Work and Energy Revision Notes for VCE SSCE Physics

Work and Energy

Introduction

Work and energy are fundamental concepts in physics that describe how forces transfer energy between objects and how energy transforms between different forms. Understanding these principles helps explain everything from simple machines to complex systems like bungee jumping.

When a bungee jumper leaps from a platform, they experience a fascinating transformation of energy. Initially, they possess maximum gravitational potential energy. As they fall, this converts to kinetic energy (energy of motion). When the cord begins to stretch, energy transfers into elastic potential energy in the cord. This interplay between different energy forms demonstrates the law of conservation of energy in action.

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The bungee jumping example demonstrates how energy continuously transforms between different forms. At any moment, the total energy remains constant (assuming negligible air resistance), but it shifts between gravitational potential, kinetic, and elastic potential energy throughout the jump.

Work

What is work?

In physics, work represents the amount of energy transferred between objects or transformed from one form to another through the action of a force.

When a force causes an object to move in the direction of that force, work is done. The amount of work depends on both the magnitude of the force and the distance moved.

Formula for work

When a force $F$ acts in the same direction as the displacement $s$ , work is calculated as:

$W = Fs$

Where:

$W$ = Work (measured in joules, J)
$F$ = Force (measured in newtons, N)
$s$ = Displacement (measured in metres, m)

chatImportant

One joule of work is done when a force of $1$ newton causes a displacement of $1$ metre in the same direction as the force.

Work when force acts at an angle

Often, forces don't act exactly in the direction of movement. For example, when pulling a cart with a rope at an angle, only the component of force parallel to the motion does work.

When force $F$ acts at an angle $\theta$ to the direction of displacement $s$ :

$W = Fs \cos \theta$

Where:

$W$ = Work (J)
$F$ = Net force (N)
$s$ = Displacement (m)
$\theta$ = Angle between the force and displacement direction (degrees)

infoNote

Work is a scalar quantity, meaning it has magnitude but no direction. This is different from vector quantities like force or displacement which have both magnitude and direction.

Special case: perpendicular forces

An important principle states that when a force acts perpendicular to the direction of displacement, no work is done.

This explains why objects in uniform circular motion maintain constant kinetic energy. The centripetal force acts perpendicular to the velocity vector, so despite a force being present, no energy is transferred to increase or decrease the object's speed.

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Worked Example: Calculating Work

Question: A person pushes a $1500$ kg car a distance of $12.6$ m at constant velocity with a force of $900$ N in the direction of displacement. A friction force of $900$ N opposes the motion.

a) Calculate the work done on the car by the person.

b) What is the increase in kinetic energy of the car?

c) Determine the energy lost due to friction.

Solution:

a) Using the work formula:

$W = Fs \cos \theta$

Since the force is parallel to displacement, $\theta = 0°$ :

$W = (900)(12.6) \cos(0)$

$W = 900 \times 12.6$

$W = 1.13 \times 10^4 \text{ J}$

b) Zero. The kinetic energy doesn't increase because the car travels at constant velocity.

c) Since the car gains no energy, all the work done must equal the energy lost to friction: $1.13 \times 10^4$ J.

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Worked Example: Work at an Angle

Question: A child pulls a cart $250$ m using a string at $31.0°$ to the horizontal. The string has constant tension of $29.4$ N. Calculate the work done.

Solution:

Using the angled force formula:

$W = Fs \cos \theta$

$W = (29.4)(250) \cos(31)$

$W = 6.30 \times 10^3 \text{ J}$

Exam tip: Always identify whether the force acts at an angle to displacement. If so, you must use the $\cos \theta$ component.

Calculating work from force-distance graphs

Force-distance graphs provide a visual way to analyse how force varies with distance. A powerful principle states that the area under a force-distance graph equals the work done.

Constant force

When a constant force of $100$ N pushes a box for $7.00$ m:

The work done equals the rectangular area under the graph:

$\text{Work} = \text{area under graph}$

$\text{Work} = 100 \times 7$

$\text{Work} = 700 \text{ J}$

Variable force

When force changes with distance, the area calculation becomes more interesting. Consider a force decreasing linearly from $80$ N to zero over $4$ m:

The area forms a triangle:

$\text{Work} = \text{area under graph}$

$\text{Work} = 0.5 \times \text{base} \times \text{height}$

$\text{Work} = 0.5 \times 4 \times 80$

$\text{Work} = 160 \text{ J}$

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Worked Example: Drawing and Using Force-Distance Graphs

Question: A person pushes a car $300$ m. Initially they apply $950$ N, reducing this at a constant rate until at $300$ m there's no net force. Draw a force-distance graph and calculate the work done.

