Ch 13 · 1.7.2 Work · O Level Physics 5054
Topic 13 of 17 in Motion, Forces and Energy — O Level Physics 5054 — 19 lessons.
Introduction
In physics, "work" has an exact meaning. Lifting a box onto a shelf is work; holding it still for an hour is not, however tired your arms get. Work done is the energy transferred when a force moves something. (Syllabus section 1.7.2.)
Key ideas
W = Fd
🧑🏫 Teacher's explanation
Think of 'work' not as a chore you do at home, but as a measure of how much energy you have successfully transferred from one place to another using a force. If you push a wall with all your might but it doesn't budge, you might feel exhausted, but in physics terms, you have done zero work because the wall didn't move.
📖 Definitions
🔍 Worked example — Easy
A person pushes a crate with a constant horizontal force of 50 N over a distance of 4 m. Calculate the work done.
🔍 Worked example — Standard
A weightlifter lifts a barbell of mass 80 kg vertically upwards through a height of 2.0 m. Calculate the work done against gravity. (Use g = 10 m/s^2 for O Level).
🔍 Worked example — Exam-style
A car engine provides a driving force of 2500 N. The car travels 500 m along a straight road. However, there is a constant resistive force (friction/air resistance) of 500 N acting against the motion. Calculate the net work done by the engine to overcome the resistive force only.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked example
Example 1: pushing a box. A horizontal force of 40 N pushes a box 3.0 m across a floor.
🔍 Worked example — Easy
A child pushes a toy car with a horizontal force of 15 N over a distance of 4.0 m. Calculate the work done.
🔍 Worked example — Standard
A weightlifter lifts a barbell of mass 80 kg vertically upwards through a height of 2.0 m. Calculate the work done by the weightlifter. (Use g = 10 m/s^2 for O Level calculations).
🔍 Worked example — Exam-style
A machine applies a constant force of 500 N to move a crate up a ramp. The crate moves 5.5 m along the ramp. However, due to friction, 2000 J of energy is lost as heat. Calculate the net work done on the crate.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) 25 × 8.0 = 200\ J. (2) None: the barbell does not move, so the distance moved is zero. (3) W = mg × h = 60 × 9.8 × 3.5 = 2058\ J ≈ 2100\ J.
Summary
All chapters in Motion, Forces and Energy
- 1.1 Physical Quantities and Measurement Techniques
- 1.2 Motion
- 1.3 Mass and Weight
- 1.4 Density
- 1.5.1 Balanced and Unbalanced Forces
- 1.5.2 Friction
- 1.5.3 Elastic Deformation
- 1.5.4 Circular Motion
- 1.5.5 Turning Effect of Forces
- 1.5.6 Centre of Gravity
- 1.6 Momentum
- 1.7.1 Energy
- 1.7.2 Work
- 1.7.3 Energy Resources
- 1.7.4 Efficiency
- 1.7.5 Power
- 1.8 Pressure