gravitational potential → elastic potential

At maximum extension, the cord contains substantial elastic potential energy.

The jumper then rebounds as some of that energy is transferred back into kinetic and gravitational potential energy.

Pole Vaulting

A pole vaulter runs toward the bar with kinetic energy.

As the pole bends:

kinetic energy → elastic potential energy

As the pole straightens:

elastic potential energy → kinetic + gravitational potential energy

This helps raise the athlete over the bar.

 

 

 

5

Vehicle Suspension

Vehicle suspension systems use springs.

When a wheel passes over a bump:

• the spring compresses
• elastic potential energy is stored
• the spring then expands

However, a vehicle also contains dampers or shock absorbers.

These reduce repeated bouncing by transferring mechanical energy into thermal energy.

Without damping, the vehicle could continue oscillating for much longer.

Elastic Limit

Springs do not behave elastically under all conditions.

The elastic limit is the point beyond which an object may no longer return completely to its original shape.

If a spring is stretched too far:

• it may become permanently deformed
• Hooke's law may no longer apply
• the simple elastic energy equation may no longer accurately describe its behaviour

Therefore, the equation:

Eₑ = ½kx²

is most appropriate when the spring behaves according to Hooke's law.

Limit of Proportionality

The limit of proportionality is the point beyond which force is no longer directly proportional to extension.

Before this point:

F ∝ x

and:

F = kx

After this point, the force-extension graph begins to curve.

The spring constant can no longer be treated as constant in the same simple way.

Force-Extension Graph Beyond the Linear Region

In the Hooke's law region:

force-extension graph = straight line

Beyond the limit of proportionality:

graph becomes curved

The elastic energy stored is still related to the area under the force-extension graph, but the simple equation:

Eₑ = ½kx²

may no longer apply.

This is an important limitation of the equation.

Comparing Energy Equations

Three important mechanical energy equations are:

Kinetic Energy

Eₖ = ½mv²

Depends on:

• mass
• speed²

Gravitational Potential Energy

ΔEₚ = mgΔh

Depends on:

• mass
• gravitational field strength
• vertical height change

Elastic Potential Energy

Eₑ = ½kx²

Depends on:

• spring stiffness
• deformation²

Notice the similarity:

Eₖ ∝ v²

and:

Eₑ ∝ x²

Both contain a squared quantity.

Graph of Elastic Energy Against Extension

Since:

Eₑ ∝ x²

a graph of elastic potential energy against extension is curved.

At:

x = 0

Eₑ = 0

As x increases, Eₑ increases more and more rapidly.

Therefore:

extension × 2 → energy × 4

extension × 3 → energy × 9

Graph of Elastic Energy Against Spring Constant

If extension remains constant:

Eₑ ∝ k

Therefore, a graph of elastic potential energy against spring constant is a straight line through the origin.

Double k:

double Eₑ

Triple k:

triple Eₑ

Percentage Changes

Suppose extension increases by 20%.

New extension:

1.2x

Because:

Eₑ ∝ x²

the new energy is:

(1.2)²Eₑ

= 1.44Eₑ

Therefore, a 20% increase in extension produces a 44% increase in elastic potential energy.

Another Percentage Example

Suppose extension decreases by 50%.

New extension:

0.5x

Therefore:

new Eₑ = (0.5)²Eₑ

new Eₑ = 0.25Eₑ

Halving the extension reduces the stored elastic potential energy to 25% of its original value.

Common Mistakes

Forgetting to Square the Extension

Incorrect:

Eₑ = ½kx

Correct:

Eₑ = ½kx²

Squaring the Spring Constant

Incorrect:

Eₑ = ½k²x

Only x is squared.

Forgetting the ½

Incorrect:

Eₑ = kx²

Correct:

Eₑ = ½kx²

Using Centimetres Instead of Metres

If:

x = 20 cm

use:

x = 0.20 m

before calculating.

Using the Total Length of the Spring

x is the change from the natural length, not the total spring length.

For example:

Natural length = 30 cm
Stretched length = 40 cm

Then:

x = 40 − 30 = 10 cm

or:

x = 0.10 m

Assuming Doubling Extension Doubles Energy

Incorrect:

x × 2 → Eₑ × 2

Correct:

x × 2 → Eₑ × 4

because x is squared.

Confusing Force with Energy

Force is measured in:

newtons (N)

Energy is measured in:

joules (J)

Confusing k with Kinetic Energy

In the elastic energy equation:

k = spring constant

It does not mean kinetic energy.

Assuming the Equation Works for Any Amount of Stretching

The equation:

Eₑ = ½kx²

assumes the spring follows Hooke's law.

If the spring is stretched beyond its proportional or elastic behaviour, the simple equation may no longer apply.

Check Your Understanding

1. Define elastic potential energy.
2. Give four examples of systems that can store elastic potential energy.
3. What is meant by elastic deformation?
4. Explain the difference between elastic and inelastic deformation.
5. State the equation for elastic potential energy.
6. State the units of Eₑ, k, and x.
7. What does the spring constant tell us about a spring?
8. Calculate the elastic potential energy stored in a spring with k = 100 N/m stretched by 0.20 m.
9. A spring has k = 300 N/m and is compressed by 0.10 m. Calculate the stored energy.
10. A spring is stretched by 5 cm. What value of x should be used in the equation?
11. A spring has k = 500 N/m and is stretched by 8 cm. Calculate the stored elastic energy.
12. Explain what happens to stored energy if the spring constant doubles while extension remains constant.
13. Explain what happens to stored energy if extension doubles.
14. What happens to elastic potential energy if extension triples?
15. A spring stores 4 J at extension x. How much energy would it store at extension 2x?
16. A spring stores 6 J at extension x. How much would it store at extension 3x?
17. A spring stores 18 J when stretched by 0.30 m. Calculate its spring constant.
18. A spring with k = 400 N/m stores 8 J. Calculate its extension.
19. Explain how a bow stores and releases elastic potential energy.
20. Describe the energy transfers involved when someone uses a trampoline.
21. Describe the energy transfers during a bungee jump.
22. Explain how a pole vault demonstrates elastic energy storage.
23. What does the area under a force-extension graph represent?
24. Explain why the equation Eₑ = ½kx² contains the factor ½.
25. What is the limit of proportionality?
26. Why might Eₑ = ½kx² become inaccurate for a severely stretched spring?
27. Compare the effect of doubling speed on kinetic energy with doubling extension on elastic potential energy.
28. A spring launcher stores 25 J. Ignoring losses, how much kinetic energy could be transferred to the projectile?
29. A spring transfers 40 J to a 0.50 kg projectile. Calculate the projectile's speed, ignoring energy losses.
30. Explain why elastic potential energy is useful in mechanical systems.

Key Terms

Elastic potential energy — Energy stored when an elastic object is stretched or compressed.

Deformation — Change in the shape or size of an object.

Elastic deformation — Deformation that disappears when the deforming force is removed.

Inelastic deformation — Permanent deformation that remains after the force is removed.

Extension — Increase in an object's length compared with its natural length.

Compression — Decrease in an object's length compared with its natural length.

Natural length — Length of an elastic object when no deforming force acts on it.

Spring constant (k) — Measure of the stiffness of a spring, measured in N/m.