Solution:

The graph forms a triangle. Calculate the area:

$\text{Work} = \text{area under graph}$

$\text{Work} = 0.5 \times 950 \times 300$

$\text{Work} = 1.425 \times 10^5 \text{ J}$

Exam tip: For force-distance graphs, identify the shape (rectangle, triangle, trapezoid) and use the appropriate area formula.

Practice questions - Set 1

Question 1: A person moves a couch $10$ cm to the right by applying $496$ N parallel to the motion at constant velocity. A friction force of $496$ N opposes this.

a) Calculate the work done to move the couch.

b) Determine the change in kinetic energy.

c) Determine the energy lost to friction.

Question 2: A person pulls a golf cart $324$ m horizontally. They pull with an average force of $250$ N at $60°$ to the horizontal. Calculate the work done.

Question 3: Students test two materials A and B by compressing them $10$ mm. Study the force-displacement graph below:

Which material required more energy to compress by $10$ mm?

Energy transfers and transformations

Energy cannot be created or destroyed, only transferred between objects or transformed between forms. This is the law of conservation of energy.

Systems: open and closed

A system represents a collection of interacting objects being studied.

A closed system doesn't allow energy or mass to transfer to the surroundings. In closed systems, total energy remains constant.

An open system allows energy or mass transfer to the surroundings. Energy can be lost as heat, sound, or material deformation.

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Understanding Systems:

Subatomic collisions in the Large Hadron Collider approximate closed systems where energy stays within the collision. In contrast, car crashes are open systems where energy dissipates as heat, sound, and material deformation to the surroundings.

Friction and energy

When friction opposes motion, it converts kinetic energy into thermal energy and sound. This energy isn't lost from the universe, but it's transferred to the surroundings.

Consider pushing a box $10$ m across carpet with $50$ N force at constant velocity (friction also equals $50$ N):

$W = Fs$

$W = (50)(10)$

$W = 500 \text{ J}$

When you stop pushing, the box becomes stationary (zero kinetic energy). The $500$ J has been transformed into thermal energy in both the carpet and box, plus some sound energy.

chatImportant

We don't say friction does "negative work" because the carpet isn't giving energy to the box. Instead, friction converts the box's energy into other forms - primarily thermal energy and sound.

Gravitational potential energy

Gravitational potential energy represents the energy an object possesses due to its position in a gravitational field.

When a weightlifter raises a weight, they do work against gravity. This work becomes stored as gravitational potential energy.

Formula for gravitational potential energy

$E_g = mg\Delta h$

Where:

$E_g$ = Gravitational potential energy (J)
$m$ = Mass (kg)
$g$ = Gravitational field strength ( $9.8$ N kg $^{-1}$ at Earth's surface)
$\Delta h$ = Change in height (m)

infoNote

We can either calculate the change in gravitational potential energy between two points, or define a reference point of zero gravitational potential energy and calculate relative to that point. The choice of reference point is arbitrary - what matters is the change in height.

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Worked Example: Gravitational Potential Energy

Question: An $80$ kg builder climbs a $4.6$ m ladder. Calculate the change in gravitational potential energy.

Solution:

$E_g = mg\Delta h$

$E_g = (80)(9.8)(4.6)$

$E_g = 3606.4 \text{ J}$

The builder gains approximately $3600$ J of gravitational potential energy.

Kinetic energy

Kinetic energy describes the energy an object possesses due to its motion relative to another object.

Formula for kinetic energy

$E_k = \frac{1}{2}mv^2$

Where:

$E_k$ = Kinetic energy (J)
$m$ = Mass (kg)
$v$ = Velocity (m s $^{-1}$ )

Converting speed units

Speed is often given in kilometres per hour (km h $^{-1}$ ) but physics formulas require metres per second (m s $^{-1}$ ).

chatImportant

Conversion rules:

km h $^{-1}$ to m s $^{-1}$ : divide by $3.6$
m s $^{-1}$ to km h $^{-1}$ : multiply by $3.6$

Example: Highway speed limits of $100$ km h $^{-1}$ equal approximately $27.7$ m s $^{-1}$ .

