Kinetic and gravitational potential energy
Calculate the energy of motion and height. Discover why speed matters so much, then use conservation to predict a falling object’s speed.
This whole lesson is Supplement / Extended. Core-only learners can continue using the qualitative stores and conservation ideas in Lesson 14. This is Cambridge IGCSE, not Egyptian Baccalaureate / Thanaweya Amma or Year 3.
Kinetic energy: motion has an energy cost
Kinetic energy is the energy of a moving object: Eₖ = ½mv². Use mass m in kilograms and speed v in metres per second; the answer is in joules. The square applies only to speed. Energy is a scalar: opposite directions do not produce negative kinetic energies.
Worked example
A 250 g ball moves at 4.0 m/s. Convert first: 250 ÷ 1000 = 0.250 kg. Square speed: 4² = 16. Then Eₖ = ½ × 0.250 × 16 = 2.0 J. Do not use 250 kg or square the whole product.
Think first: A 2 kg trolley moves at 3 m/s. Is its kinetic energy 3 J or 9 J?
9 J: ½ × 2 × 3² = 9. The 3 J answer forgot to square speed.
From an energy account to a speed
- Identify the starting and ending positions. Find the vertical difference, not the path length.
- Convert mass to kg and height to m. Write the stated value of g.
- State the assumptions: starts from rest, no resistance, and no energy in rotation or other stores.
- Calculate gravitational potential energy lost: mg × distance fallen. Set it equal to kinetic energy gained under those assumptions.
- Rearrange v = √(2Eₖ/m). Round only the final result and add m/s.
- Substitute back into ½mv² and check the energy account.
Worked example: a model cart partway down a track
A 0.50 kg non-rotating model starts from rest 5.0 m above the lowest point. Find its speed when it is 3.0 m above that point, ignoring resistance. Drop = 5.0 − 3.0 = 2.0 m. Energy lost = 0.50 × 9.8 × 2.0 = 9.8 J. Hence Eₖ = 9.8 J, and v = √(2 × 9.8 ÷ 0.50) = 6.26 m/s (2 decimal places). Do not use the 3.0 m height remaining as the drop. Remaining gravitational potential energy is 14.7 J; 14.7 + 9.8 = the initial 24.5 J.
Swap height for speed
Predict the speed at a chosen height
A non-rotating object starts from rest and falls vertically. This is a mathematical model, not an instruction to drop heavy objects. Ignore air resistance; g = 9.8 N/kg. Ground is the zero gravitational-potential-energy level.
Current height = 3.0 m. Vertical distance fallen = 2.0 m.
Investigate: reveal an answer, then change only mass. Predict whether speed changes. Next lower the current-height slider: which store decreases? A changed control clears the previous prediction and answer.
Practise, then explain
Four multiple-choice questions, four numerical responses and four written tasks. This whole lesson and all its questions are Supplement / Extended. Numerical answers are checked automatically; written answers use a self-assessment guide. The suggested marks are for this original practice, not an official Cambridge mark scheme.
Practice stays only while this page is open. Refreshing or leaving clears these responses. Copy useful written work into your notes before leaving.
The same trolley doubles its speed. What happens to its kinetic energy?
Optional hint
Only speed changes. Compare the square of the new speed with the old square.
Source notes and common traps
The original PDF is preserved. Its worked fall and pendulum examples quote a positive amount of gravitational potential energy lost. Strictly, ΔEₚ = mgΔh is negative for a fall if Δh means final height minus initial height; the amount lost is positive. We distinguish those meanings here.
The source describes real speeds as “slightly less”. Resistance can reduce speed by a small or a large amount; it depends on the situation. We state no-resistance assumptions in each ideal model rather than silently applying them to real devices. Figures 15.1–15.2 use vertical level differences, not track or pendulum-path lengths.
Remember: square speed, convert grams to kilograms, use vertical height and finish with the square root. Kinetic energy is not momentum; its unit is J, not kg m/s.
Source alignment: supplied Physics-0625 Volume A, PDF pp. 108–112 / printed pp. 100–104, including Figures 15.1–15.2, and the official Cambridge 2026–2028 syllabus, section 1.7.1. Original learning text, model and practice; source-download access is unchanged.