Hooke's law — Relationship stating that force is proportional to extension for an ideal spring within its proportional region.

Restoring force — Force produced by an elastic object that acts toward its equilibrium position.

Equilibrium position — Position of an elastic system when it is not displaced from its normal state.

Limit of proportionality — Point beyond which force is no longer directly proportional to extension.

Elastic limit — Point beyond which an object may become permanently deformed.

Force-extension graph — Graph showing how the force applied to an elastic object changes with its deformation.

Work done — Energy transferred when a force causes displacement.

Joule (J) — SI unit of energy.

Newton (N) — SI unit of force.

Energy transfer — Movement of energy between stores or systems.

Key Takeaways

• Elastic potential energy is stored when an elastic object is stretched or compressed.
• Springs, elastic bands, bows, trampolines, bungee cords, and flexible poles can store elastic potential energy.
• For an ideal spring obeying Hooke's law, elastic potential energy is calculated using Eₑ = ½kx².
• Eₑ is measured in joules.
• k is the spring constant and is measured in N/m.
• x is the extension or compression measured from the object's natural position.
• x must be expressed in metres when using standard SI units.
• A larger spring constant means a stiffer spring.
• Doubling the spring constant doubles the elastic energy if deformation remains constant.
• Elastic potential energy depends on the square of the deformation.
• Doubling extension produces four times the elastic potential energy.
• Tripling extension produces nine times the elastic potential energy.
• Halving extension reduces the stored energy to one-quarter.
• The area under a force-extension graph represents the work done and therefore the elastic potential energy stored.
• Elastic potential energy can be transferred into kinetic energy when a spring or other elastic object is released.
• Elastic energy can also be transferred into gravitational potential energy.
• Real systems also transfer some energy into thermal energy and sound.
• Hooke's law only applies over the region in which force is proportional to extension.
• If an elastic object is stretched too far, it may become permanently deformed.
• The simple equation Eₑ = ½kx² should therefore be used only when the spring behaves appropriately.

The central relationship is:

Eₑ = ½kx²

So:

spring constant × 2 → elastic energy × 2

but:

extension × 2 → elastic energy × 4

and:

extension × 3 → elastic energy × 9

Like kinetic energy, elastic potential energy contains a squared quantity, so relatively small increases in deformation can produce much larger increases in stored energy.

4. Conservation of Mechanical Energy

Conservation of Mechanical Energy

Mechanical energy is the energy associated with an object's motion and position.

The main mechanical energy stores considered here are:

• kinetic energy — energy associated with motion
• gravitational potential energy — energy associated with position in a gravitational field
• elastic potential energy — energy stored when an elastic object is stretched or compressed

In an ideal system where no mechanical energy is transferred to the surroundings:

total mechanical energy remains constant

This is the principle of conservation of mechanical energy.

Mechanical energy can change from one form to another, but the total remains the same.

 

 

 

6

Mechanical Energy

The total mechanical energy of a system can include:

Emechanical = Eₖ + Eₚ + Eₑ

where:

• Eₖ = kinetic energy
• Eₚ = gravitational potential energy
• Eₑ = elastic potential energy

The equations we have already studied are:

Eₖ = ½mv²

ΔEₚ = mgΔh

Eₑ = ½kx²

These three equations can be combined using conservation of mechanical energy.

---

Conservation of Mechanical Energy

Consider a system at two different times.

If there are no energy losses:

initial mechanical energy = final mechanical energy

Therefore:

Eₖ(initial) + Eₚ(initial) + Eₑ(initial) = Eₖ(final) + Eₚ(final) + Eₑ(final)

Not every problem contains all three energy stores.

For example, a falling object might involve only:

gravitational potential energy ↔ kinetic energy

A spring launcher might involve:

elastic potential energy ↔ kinetic energy

A spring launching an object upward might involve:

elastic potential → kinetic → gravitational potential

---

Conservation of Energy vs Conservation of Mechanical Energy

These ideas are related but not identical.

The conservation of energy states that energy cannot be created or destroyed.

Energy may be:

• transferred
• transformed
• stored in different ways

The total energy of a closed system remains constant.

Mechanical energy, however, can decrease when energy is transferred from mechanical stores into other forms.

For example:

kinetic energy → thermal energy

because of friction.

Total energy is still conserved, but mechanical energy is not.

This distinction is extremely important.

---

Energy Transformations

Mechanical energy can move between different stores.

Falling Object

gravitational potential → kinetic

Object Thrown Upward

kinetic → gravitational potential

Spring Launcher

elastic potential → kinetic

Object Compressing a Spring

kinetic → elastic potential

Roller Coaster

gravitational potential ↔ kinetic

Pendulum

gravitational potential ↔ kinetic

These transformations can occur repeatedly.

---

Falling Objects

Consider a ball held above the ground.

At the top:

• gravitational potential energy is high
• kinetic energy is zero if the ball is stationary

As the ball falls:

• gravitational potential energy decreases
• kinetic energy increases

Ignoring air resistance:

GPE lost = KE gained

Therefore:

mgΔh = ½mv²

 

 

 

5

Worked Example: Falling Ball

A 2 kg ball falls from rest through a vertical distance of 5 m.

Use:

g = 9.8 N/kg

Initial kinetic energy:

Eₖ = 0 J

Gravitational potential energy lost:

ΔEₚ = mgΔh

ΔEₚ = 2 × 9.8 × 5

ΔEₚ = 98 J

Ignoring air resistance:

KE gained = 98 J

Therefore, just before reaching the lower position:

Eₖ = 98 J

---

Finding Speed Using Energy

We can use conservation of energy to calculate the ball's speed.

We know:

mgΔh = ½mv²

For a fall of 5 m:

2 × 9.8 × 5 = 0.5 × 2 × v²

98 = v²

Therefore:

v = √98

v ≈ 9.9 m/s

The ball reaches approximately 9.9 m/s.

---

Why Mass Cancels

Starting with:

mgΔh = ½mv²

m appears on both sides.

Cancel m:

gΔh = ½v²

Therefore:

v = √(2gΔh)

This means that, ignoring air resistance, objects dropped from the same height reach the same speed regardless of their mass.

A heavier object begins with more gravitational potential energy, but it also requires proportionally more kinetic energy to reach the same speed.

---

Energy at Different Points During a Fall

Suppose an object starts with 100 J of GPE and 0 J of KE.

Ignoring air resistance:

Position	GPE	KE	Total Mechanical Energy
Top	100 J	0 J	100 J
Partway down	75 J	25 J	100 J
Lower	40 J	60 J	100 J
Bottom	0 J	100 J	100 J

GPE decreases while KE increases.

But:

total mechanical energy = 100 J throughout

---

Objects Moving Upward

The transformation works in the opposite direction when an object moves upward.

Suppose a ball is thrown vertically upward.

Initially:

• kinetic energy is high
• gravitational potential energy is relatively low

As it rises:

kinetic → gravitational potential

The ball slows down.

At maximum height:

v = 0

so:

Eₖ = 0

The initial kinetic energy has been transferred into gravitational potential energy, assuming air resistance is negligible.

---

Worked Example: Ball Thrown Upward

A 0.50 kg ball is thrown upward at 10 m/s.

