13  Work, Energy and Power

13.1 Syllabus inquiry question

  • How is energy transferred and transformed in motion?
Feynman Insight

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.

13.2 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.

13.3 Content

13.3.1 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

Key Cases
  • Force parallel to motion (\(\theta = 0°\)): \(W = Fd\) (positive work)
  • Force opposite to motion (\(\theta = 180°\)): \(W = -Fd\) (negative work)
  • Force perpendicular to motion (\(\theta = 90°\)): \(W = 0\) (no work done)

13.3.2 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 \(v^2\))

13.3.3 Interactive: Energy Bar Chart - Motion

Visualise kinetic energy as an object speeds up:

13.3.4 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/s² on Earth) - \(h\) = height above reference point (m)

Reference Point

GPE depends on the chosen reference height. Usually, we set GPE = 0 at the lowest point in a problem.

13.3.5 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.

13.3.6 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).

13.3.7 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!

13.3.8 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).

13.3.9 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

13.3.10 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.

13.3.11 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.

13.4 Worked Examples

13.4.1 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 \(\theta = 0°\)

  2. Work: \(W = Fd\cos\theta = 25 \times 8.0 \times \cos0° = 25 \times 8.0 \times 1 = 200\) J

  3. Positive work—energy is transferred to the sled

13.4.2 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

13.4.3 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

13.4.4 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

13.4.5 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: \[v = \sqrt{2g\Delta h} = \sqrt{2 \times 9.8 \times 12} = 15.3 \text{ m/s}\]

13.5 Common Misconceptions

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.

13.6 Practice Questions

13.6.1 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

13.6.2 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

13.6.3 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%

13.7 Multiple Choice Questions

Test your understanding with these interactive questions:

13.8 Quick Quiz: Energy Transformations

13.9 Extended Response Practice

13.10 Summary

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

13.11 Self-Assessment

Check your understanding:

After studying this section, you should be able to:

13.12 Module 2 Complete

Congratulations on completing Module 2: Dynamics!

What you’ve learned
  • 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