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Worked Example: Calculating Kinetic Energy

Question: A tennis ball of mass $58$ g travels at $27$ m s $^{-1}$ . Calculate its kinetic energy.

Solution:

First convert mass to kilograms: $58$ g $= 0.058$ kg

$E_k = \frac{1}{2}mv^2$

$E_k = \frac{1}{2} \times 0.058 \times 27^2$

$E_k = 21.1 \text{ J}$

The kinetic energy-velocity relationship

The relationship between kinetic energy and velocity is quadratic, not linear. This means doubling velocity quadruples kinetic energy.

chatImportant

Important insight: Changing velocity from $0$ to $1$ m s $^{-1}$ requires much less energy than changing from $3$ to $4$ m s $^{-1}$ for the same mass object.

For a $2$ kg object:

$0 \rightarrow 1$ m s $^{-1}$ : requires $1$ J
$3 \rightarrow 4$ m s $^{-1}$ : requires $7$ J

This quadratic relationship explains why high-speed collisions are so much more dangerous than low-speed ones.

Relationship between work and kinetic energy

When work is done on a point-like object with no other forces present, the work changes the object's kinetic energy:

$W = \Delta E_k = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

This can be proven using kinematic equations and Newton's second law.

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Worked Example: Change in Kinetic Energy

Question: A cyclist travels at $63$ km h $^{-1}$ then brakes to $54$ km h $^{-1}$ . The combined mass of cyclist and bicycle is $93.2$ kg. Calculate the decrease in kinetic energy.

Solution:

First convert speeds to m s $^{-1}$ :

$v = \frac{54}{3.6} = 15 \text{ m s}^{-1}$

$u = \frac{63}{3.6} = 17.5 \text{ m s}^{-1}$

Calculate the change:

$W = \Delta E_k$

$W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

$W = \frac{1}{2}(93.2)(15)^2 - \frac{1}{2}(93.2)(17.5)^2$

$W = -3.79 \times 10^3 \text{ J}$

The negative sign indicates energy was lost from the bicycle to the surroundings as heat and sound.

Practice questions - Set 2

Question 1: When a diver climbs to the top of a $10$ m diving board, their gravitational potential energy increases by $5.68 \times 10^3$ J. What is the diver's mass?

Question 2: Calculate the kinetic energy of a $465$ kg motorbike travelling at $117$ km h $^{-1}$ .

Question 3: A $70$ kg runner travelling at $4$ m s $^{-1}$ has $560$ J of kinetic energy. If they doubled their kinetic energy to $1120$ J, what would their new speed be?

Conservation of energy: gravitational potential and kinetic energy

In many physics problems, energy transforms between gravitational potential energy and kinetic energy. Understanding these transformations is crucial for solving energy problems.

Energy transformation on a ramp

Consider a ball starting at point A on a ramp with no initial speed:

At point A (top):

Maximum gravitational potential energy
Zero kinetic energy

At point B (bottom):

Zero gravitational potential energy (reference point)
Maximum kinetic energy (all GPE converted to KE)

At point C (halfway up):

Half the total energy as gravitational potential energy
Half the total energy as kinetic energy

At point D (same height as A):

Maximum gravitational potential energy again
Zero kinetic energy

If no friction or air resistance exists, the ball reaches the same height as it started.

Conservation equation

For closed systems where only gravitational potential energy and kinetic energy transform:

$E_{k \text{ initial}} + E_{g \text{ initial}} = E_{k \text{ final}} + E_{g \text{ final}}$

Written with formulas:

$\frac{1}{2}mv_i^2 + mg\Delta h_i = \frac{1}{2}mv_f^2 + mg\Delta h_f$

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Worked Example: Energy Transformations

Question: A swimmer stands on a $10$ m diving platform before stepping off. Calculate their velocity when hitting the water. Ignore air resistance.

Solution:

All gravitational potential energy converts to kinetic energy:

$\frac{1}{2}mv_i^2 + mg\Delta h_i = \frac{1}{2}mv_f^2 + mg\Delta h_f$

Initial velocity is zero, final height is zero:

$0 + (9.8)(10) = \frac{1}{2}v_f^2 + (9.8)(0)$

$v_f = 14 \text{ m s}^{-1}$

Note: Mass cancels out - the final velocity doesn't depend on the diver's mass.