Initial kinetic energy:

Eₖ = ½mv²

Eₖ = 0.5 × 0.50 × 10²

Eₖ = 25 J

At maximum height:

Eₖ = 0

Ignoring air resistance:

GPE gained = 25 J

Use:

ΔEₚ = mgΔh

25 = 0.50 × 9.8 × Δh

25 = 4.9Δh

Therefore:

Δh = 25/4.9

Δh ≈ 5.1 m

The ball rises approximately 5.1 m.

---

Roller Coasters

Roller coasters are excellent examples of mechanical energy transformations.

At the top of a hill:

• GPE is high
• KE may be relatively low

As the coaster descends:

GPE → KE

The coaster speeds up.

At the bottom:

• GPE is relatively low
• KE is high

As the coaster climbs another hill:

KE → GPE

The coaster slows down.

 

 

 

6

Worked Example: Roller Coaster

A 500 kg roller-coaster car starts from rest at the top of a 20 m hill.

Ignore friction.

Use:

g = 9.8 N/kg

Initial GPE relative to the bottom:

Eₚ = mgh

Eₚ = 500 × 9.8 × 20

Eₚ = 98,000 J

Initial KE:

Eₖ = 0 J

Total mechanical energy:

98,000 J

At the bottom:

Eₚ = 0 J

Therefore:

Eₖ = 98,000 J

---

Finding the Roller Coaster's Speed

At the bottom:

Eₖ = ½mv²

Therefore:

98,000 = 0.5 × 500 × v²

98,000 = 250v²

v² = 392

v ≈ 19.8 m/s

The coaster reaches approximately:

19.8 m/s

if friction and air resistance are ignored.

---

Energy at an Intermediate Height

Suppose the same coaster reaches another point 8 m above the bottom.

Total mechanical energy remains:

98,000 J

GPE at 8 m:

Eₚ = 500 × 9.8 × 8

Eₚ = 39,200 J

Therefore:

Eₖ = 98,000 − 39,200

Eₖ = 58,800 J

Now calculate speed:

58,800 = 0.5 × 500 × v²

58,800 = 250v²

v² = 235.2

v ≈ 15.3 m/s

---

Why Later Roller-Coaster Hills Are Usually Lower

In an ideal frictionless system, a roller coaster released from rest could rise again to its original height.

In reality, some mechanical energy is continually transferred to:

• thermal energy through friction
• thermal energy through air resistance
• sound

Therefore, less mechanical energy remains available.

Without an additional energy input:

later hill < original hill

in most real roller-coaster designs.

---

Pendulums

A pendulum continually transfers energy between gravitational potential energy and kinetic energy.

At the highest point:

• GPE is maximum
• KE is zero momentarily

As the pendulum swings downward:

GPE → KE

At the lowest point:

• GPE is minimum
• KE is maximum
• speed is maximum

As it swings upward:

KE → GPE

 

 

 

6

Pendulum Energy Positions

Consider three positions:

Position A — Highest Point

GPE: maximum

KE: zero

Position B — Lowest Point

GPE: minimum

KE: maximum

Position C — Opposite Highest Point

GPE: maximum

KE: zero

In an ideal system:

total mechanical energy remains constant

---

Why a Real Pendulum Eventually Stops

A real pendulum does not swing forever.

Mechanical energy is gradually transferred by:

• air resistance
• friction at the pivot

into:

• thermal energy
• sound

Therefore, the pendulum reaches a slightly lower height on each swing.

Eventually, it stops.

Energy has not disappeared.

It has been transferred to the surroundings.

---

Elastic Potential Energy

Springs introduce another mechanical energy store:

elastic potential energy

For an ideal spring:

Eₑ = ½kx²

A compressed or stretched spring can transfer its stored elastic energy into kinetic energy.

---

Spring Launcher

Consider a compressed spring launching a cart.

Before release:

• elastic potential energy is high
• kinetic energy is zero

During release:

elastic potential → kinetic

When the spring reaches its natural length:

• elastic potential energy may be zero
• kinetic energy is maximum

assuming no energy losses.

 

 

 

5

Worked Example: Spring Launcher

A spring has:

k = 400 N/m

and is compressed by:

x = 0.20 m

Calculate the elastic potential energy.

Eₑ = ½kx²

Eₑ = 0.5 × 400 × 0.20²

Eₑ = 8 J

The spring launches a 0.25 kg cart.

Ignoring friction:

elastic energy lost = kinetic energy gained

Therefore:

Eₖ = 8 J

Use:

Eₖ = ½mv²

8 = 0.5 × 0.25 × v²

8 = 0.125v²

v² = 64

v = 8 m/s

The cart leaves the spring at 8 m/s.

---

Spring Launching an Object Upward

A spring can transfer energy through several mechanical stores.

Initially:

elastic potential energy

After release:

kinetic energy

As the object rises:

gravitational potential energy

The overall transformation is:

elastic potential → kinetic → gravitational potential

---

Worked Example: Spring to Height

A spring stores 30 J of elastic potential energy and launches a 0.60 kg object vertically upward.

Ignoring losses, at maximum height:

Eₖ = 0

Therefore:

elastic energy = gained GPE

30 = mgΔh

Using:

g = 9.8 N/kg

30 = 0.60 × 9.8 × Δh

30 = 5.88Δh

Δh ≈ 5.10 m

The object rises approximately 5.1 m above its starting point.

---

Multiple Energy Stores

Some problems involve all three mechanical energy stores.

For example, imagine a cart:

1. begins at the top of a hill
2. rolls downward
3. compresses a spring

Initially:

gravitational potential energy

During motion:

kinetic energy

At maximum spring compression:

elastic potential energy

The transformation is:

GPE → KE → elastic PE

Ignoring losses:

initial GPE = final elastic PE

Therefore:

mgh = ½kx²

---

Worked Example: Hill to Spring

A 2 kg cart begins from rest 5 m above a spring.

At the bottom, it compresses a spring with:

k = 1000 N/m

Ignore friction.

Initial GPE:

Eₚ = mgh

Eₚ = 2 × 9.8 × 5

Eₚ = 98 J

At maximum compression, the cart momentarily stops.

Therefore:

Eₖ = 0

All available mechanical energy is stored elastically:

98 = ½ × 1000 × x²

98 = 500x²

x² = 0.196

x ≈ 0.443 m

The spring compresses approximately 0.44 m.

---

A General Problem-Solving Method

Conservation-of-energy problems become much easier if you use a consistent process.

Identify the Initial State

Ask:

• Is the object moving?
• Is it elevated?
• Is a spring stretched or compressed?

Determine which energy stores are present.

Identify the Final State

Again determine:

• kinetic energy
• gravitational potential energy
• elastic potential energy

Write Conservation of Energy

For an ideal system:

initial mechanical energy = final mechanical energy

Remove Zero Terms

For example:

If the object starts from rest:

initial KE = 0

If the spring is initially undeformed:

initial elastic PE = 0

If the reference height is chosen as zero:

GPE = 0

Substitute Equations

Use:

Eₖ = ½mv²

ΔEₚ = mgΔh

Eₑ = ½kx²

Solve

Rearrange for the unknown quantity.

