Work, Energy, and Power
Orientation
Lesson goal: build accurate physics fluency for work, energy, and power and use that fluency to support clear HSC-style scientific writing.
This page is materialised into the MentorMind course shell from existing teaching, textbook, and eduKG material. Use it as the main lesson surface; use the tutor for targeted repair, worked examples, and concise writing feedback.
Source Lesson Material
Syllabus inquiry question
- How is energy transferred and transformed in motion?
From The Feynman Lectures on Physics, Vol I, Chapter 4:
Energy accounting does not depend on the path taken. What matters is the initial and final states and the work done by non-conservative forces.
Learning Objectives
- Define work, kinetic energy, and gravitational potential energy.
- Apply the work-energy theorem.
- Calculate power and efficiency.
- Interpret energy changes in simple systems.
Content
Work
Work is the energy transferred by a force acting through a displacement:
$$W = Fd\cos\theta$$
where:
- $F$ = magnitude of force (N)
- $d$ = displacement (m)
- $\theta$ = angle between force and displacement
SI unit: Joule (J), where 1 J = 1 N·m
- Force parallel to motion ($\theta = 0 degrees$): $W = Fd$ (positive work)
- Force opposite to motion ($\theta = 180 degrees$): $W = -Fd$ (negative work)
- Force perpendicular to motion ($\theta = 90 degrees$): $W = 0$ (no work done)
Kinetic Energy
Kinetic energy is the energy of motion:
$$KE = \frac{1}{2}mv^2$$
- Always positive (mass and $v^2$ are positive)
- Doubles when speed doubles? No! Quadruples (because $v^2$)
Interactive: Energy Bar Chart - Motion
Visualise kinetic energy as an object speeds up:
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy stored due to height:
$$GPE = mgh$$
where:
- $m$ = mass (kg)
- $g$ = gravitational field strength (9.8 m/s^2 on Earth)
- $h$ = height above reference point (m)
GPE depends on the chosen reference height. Usually, we set GPE = 0 at the lowest point in a problem.
Work-Energy Theorem
The net work done on an object equals its change in kinetic energy:
$$W_{net} = \Delta KE = KE_f - KE_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$
This powerful theorem connects force, displacement, and speed change.
Interactive: Energy Conservation - Falling Object
As an object falls, GPE converts to KE:
Key observation: Total mechanical energy (KE + GPE) remains constant (in the absence of friction).
Conservation of Mechanical Energy
In the absence of friction and air resistance:
$$KE_i + GPE_i = KE_f + GPE_f$$
Or equivalently: $$\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f$$
This is one of the most useful equations in physics!
Interactive: Pendulum Energy
A pendulum swings back and forth, continuously converting between KE and GPE:
At the highest points, all energy is GPE (momentarily stationary). At the lowest point, all energy is KE (maximum speed).
Power
Power is the rate of doing work or transferring energy:
$$P = \frac{W}{t} = \frac{E}{t}$$
For constant force and velocity: $$P = Fv$$
SI unit: Watt (W), where 1 W = 1 J/s
Efficiency
Efficiency measures how much input energy becomes useful output:
$$\eta = \frac{E_{useful}}{E_{input}} \times 100% = \frac{P_{output}}{P_{input}} \times 100%$$
No machine is 100% efficient-some energy is always lost to friction, heat, sound, etc.
Interactive: Energy with Friction
When friction is present, some mechanical energy is lost:
Friction does negative work, removing mechanical energy and converting it to thermal energy.
Worked Examples
Example 1: Work by a force
A 25 N force pulls a sled 8.0 m on level ground, parallel to motion.
Solution:
-
Force and displacement are parallel, so $\theta = 0 degrees$
-
Work: $W = Fd\cos\theta = 25 \times 8.0 \times \cos0 degrees = 25 \times 8.0 \times 1 = 200$ J
-
Positive work-energy is transferred to the sled
Example 2: Speed from energy conservation
A 2.0 kg object falls 5.0 m from rest. Find its speed at the bottom (ignore air resistance).
Solution:
-
Use energy conservation: $GPE_i = KE_f$
-
$mgh = \frac{1}{2}mv^2$ (mass cancels!)
-
$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5.0} = \sqrt{98} = 9.9$ m/s
Example 3: Work-energy theorem
A 1.2 kg cart speeds up from 2.0 m/s to 6.0 m/s. Find the net work done.
Solution:
-
Initial KE: $KE_i = \frac{1}{2} \times 1.2 \times 2.0^2 = 2.4$ J
-
Final KE: $KE_f = \frac{1}{2} \times 1.2 \times 6.0^2 = 21.6$ J
-
Net work: $W_{net} = KE_f - KE_i = 21.6 - 2.4 = 19.2$ J
Example 4: Power and efficiency
A motor lifts a 150 N load at 0.60 m/s using 200 W of electrical power.
Solution:
-
Mechanical power output: $P_{out} = Fv = 150 \times 0.60 = 90$ W
-
Efficiency: $\eta = \frac{P_{out}}{P_{in}} \times 100% = \frac{90}{200} \times 100% = 45%$
-
The motor is 45% efficient-55% of input power is lost to friction and heat
Example 5: Roller coaster speed
A 500 kg roller coaster car starts from rest at 20 m high. Find its speed at 8 m high (ignore friction).
