Work, Energy, and Power

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.

• 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.

• Define work, kinetic energy, and gravitational potential energy.
• Apply the work-energy theorem.
• Calculate power and efficiency.
• Interpret energy changes in simple systems.

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.

Example 1: Work by a force

A 25 N force pulls a sled 8.0 m on level ground, parallel to motion.

Solution:

1. Force and displacement are parallel, so $\theta = 0 degrees$

2. Work: $W = Fd\cos\theta = 25 \times 8.0 \times \cos0 degrees = 25 \times 8.0 \times 1 = 200$ J

3. 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:

1. Use energy conservation: $GPE_i = KE_f$

2. $mgh = \frac{1}{2}mv^2$ (mass cancels!)

3. $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5.0} = \sqrt{98} = 9.9$ 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:

1. Initial KE: $KE_i = \frac{1}{2} \times 1.2 \times 2.0^2 = 2.4$ J

2. Final KE: $KE_f = \frac{1}{2} \times 1.2 \times 6.0^2 = 21.6$ J

3. 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:

1. Mechanical power output: $P_{out} = Fv = 150 \times 0.60 = 90$ W

2. Efficiency: $\eta = \frac{P_{out}}{P_{in}} \times 100% = \frac{90}{200} \times 100% = 45%$

3. 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:

1. Energy conservation: $KE_i + GPE_i = KE_f + GPE_f$

2. Initial: $KE_i = 0$, $GPE_i = mgh_i = 500 \times 9.8 \times 20 = 98000$ J

3. Final: $GPE_f = mgh_f = 500 \times 9.8 \times 8 = 39200$ J

4. $KE_f = 98000 - 39200 = 58800$ J

5. 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: $$v = \sqrt{2g\Delta h} = \sqrt{2 \times 9.8 \times 12} = 15.3 \text{ m/s}$$

• 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.

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%

Test your understanding with these interactive questions:

• 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: $\eta = (E_{out}/E_{in}) \times 100%$

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

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

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.