Check

Ask:

• Are the units correct?
• Is the answer physically reasonable?
• Did I include energy losses if the question mentions them?

---

Energy Bar Charts

Energy bar charts can help visualize conservation.

Suppose a falling object initially has:

100 J GPE

At the top:

GPE: 100 J
KE: 0 J

Halfway:

GPE: 50 J
KE: 50 J

At the bottom:

GPE: 0 J
KE: 100 J

The sizes of the individual stores change, but their total remains:

100 J

---

Friction

Friction transfers mechanical energy into thermal energy.

Consider a sliding object.

Initially it has kinetic energy.

As friction slows it:

kinetic energy → thermal energy

The object's mechanical energy decreases.

However:

total energy is still conserved

The "missing" mechanical energy has been transferred to the thermal energy stores of the object and surroundings.

---

Work Done Against Friction

If friction acts with approximately constant force:

energy transferred by friction = Fd

where:

• F = frictional force in N
• d = distance traveled in m

For a system involving friction:

initial mechanical energy = final mechanical energy + energy transferred thermally

A useful form is:

Emechanical(initial) = Emechanical(final) + Fd

---

Worked Example: Friction

A cart initially has:

500 J of mechanical energy

As it travels, friction transfers:

120 J

to thermal energy.

Remaining mechanical energy:

500 − 120 = 380 J

The cart now has:

380 J of mechanical energy

But total energy remains:

380 J mechanical + 120 J thermal = 500 J

---

Worked Example: Roller Coaster with Friction

A roller-coaster car begins with:

100,000 J of GPE

During the descent, 15,000 J is transferred to thermal energy through friction and air resistance.

Mechanical energy at the bottom:

100,000 − 15,000 = 85,000 J

If GPE at the bottom is zero:

Eₖ = 85,000 J

The coaster therefore moves more slowly than it would in an ideal frictionless system.

---

Air Resistance

Air resistance, or drag, is another force that transfers mechanical energy.

For a falling object:

Without air resistance:

GPE → KE

With air resistance:

GPE → KE + thermal energy

Therefore, the object gains less kinetic energy than predicted by a frictionless calculation.

 

 

 

6

Mechanical Energy Is Not Always Conserved

Mechanical energy is conserved when only conservative forces, such as gravity and ideal spring forces, transfer energy within the system.

Mechanical energy is not conserved when significant energy is transferred by forces such as:

• friction
• air resistance
• other forms of drag

However, total energy remains conserved.

This distinction should always be stated clearly.

---

Efficiency

Real machines and systems are not perfectly efficient.

Some energy is transferred into less useful forms.

Efficiency can be calculated using:

efficiency = useful energy output / total energy input

As a percentage:

efficiency (%) = useful energy output / total energy input × 100

---

Worked Example: Spring Launcher Efficiency

A spring initially stores:

50 J

The launched object receives:

40 J of kinetic energy

Efficiency:

efficiency = 40/50 × 100

efficiency = 80%

Energy transferred elsewhere:

50 − 40 = 10 J

That energy may have been transferred into:

• thermal energy
• sound
• vibration

---

Real-World Energy Transfers

Mechanical energy transformations occur constantly.

Hydroelectric Dam

gravitational potential → kinetic → mechanical → electrical

Bow and Arrow

elastic potential → kinetic

Bungee Jump

gravitational potential → kinetic → elastic potential

Trampoline

gravitational potential → kinetic → elastic potential → kinetic → gravitational potential

Pendulum

gravitational potential ↔ kinetic

Roller Coaster

gravitational potential ↔ kinetic

Spring Launcher

elastic potential → kinetic

---

Bungee Jumping

A bungee jump demonstrates all three mechanical energy stores.

At the platform:

• GPE is high
• KE is zero
• elastic PE is zero if the cord is slack

During the fall:

GPE → KE

When the cord stretches:

GPE + KE → elastic PE

Near maximum extension:

• speed becomes zero momentarily
• KE becomes zero
• elastic PE is large

The stretched cord then accelerates the jumper upward.

 

 

 

5

Pendulum with Friction

An ideal pendulum would continue swinging to the same height forever.

A real pendulum reaches slightly lower heights on successive swings.

Why?

Mechanical energy is transferred into:

• thermal energy at the pivot
• thermal energy through air resistance
• sound

Eventually:

mechanical energy → mostly thermal energy

and the pendulum stops.

---

Conservation of Energy Does Not Mean Constant Speed

A common misconception is that conservation of energy means an object's speed remains constant.

It does not.

For a roller coaster:

GPE decreases → KE increases → speed increases

Total mechanical energy can remain constant even while speed changes dramatically.

What remains constant is the sum of the mechanical energy stores, not each individual store.

---

Conservation Does Not Mean Each Energy Store Is Constant

Suppose:

Initial:

GPE = 100 J

KE = 0 J

Later:

GPE = 30 J

KE = 70 J

The individual stores changed.

But:

100 + 0 = 30 + 70

Therefore:

100 J = 100 J

Mechanical energy has been conserved.

---

Reference Heights

Gravitational potential energy depends on the chosen reference height.

Often we choose:

h = 0

at the lowest point of the motion.

The choice of reference does not change the physics as long as it is used consistently.

What matters in energy calculations is usually the change in height.

---

Common Mistakes

Saying Energy Is "Used Up"

Energy is not destroyed.

It is transferred or transformed.

---

Saying Friction Destroys Energy

Friction transfers mechanical energy into thermal energy.

Total energy remains conserved.

---

Assuming Mechanical Energy Is Always Conserved

Mechanical energy is conserved only when energy is not being transferred from mechanical stores into other forms.

---

Confusing Mechanical Energy with Total Energy

Mechanical energy includes:

• kinetic
• gravitational potential
• elastic potential

Thermal energy is not mechanical energy.

---

Forgetting Zero-Energy Conditions

If an object starts from rest:

Eₖ = 0

If a spring is undeformed:

Eₑ = 0

Using a chosen reference level:

Eₚ = 0

Recognizing these conditions can simplify calculations greatly.

---

Forgetting That Speed Is Squared

Kinetic energy:

Eₖ = ½mv²

not:

Eₖ = ½mv

---

Forgetting That Spring Deformation Is Squared

Elastic energy:

Eₑ = ½kx²

not:

Eₑ = ½kx

---

Using Path Length Instead of Vertical Height

For gravitational potential energy, use vertical height change.

---

Ignoring Friction When the Question Includes It

If friction transfers energy, include that transfer in the energy balance.