Solution:
-
Energy conservation: $KE_i + GPE_i = KE_f + GPE_f$
-
Initial: $KE_i = 0$, $GPE_i = mgh_i = 500 \times 9.8 \times 20 = 98000$ J
-
Final: $GPE_f = mgh_f = 500 \times 9.8 \times 8 = 39200$ J
-
$KE_f = 98000 - 39200 = 58800$ J
-
Speed: $v = \sqrt{\frac{2 \times KE_f}{m}} = \sqrt{\frac{2 \times 58800}{500}} = \sqrt{235.2} = 15.3$ m/s
Alternatively, using height difference: $$v = \sqrt{2g\Delta h} = \sqrt{2 \times 9.8 \times 12} = 15.3 \text{ m/s}$$
Common Misconceptions
-
Misconception: Work is done whenever a force exists. Correction: Work requires displacement. No displacement = no work.
-
Misconception: Power and energy are the same. Correction: Energy is a quantity (J); power is a rate (J/s = W).
-
Misconception: Potential energy depends on path. Correction: GPE depends only on height. The path taken doesn't matter.
-
Misconception: Friction destroys energy. Correction: Friction converts mechanical energy to thermal energy. Total energy is still conserved.
-
Misconception: Doubling speed doubles kinetic energy. Correction: KE depends on $v^2$. Doubling speed quadruples KE.
Practice Questions
Easy (2 marks)
A 10 N force moves an object 3.0 m in the direction of the force. Calculate the work done.
- Use $W = Fd$ (since force is parallel to displacement) (1)
- Correct value: $W = 10 \times 3.0 = 30$ J with units (1)
Answer: 30 J
Medium (4 marks)
A 1.2 kg cart speeds up from 2.0 m/s to 6.0 m/s. Find the net work done on the cart.
- Initial KE: $KE_i = \frac{1}{2} \times 1.2 \times 2.0^2 = 2.4$ J (1)
- Final KE: $KE_f = \frac{1}{2} \times 1.2 \times 6.0^2 = 21.6$ J (1)
- Work-energy theorem: $W = \Delta KE$ (1)
- Correct answer: $W = 21.6 - 2.4 = 19.2$ J (1)
Answer: 19.2 J
Hard (5 marks)
A pump raises 500 kg of water by 4.0 m in 60 s. Determine the output power and efficiency if the electrical input is 600 W.
- GPE gained: $E = mgh = 500 \times 9.8 \times 4.0 = 19600$ J (1)
- Work done equals energy gained (1)
- Output power: $P_{out} = W/t = 19600/60 = 327$ W (1)
- Efficiency formula applied correctly (1)
- Correct efficiency: $\eta = 327/600 \times 100% = 54.4%$ (1)
Answer: Output power = 327 W; Efficiency = 54%
Multiple Choice Questions
Test your understanding with these interactive questions:
Quick Quiz: Energy Transformations
Extended Response Practice
Summary
- Work: $W = Fd\cos\theta$ (units: J)
- Kinetic energy: $KE = \frac{1}{2}mv^2$
- Gravitational PE: $GPE = mgh$
- Work-energy theorem: $W_{net} = \Delta KE$
- Conservation: $KE_i + GPE_i = KE_f + GPE_f$ (no friction)
- Power: $P = W/t = Fv$ (units: W)
- Efficiency: $\eta = (E_{out}/E_{in}) \times 100%$
Self-Assessment
Check your understanding:
After studying this section, you should be able to:
- Calculate work using $W = Fd\cos\theta$
- Apply the work-energy theorem
- Use conservation of mechanical energy
- Calculate power and efficiency
- Explain energy transformations in everyday situations
Module 2 Complete
Congratulations on completing Module 2: Dynamics!
- Forces and their effects on motion
- Newton's three laws of motion
- Friction and inclined plane analysis
- Momentum and impulse in collisions
- Work, energy, and power relationships
Scientific Writing And Exam Support
When answering questions from this lesson, separate:
- the physical quantity being discussed,
- the model or law being applied,
- the mathematical relationship, including units,
- the conclusion in words.
For explanation questions, write in the pattern: claim -> physics reason -> consequence. For calculation questions, state the formula, substitute with units, calculate, then interpret the answer.
Tutor Context
Use this lesson context when the student asks about work, energy, and power, related calculations, representations, or scientific writing. Prefer a short diagnostic before re-teaching. Check whether the student is confusing closely related categories such as force, velocity, acceleration, field, energy, momentum, model evidence, or mathematical representation.
Useful tutor diagnostic:
Which quantity is changing here, what causes that change, and what unit should the final answer use?
Maintenance Loop
One-minute retrieval:
- State the key law, model, or relationship used in this lesson.
- Identify one common misconception that would lead to a wrong answer.
- Write one sentence that links the calculation or evidence back to the physical meaning.
Source Trace
This content record was materialised from:
edu/physics-prep/hsc/physics/textbook/module-2/sections/m2-5-work-energy-power.qmdedu/physics-prep/hsc/physics/eduKG/lessons/program/year-11/module-2/T2W3L1-m2-5-work-energy-power.qmd/Users/philiphaynes/devel/teaching/teaching/cromer/2026/classes/11PHY5/lessons/mod-2/lesson-4-work-energy-power.qmd