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.
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:
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.
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.