---

Check Your Understanding

1. Define mechanical energy.
2. State the principle of conservation of mechanical energy.
3. Name the three mechanical energy stores considered in this topic.
4. State the equations for kinetic, gravitational potential, and elastic potential energy.
5. Explain the difference between conservation of total energy and conservation of mechanical energy.
6. A ball loses 200 J of GPE while falling. Ignoring air resistance, how much KE does it gain?
7. A 2 kg object falls 10 m from rest. Calculate the GPE lost using g = 9.8 N/kg.
8. Use conservation of energy to determine its kinetic energy after falling 10 m.
9. Calculate its speed.
10. Explain why mass cancels when calculating the speed of an object falling from rest through a known height.
11. A roller coaster begins from rest 25 m above the bottom. Calculate its speed at the bottom, ignoring friction.
12. A coaster begins with 120,000 J of mechanical energy. At one point it has 45,000 J of GPE. How much KE does it have?
13. Describe the energy transformations during one complete swing of a pendulum.
14. Where does a pendulum have maximum kinetic energy?
15. Where does it have maximum gravitational potential energy?
16. Why does a real pendulum eventually stop?
17. A spring stores 18 J. It launches a 0.25 kg object. Calculate the object's speed, ignoring losses.
18. A spring with k = 500 N/m is compressed by 0.20 m. Calculate its elastic potential energy.
19. If all this energy is transferred to a 0.40 kg cart, calculate the cart's speed.
20. Describe the energy transformations during a bungee jump.
21. A system begins with 800 J of mechanical energy. Friction transfers 150 J into thermal energy. How much mechanical energy remains?
22. Explain why total energy is still conserved in Question 21.
23. A roller coaster begins with 200 kJ of GPE and loses 30 kJ to friction. How much kinetic energy could it have at the bottom?
24. A 3 kg object is launched vertically upward with 150 J of kinetic energy. Ignoring air resistance, calculate its maximum height above the starting point.
25. A 2 kg cart rolls from a height of 4 m and compresses a spring with k = 800 N/m. Calculate the maximum spring compression, ignoring friction.
26. Explain why conservation of mechanical energy does not mean speed remains constant.
27. Explain why later roller-coaster hills are generally lower than the first major hill.
28. A spring launcher receives 60 J of stored elastic energy but produces only 48 J of projectile kinetic energy. Calculate its efficiency.
29. Where did the remaining energy in Question 28 go?
30. Give one real-world example involving transformations among all three mechanical energy stores.

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Key Terms

Mechanical energy — Energy associated with motion and position, including kinetic and potential energy stores.

Conservation of mechanical energy — Principle that total mechanical energy remains constant when no mechanical energy is transferred to other forms.

Conservation of energy — Principle that energy cannot be created or destroyed but can be transferred between stores and systems.

Kinetic energy — Energy associated with motion.

Gravitational potential energy — Energy associated with position in a gravitational field.

Elastic potential energy — Energy stored when an elastic object is stretched or compressed.

Energy transformation — Change from one form or store of energy to another.

Energy transfer — Movement of energy between objects, systems, or stores.

Friction — Force opposing relative motion between surfaces that can transfer mechanical energy into thermal energy.

Air resistance — Drag force acting on an object moving through air.

Drag — Resistive force acting on an object moving through a fluid.

Thermal energy — Energy associated with the microscopic motion and interactions of particles.

Efficiency — Fraction of input energy transferred into the desired useful output.

Conservative force — Force for which energy can be stored as potential energy and recovered, such as gravity or an ideal spring force.

Energy dissipation — Spreading of energy into less concentrated, often less useful stores such as thermal energy in the surroundings.

Reference height — Chosen level from which gravitational height is measured.

Closed system — System that does not exchange energy with its surroundings.

---

Key Takeaways

• Mechanical energy includes kinetic, gravitational potential, and elastic potential energy.
• In an ideal system with no mechanical energy losses, total mechanical energy remains constant.
• Mechanical energy can be transformed between different stores without being destroyed.
• Falling objects transform gravitational potential energy into kinetic energy.
• Objects moving upward transform kinetic energy into gravitational potential energy.
• Springs can transform elastic potential energy into kinetic energy.
• Moving objects can compress springs and transfer kinetic energy into elastic potential energy.
• Roller coasters continuously exchange gravitational potential and kinetic energy.
• Pendulums continuously exchange gravitational potential and kinetic energy.
• Spring launchers transfer elastic potential energy into kinetic energy.
• Systems such as bungee jumps can involve all three mechanical energy stores.
• Conservation of mechanical energy can be used to calculate unknown speeds, heights, spring compressions, and energies.
• A useful general relationship is initial mechanical energy = final mechanical energy for an ideal system.
• Kinetic energy is calculated using Eₖ = ½mv².
• Changes in gravitational potential energy are calculated using ΔEₚ = mgΔh.
• Elastic potential energy is calculated using Eₑ = ½kx².
• Friction and air resistance transfer mechanical energy into thermal energy and sometimes sound.
• When friction is present, mechanical energy may decrease, but total energy is still conserved.
• Real roller coasters, pendulums, and spring systems therefore do not perfectly conserve mechanical energy.
• Energy is never "used up" or destroyed; it is transferred to other stores.
• Conservation of energy does not mean that speed, height, kinetic energy, or potential energy remains constant.
• It is the total energy accounting that must balance.

The central idea can be summarized as:

GPE ↔ KE ↔ Elastic PE

For an ideal system:

Eₖ + Eₚ + Eₑ = constant

For a real system with friction:

initial mechanical energy = final mechanical energy + energy transferred to the surroundings

So when solving energy problems, the key question is not:

"Where did the energy go?"

but:

"Which energy store did it transfer into?"

 

 

 

5. Energy Transformations and Calculations

Energy Transformations and Calculations

Most real physical systems involve several energy transformations, not just one.

A roller coaster, bouncing ball, spring launcher, hydroelectric dam, bungee jumper, or vehicle may transfer energy through several stores before the process is complete.

The key skill is learning to track the energy:

Where is the energy initially? → Where does it go? → How much remains?

Energy is never created or destroyed. It is transferred between stores and, in real systems, some energy is often dissipated into the surroundings.

 

 

 

4

The Mechanical Energy Stores

Three important mechanical energy stores are:

Kinetic Energy

Energy associated with motion.

Eₖ = ½mv²

where:

• Eₖ = kinetic energy (J)
• m = mass (kg)
• v = speed (m/s)

Gravitational Potential Energy

Energy associated with position in a gravitational field.

ΔEₚ = mgΔh

where:

• ΔEₚ = change in gravitational potential energy (J)
• m = mass (kg)
• g = gravitational field strength (N/kg)
• Δh = vertical height change (m)

Elastic Potential Energy

Energy stored when an elastic object is stretched or compressed.

Eₑ = ½kx²

where:

• Eₑ = elastic potential energy (J)
• k = spring constant (N/m)
• x = extension or compression (m)

These equations can be combined to analyze complex systems.

Energy Transformations

An energy transformation occurs when energy changes from one store to another.

Examples include:

gravitational potential → kinetic

kinetic → gravitational potential

elastic potential → kinetic

kinetic → elastic potential

gravitational potential → elastic potential

kinetic → thermal

A physical system may involve several transformations in sequence.

Energy Flow Diagrams

An energy flow diagram shows how energy moves through a system.

For a falling object:

gravitational potential energy → kinetic energy

For a spring launcher:

elastic potential energy → kinetic energy

For a ball thrown upward:

kinetic energy → gravitational potential energy

These simple diagrams become particularly useful when several transformations occur.

Multi-Step Energy Flow

Consider a cart released from the top of a hill that rolls downward and compresses a spring.

The energy flow is:

gravitational potential → kinetic → elastic potential

If friction is present:

gravitational potential → kinetic → elastic potential + thermal energy

The thermal energy represents mechanical energy transferred to the surroundings.

Useful and Dissipated Energy

Real systems are rarely perfectly efficient.