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Worked Example: Energy Transformation Down a Slope

Question: A snowboarder passes point A ( $50$ m above ground) at $1.20$ m s $^{-1}$ . Calculate their velocity at point B ( $13.6$ m above ground). Assume friction and air resistance are negligible.

Solution:

$\frac{1}{2}mv_i^2 + mg\Delta h_i = \frac{1}{2}mv_f^2 + mg\Delta h_f$

Substitute values (mass cancels):

$\frac{1}{2}(1.2)^2 + (9.8)(50) = \frac{1}{2}v_f^2 + (9.8)(13.6)$

$v_f = 26.7 \text{ m s}^{-1}$

Exam tip: When using conservation of energy, clearly identify your reference point for zero gravitational potential energy (usually ground level). Remember that mass often cancels out when dealing with energy per unit mass.

Understanding how variables relate to each other

An important physics skill involves recognizing how variables in formulas relate through proportionality.

Direct proportionality

When one variable is directly proportional to another, as one increases, the other increases. We use the symbol $\propto$ to represent "is proportional to".

Example: In the gravitational potential energy formula $E_g = mg\Delta h$ :

$E_g \propto m$ (GPE is directly proportional to mass)
$E_g \propto \Delta h$ (GPE is directly proportional to height change)

For direct proportionality: $k = \frac{A}{B}$ (constant)

Inverse proportionality

When one variable is inversely proportional to another, as one increases, the other decreases.

Example: In the centripetal force formula $F_c = \frac{mv^2}{r}$ :

$F_c \propto \frac{1}{r}$ (centripetal force is inversely proportional to radius)

For inverse proportionality: $k = AB$ (constant)

Powers and proportionality

Variables with powers follow special proportionality rules:

$F_c \propto v^2$ means centripetal force is proportional to velocity squared
Doubling velocity quadruples the force

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Worked Example: Using Proportionality

Question: A car turns a corner at constant speed $v$ . If it takes the same corner at speed $2v$ , by what factor does centripetal force increase?

Solution:

Since $F_c \propto v^2$ :

$k = \frac{F_c}{v^2}$

For the new situation:

$k = \frac{xF_c}{(2v)^2} = \frac{xF_c}{4v^2}$

Since $k$ is constant:

$\frac{F_c}{v^2} = \frac{xF_c}{4v^2}$

$x = 4$

The centripetal force increases by a factor of $4$ .

Exam tip: Understanding proportionality helps you solve problems without needing to know all values. It's particularly useful for multiple-choice questions.

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Key Points to Remember:

Work measures energy transfer when a force causes displacement: $W = Fs\cos\theta$ . When force is perpendicular to displacement, no work is done.
The area under a force-distance graph equals the work done. Use appropriate geometric formulas (rectangle, triangle, trapezoid) to calculate this area.
Gravitational potential energy depends on height: $E_g = mg\Delta h$ . Always define your reference point (usually ground level = zero GPE).
Kinetic energy depends on velocity squared: $E_k = \frac{1}{2}mv^2$ . Doubling velocity quadruples kinetic energy due to this quadratic relationship.
Energy is conserved in closed systems. It can transform between different forms (gravitational potential ↔ kinetic) but the total remains constant: $E_{k \text{ initial}} + E_{g \text{ initial}} = E_{k \text{ final}} + E_{g \text{ final}}$
Always convert units properly: km h $^{-1}$ to m s $^{-1}$ by dividing by $3.6$ ; masses to kilograms; distances to metres.
Understanding variable relationships (direct and inverse proportionality) helps solve problems efficiently without needing all numerical values.

Work and Energy (VCE SSCE Physics): Revision Notes

Work and Energy

Introduction

Work

What is work?

Formula for work

Work when force acts at an angle

Special case: perpendicular forces

Calculating work from force-distance graphs

Constant force

Variable force

Practice questions - Set 1

Energy transfers and transformations

Systems: open and closed

Friction and energy

Gravitational potential energy

Formula for gravitational potential energy

Kinetic energy

Formula for kinetic energy

Converting speed units

The kinetic energy-velocity relationship

Relationship between work and kinetic energy

Practice questions - Set 2

Conservation of energy: gravitational potential and kinetic energy

Energy transformation on a ramp

Conservation equation

Understanding how variables relate to each other

Direct proportionality

Inverse proportionality

Powers and proportionality

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