Some energy is usually transferred into less useful forms, especially:

• thermal energy
• sound
• vibration

This is often described as energy dissipation.

For example:

chemical → kinetic + thermal + sound

Energy has not disappeared.

It has simply become more spread out in the surroundings and is often less useful for performing the intended task.

Conservation of Energy

The fundamental principle is:

total energy before = total energy after

For an ideal mechanical system:

initial mechanical energy = final mechanical energy

For a real system:

initial energy = useful final energy + dissipated energy

This provides the foundation for solving multi-step energy problems.

A Systematic Problem-Solving Method

Complex energy problems become much easier when solved systematically.

Identify the System

Determine which objects are involved.

For example:

• ball + Earth
• cart + track + Earth
• cart + spring
• roller coaster + Earth

Identify the Initial Energy Stores

Ask:

• Is anything moving?
• Is anything elevated?
• Is anything stretched or compressed?

Identify the Final Energy Stores

Ask the same questions at the final position.

Identify Energy Losses

Look for:

• friction
• air resistance
• sound
• deformation

Draw an Energy Flow

For example:

GPE → KE → elastic PE + thermal

Write the Energy Equation

Then substitute numerical values.

This method prevents many calculation mistakes.

Worked Example: Hill to Kinetic Energy

A 500 kg roller-coaster car begins from rest 30 m above the bottom of a hill.

Ignore friction.

Use:

g = 9.8 N/kg

Initial GPE:

Eₚ = mgh

Eₚ = 500 × 9.8 × 30

Eₚ = 147,000 J

Initial KE:

0 J

At the bottom:

GPE → KE

Therefore:

Eₖ = 147,000 J

Now:

Eₖ = ½mv²

147,000 = 0.5 × 500 × v²

147,000 = 250v²

v² = 588

v ≈ 24.2 m/s

The coaster reaches approximately 24.2 m/s.

 

 

 

7

Worked Example: Roller Coaster with Friction

Now suppose the same coaster loses 27,000 J of mechanical energy because of friction and air resistance.

Initial mechanical energy:

147,000 J

Energy dissipated:

27,000 J

Remaining mechanical energy:

147,000 − 27,000 = 120,000 J

At the bottom:

Eₖ = 120,000 J

Therefore:

120,000 = 0.5 × 500 × v²

120,000 = 250v²

v² = 480

v ≈ 21.9 m/s

Without friction:

24.2 m/s

With friction:

21.9 m/s

The coaster is slower because some mechanical energy has been transferred into the surroundings.

Energy Flow for the Roller Coaster

The idealized system is:

GPE → KE

The real system is:

GPE → KE + thermal + sound

Notice that total energy is conserved in both cases.

What changes is the amount of energy remaining in the mechanical energy stores.

Worked Example: Hill to Spring

A 4 kg cart starts from rest at a height of 5 m.

It rolls down a frictionless track and compresses a spring with:

k = 800 N/m

Find the maximum compression.

Initial energy:

GPE = mgh

GPE = 4 × 9.8 × 5

GPE = 196 J

At maximum compression:

• cart is momentarily stationary
• KE = 0
• energy is stored in the spring

Therefore:

GPE → elastic PE

196 = ½kx²

196 = 0.5 × 800 × x²

196 = 400x²

x² = 0.49

x = 0.70 m

The spring compresses by 0.70 m.

Tracking the Intermediate Stage

The previous problem actually contains three stages.

Stage A — Top of Hill

GPE: 196 J

KE: 0 J

Elastic PE: 0 J

Stage B — Bottom Before Spring

GPE: 0 J

KE: 196 J

Elastic PE: 0 J

Stage C — Maximum Compression

GPE: 0 J

KE: 0 J

Elastic PE: 196 J

So:

GPE → KE → elastic PE

The energy changes form twice, but the total mechanical energy remains 196 J.

Worked Example: Hill, Friction, and Spring

Now suppose 36 J is dissipated by friction before the cart reaches maximum compression.

Initial energy:

196 J

Dissipated:

36 J

Energy remaining:

196 − 36 = 160 J

Therefore:

160 = ½ × 800 × x²

160 = 400x²

x² = 0.40

x ≈ 0.632 m

The spring compresses approximately:

0.63 m

Less energy reaches the spring because some energy has been dissipated.

Spring Launcher to Height

A spring launcher can involve:

elastic PE → KE → GPE

Suppose a spring stores 50 J and launches a 1 kg object vertically upward.

Ignoring losses, the object eventually reaches maximum height.

At maximum height:

KE = 0

Therefore:

elastic PE → GPE

50 = mgΔh

50 = 1 × 9.8 × Δh

Δh = 50/9.8

Δh ≈ 5.1 m

The object rises approximately 5.1 m.

Spring Launcher with Energy Loss

Suppose the spring still stores 50 J, but 10 J is dissipated through friction and air resistance.

Energy available for GPE:

50 − 10 = 40 J

Therefore:

40 = 1 × 9.8 × Δh

Δh ≈ 4.08 m

The object reaches only about 4.1 m.

The energy flow is:

elastic PE → kinetic → GPE + thermal + sound

Bouncing Ball

A bouncing ball involves several energy transformations.

Before falling:

gravitational potential energy

During the fall:

GPE → KE

During impact:

KE → elastic PE + thermal + sound

During rebound:

elastic PE → KE

As the ball rises:

KE → GPE

Therefore:

GPE → KE → elastic PE → KE → GPE

 

 

 

4

Why a Ball Does Not Return to Its Original Height

If all mechanical energy were recovered, the ball would return to its original height.

In reality, energy is dissipated through:

• sound
• thermal energy
• internal deformation
• air resistance

Therefore:

final GPE < initial GPE

and:

rebound height < original height

Worked Example: Bouncing Ball

A 0.50 kg ball is dropped from 4 m.

Use:

g = 9.8 N/kg

Initial GPE:

Eₚ = 0.50 × 9.8 × 4

Eₚ = 19.6 J

Suppose 5.6 J is dissipated during the fall and bounce.

Energy remaining:

19.6 − 5.6 = 14.0 J

At maximum rebound height:

KE = 0

Therefore:

14.0 = 0.50 × 9.8 × h

14.0 = 4.9h

h ≈ 2.86 m

The ball rebounds to approximately 2.9 m.

Pendulum Energy Transformations

A pendulum repeatedly transforms energy between:

GPE ↔ KE

At the highest points:

• GPE maximum
• KE minimum

At the lowest point:

• GPE minimum
• KE maximum

In a real pendulum:

GPE ↔ KE + thermal + sound

As energy is dissipated, the pendulum's maximum height gradually decreases.

 

 

 

4

Energy Transfer in a Trampoline

A trampoline involves several stores.

A person standing above the trampoline has GPE.

During the fall:

GPE → KE

As the trampoline stretches:

KE → elastic PE

As the trampoline returns to shape:

elastic PE → KE

As the person rises:

KE → GPE

Therefore:

GPE → KE → elastic PE → KE → GPE

Some energy is also dissipated as:

• thermal energy
• sound

Bungee Jumping

A bungee jump is another excellent multi-step system.

Initially:

GPE

As the jumper falls:

GPE → KE

As the cord begins stretching:

KE + GPE → elastic PE

At maximum extension:

• speed is momentarily zero
• KE = 0
• substantial energy is stored elastically

During rebound:

elastic PE → KE → GPE

 

 

 

4

Energy Flow Diagrams for Complex Systems

Energy flow diagrams can be written simply using arrows.

Roller Coaster

GPE → KE → GPE → KE

With friction:

GPE → KE + thermal + sound

Spring Launcher

elastic PE → KE

Spring Launcher Upward

elastic PE → KE → GPE

Bouncing Ball

GPE → KE → elastic PE → KE → GPE

Hydroelectric Dam

GPE of water → KE of water → mechanical energy of turbine → electrical energy

Vehicle

chemical energy → kinetic energy + thermal energy + sound

Regenerative Braking

kinetic energy → electrical energy → chemical energy in battery

These diagrams make complicated processes easier to analyze.

Sankey Diagrams

A Sankey diagram is another way of representing energy transfers.

The width of each arrow represents the amount of energy transferred.

For example, suppose a machine receives:

100 J input

and produces:

70 J useful output

with:

30 J dissipated

The energy accounting is:

100 J → 70 J useful + 30 J dissipated

The arrows in a Sankey diagram would show a larger useful-energy branch and a smaller dissipated-energy branch.

The total widths leaving the system must equal the input width because energy is conserved.

Efficiency

Efficiency describes how much input energy becomes useful output energy.

efficiency = useful energy output / total energy input

As a percentage:

efficiency (%) = useful energy output / total energy input × 100

Efficiency cannot exceed 100%.

Worked Example: Efficiency

A motor receives:

500 J

of electrical energy.

It transfers:

400 J

into useful mechanical energy.

Efficiency:

efficiency = 400/500 × 100

efficiency = 80%

Energy dissipated:

500 − 400 = 100 J

Therefore:

500 J electrical → 400 J mechanical + 100 J thermal/sound

Worked Example: Finding Useful Energy

A machine is 75% efficient and receives 2000 J of energy.

Useful energy:

useful output = efficiency × input

Convert:

75% = 0.75

Therefore:

useful output = 0.75 × 2000

useful output = 1500 J

Energy dissipated:

2000 − 1500 = 500 J

Worked Example: Finding Input Energy

A machine produces 600 J of useful mechanical energy and is 80% efficient.

0.80 = 600/input

Therefore:

input = 600/0.80

input = 750 J

Dissipated energy:

750 − 600 = 150 J

Cars and Energy Transformations

A moving vehicle involves many energy transfers.

Initially, fuel or a battery provides stored energy.

For a fuel-powered vehicle:

chemical energy → kinetic energy + thermal energy + sound

When the vehicle climbs a hill:

kinetic/chemical energy → gravitational potential energy

When braking:

kinetic energy → thermal energy

The brakes become hot because mechanical energy has been transferred into thermal energy.

 

 

 

Regenerative Braking

Some electric and hybrid vehicles recover part of the vehicle's kinetic energy during braking.

Instead of transferring nearly all of it into thermal energy in the brakes:

kinetic energy → electrical energy → stored energy in battery

The process is not perfectly efficient, so some energy is still dissipated.

This demonstrates how engineering can improve energy efficiency by reducing unwanted energy transfers.

Hydroelectric Power

Water stored behind a dam possesses gravitational potential energy.

The sequence is approximately:

GPE of water → KE of water → rotational KE of turbine → electrical energy

Some energy is dissipated through:

• turbulence
• friction
• sound
• electrical resistance

 

 

 

6

Worked Example: Hydroelectric System

Suppose 1000 kg of water falls through 20 m.

Use:

g = 9.8 N/kg

GPE lost:

ΔEₚ = mgΔh

ΔEₚ = 1000 × 9.8 × 20

ΔEₚ = 196,000 J

Suppose the system is 85% efficient.

Electrical energy produced:

0.85 × 196,000

= 166,600 J

Energy dissipated:

196,000 − 166,600

= 29,400 J

Multi-Step Problem: Spring, Motion, and Height

A spring with:

k = 600 N/m

is compressed by:

x = 0.30 m

It launches a 0.50 kg cart up a ramp.

The cart loses 5 J because of friction.

How high does it rise vertically?

Step 1: Calculate Elastic Energy

Eₑ = ½kx²

Eₑ = 0.5 × 600 × 0.30²

Eₑ = 27 J

Step 2: Subtract Dissipated Energy

27 − 5 = 22 J

Step 3: Convert Remaining Energy to GPE

22 = mgΔh

22 = 0.50 × 9.8 × Δh

22 = 4.9Δh

Δh ≈ 4.49 m

The cart rises approximately 4.5 m vertically.

The complete energy flow is:

elastic PE → KE → GPE + thermal

Multi-Step Problem: Falling onto a Spring

A 2 kg object falls through 3 m before compressing a spring.

The spring constant is:

k = 500 N/m

Ignore friction and, for this simplified example, take the 3 m as the total vertical GPE decrease being converted to elastic energy.

Initial GPE available:

Eₚ = mgh

Eₚ = 2 × 9.8 × 3

Eₚ = 58.8 J

At maximum compression:

KE = 0

Therefore:

58.8 = ½ × 500 × x²

58.8 = 250x²

x² = 0.2352

x ≈ 0.485 m

The spring compresses approximately 0.49 m.

In a more detailed vertical-spring problem, the additional downward movement while the spring compresses would also change gravitational potential energy and must be included.

Evaluating Real Systems

Real-world energy analysis involves more than calculating numbers.

We should also ask:

• Which energy transfers are useful?
• Which are unwanted?
• Where is energy dissipated?
• Can the system be made more efficient?
• What assumptions are being made?
• Are friction and air resistance significant?
• Does the mathematical model accurately represent the real system?

This turns energy calculations into an engineering tool.

Reducing Energy Dissipation

Engineers often try to reduce unwanted energy transfers.

Examples include:

Lubrication

Reduces friction between moving surfaces.

Streamlining

Reduces air resistance.

Thermal Insulation

Reduces unwanted thermal energy transfer.

Low-Resistance Bearings

Reduce friction in rotating systems.

Regenerative Braking

Recovers some kinetic energy rather than dissipating it entirely as heat.

Efficient Motors

Reduce unwanted thermal energy transfers.

However, unwanted energy transfers can rarely be eliminated completely.

When Dissipation Is Useful

Not all energy dissipation is unwanted.

Vehicle brakes deliberately transform:

kinetic energy → thermal energy

This is useful because it slows the vehicle.

Shock absorbers deliberately dissipate mechanical energy to reduce bouncing.

Protective materials may transform kinetic energy through deformation and thermal processes during impacts.

Therefore, whether an energy transfer is "useful" depends on the purpose of the system.

Choosing the Correct Equation

Use:

Eₖ = ½mv²

when motion or speed is involved.

Use:

ΔEₚ = mgΔh

when vertical height changes.

Use:

Eₑ = ½kx²

when a spring or elastic system is stretched or compressed.

Use:

efficiency = useful output/input

when useful and wasted energy are being compared.

Most challenging problems require more than one equation.

Worked Example: Three Equations in One Problem

A 1.0 kg cart starts from rest 5 m above the bottom of a track.

It descends and loses 9 J through friction before compressing a spring with:

k = 400 N/m

Find the maximum spring compression.

Initial GPE

Eₚ = mgh

Eₚ = 1 × 9.8 × 5

Eₚ = 49 J

Subtract Dissipation

49 − 9 = 40 J

Transfer to Spring

At maximum compression:

KE = 0

Therefore:

40 = ½kx²

40 = 0.5 × 400 × x²

40 = 200x²

x² = 0.20

x ≈ 0.447 m

Therefore:

x ≈ 0.45 m

Energy flow:

GPE → KE → elastic PE + thermal

Energy Accounting Table

An energy table can help organize complex problems.

Check:

49 J = 40 J + 9 J

The energy accounting balances.

Common Mistakes

Treating Energy as Disappearing

Incorrect:

"9 J of energy was lost."

Better:

"9 J was transferred from mechanical energy stores to thermal energy in the surroundings."

Ignoring Intermediate Transformations

A spring launching an object upward is not simply:

elastic → gravitational

The actual sequence includes:

elastic → kinetic → gravitational

Even if the calculation can compare the initial and final states directly.

Assuming Mechanical Energy Is Always Conserved

If friction or air resistance is significant:

mechanical energy decreases

but:

total energy remains conserved

Forgetting to Subtract Dissipated Energy

If:

Initial energy = 100 J

Dissipated energy = 20 J

then useful mechanical energy remaining is:

80 J

Using the Wrong Energy Equation

Speed → kinetic energy.

Height → gravitational potential energy.

Spring deformation → elastic potential energy.

Forgetting Squared Quantities

Remember:

Eₖ = ½mv²

and:

Eₑ = ½kx²

Both v and x are squared.

Using Centimetres for Spring Extension

Convert to metres.

20 cm = 0.20 m

Using Total Ramp Length for GPE

Use vertical height change, not distance along the ramp.

Assuming Efficiency Can Exceed 100%

An efficiency above 100% would imply more energy leaves the system than enters it.

That violates conservation of energy.

Confusing Useful and Dissipated Energy

Dissipated energy is not destroyed.

It has been transferred into less useful stores, commonly thermal energy in the surroundings.

Check Your Understanding

1. Name the three main mechanical energy stores studied in this unit.
2. Write the equations for kinetic, gravitational potential, and elastic potential energy.
3. Explain what is meant by an energy transformation.
4. Construct an energy flow diagram for a falling ball.
5. Construct an energy flow diagram for a spring launching a cart.
6. Construct an energy flow diagram for a bouncing ball.
7. Describe the energy transformations occurring during a bungee jump.
8. A 2 kg object falls through 5 m. Calculate the GPE lost using g = 9.8 N/kg.
9. Ignoring air resistance, calculate its KE after falling.
10. Calculate its speed.
11. A 3 kg cart starts 4 m above the bottom of a frictionless track. Calculate its kinetic energy at the bottom.
12. A spring with k = 500 N/m is compressed by 0.20 m. Calculate the stored elastic energy.
13. If all this energy is transferred to a 0.50 kg cart, calculate its speed.
14. A spring stores 40 J and launches a 1 kg object vertically upward. Calculate its maximum height, ignoring losses.
15. Repeat Question 14 if 8 J is dissipated.
16. A ball initially has 30 J of GPE and loses 6 J during its fall and bounce. How much mechanical energy remains?
17. Explain why the ball does not return to its original height.
18. A roller coaster starts with 250 kJ of GPE. During its descent, 40 kJ is dissipated. How much kinetic energy can it have at the bottom?
19. Explain where the 40 kJ went.
20. A machine receives 800 J and produces 600 J of useful energy. Calculate its efficiency.
21. How much energy was dissipated in Question 20?
22. A system is 70% efficient and receives 1500 J. Calculate its useful output.
23. A 5 kg cart descends 6 m and then compresses a spring with k = 1000 N/m. Calculate the maximum compression, ignoring friction.
24. Repeat Question 23 if 50 J is dissipated before maximum compression.
25. Describe the energy transformations in a hydroelectric power station.
26. Explain how regenerative braking changes the energy flow in a vehicle.
27. Why does a real pendulum gradually stop?
28. Explain the difference between conservation of mechanical energy and conservation of total energy.
29. Give an example where energy dissipation is useful.
30. Explain why tracking energy through several stages can be more useful than examining only the beginning and end of a process.

Key Terms

Energy transformation — Change of energy from one store or form to another.

Energy transfer — Movement of energy between objects, stores, or systems.

Mechanical energy — Energy associated with motion and position, including kinetic and potential energy.

Kinetic energy — Energy associated with motion.

Gravitational potential energy — Energy associated with position in a gravitational field.

Elastic potential energy — Energy stored by elastic deformation.

Conservation of energy — Principle that energy cannot be created or destroyed.

Conservation of mechanical energy — Conservation of the total kinetic and potential energy of a system when no mechanical energy is transferred elsewhere.

Energy dissipation — Transfer and spreading of energy into less useful stores, commonly thermal energy in the surroundings.

Friction — Force opposing relative motion between surfaces that can transfer mechanical energy into thermal energy.

Air resistance — Resistive force acting against motion through air.

Efficiency — Fraction of input energy transferred into the desired useful output.

Useful energy — Energy transferred in the form required for the intended purpose.

Dissipated energy — Energy transferred into less useful stores and spread through the surroundings.

Energy flow diagram — Representation using arrows to show transfers and transformations of energy.

Sankey diagram — Diagram in which arrow widths represent quantities of energy transferred.

System — Object or collection of objects chosen for analysis.

Energy accounting — Tracking energy entering, leaving, and changing stores within a system.

Key Takeaways

• Physical systems often involve several energy transformations.
• The three main mechanical energy stores are kinetic, gravitational potential, and elastic potential energy.
• Kinetic energy is calculated using Eₖ = ½mv².
• Changes in gravitational potential energy are calculated using ΔEₚ = mgΔh.
• Elastic potential energy is calculated using Eₑ = ½kx².
• Energy flow diagrams help track transformations through complex systems.
• A falling object transforms gravitational potential energy into kinetic energy.
• A spring launcher transforms elastic potential energy into kinetic energy.
• A bouncing ball can involve gravitational, kinetic, and elastic potential energy.
• A system can pass through several intermediate energy stores even when calculations compare only its initial and final states.
• Total energy is always conserved.
• Mechanical energy is conserved only when energy is not transferred out of mechanical stores.
• Friction and air resistance commonly transfer mechanical energy into thermal energy.
• Dissipated energy has not disappeared.
• Energy accounting should balance: energy in = energy out + change in stored energy.
• Efficiency measures the fraction of input energy transferred usefully.
• Real systems are less than 100% efficient.
• Engineers improve efficiency by reducing unwanted energy transfers.
• Some dissipation, such as braking or damping, can be useful.
• Multi-step problems are easiest when the initial state, final state, intermediate transformations, and dissipated energy are identified before calculations begin.

A useful approach for almost every complex energy problem is:

Identify stores → draw energy flow → calculate initial energy → account for dissipation → calculate final energy → solve for the unknown

The most important principle remains:

Energy does not disappear. Track where it goes.