Derive kinetic energy from motion equations
Connect work done by a resultant force to a change in speed and derive the kinetic-energy expression.
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 5. These original explanations and calculations support theory study; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use force, displacement, speed, mass and gravitational field strength with consistent SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which equation gives kinetic energy for mass m and speed v?
What you will learn
- Derive kinetic energy using F = ma, W = Fs and a motion equation
- Recall and use Eₖ = ½mv²
- Interpret kinetic-energy changes and compare speeds
Derive the change in kinetic energy
For a constant resultant force F acting along a straight displacement s, work done W = Fs = mas. The equation v² = u² + 2as gives as = (v² − u²)/2.
Substituting gives W = ½m(v² − u²). If the object starts from rest, u = 0 and the work increases kinetic energy by Eₖ = ½mv². More generally, the net work changes kinetic energy by the difference between final and initial values.
Speed has a squared effect
Kinetic energy depends on mass and the square of speed. Doubling an object's speed at the same mass multiplies its kinetic energy by four. The formula uses speed magnitude, so reversing direction without changing speed does not change kinetic energy.
Use kg and m s⁻¹ to obtain joules. If a force slows an object, its kinetic energy decreases; the work done by the resultant force is negative relative to the displacement, while energy may be transferred to other stores.
Pause and explain
At fixed mass, what happens to kinetic energy if speed doubles?
Worked example
Find the kinetic energy of a 1200 kg car moving at 20 m s⁻¹.
Show the worked solution
- Use Eₖ = ½mv².
- Eₖ = 0.5 × 1200 × 20².
- Square the speed before multiplying.
Answer 2.4 × 10⁵ J, or 240 kJ.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Calculate the kinetic energy of a 0.50 kg ball moving at 8.0 m s⁻¹.
Give me a hint
Square the speed, multiply by mass, then take half.
Compare my reasoning
- Eₖ = ½mv².
- Eₖ = 0.5 × 0.50 × 8.0².
- Evaluate and report joules.
16 J.
Look for these in your work
- Correct formula
- Speed squared
- Energy unit stated
A 2.0 kg trolley increases speed from 3.0 m s⁻¹ to 5.0 m s⁻¹. Find its change in kinetic energy.
Give me a hint
Calculate final kinetic energy minus initial kinetic energy.
Compare my reasoning
- Eᵢ = ½(2.0)(3.0²) = 9.0 J.
- E_f = ½(2.0)(5.0²) = 25 J.
- ΔEₖ = 25 − 9.0.
16 J increase in kinetic energy.
Look for these in your work
- Initial and final values calculated
- Final minus initial used
- Increase stated
A 1000 kg car has 250 kJ of kinetic energy. Find its speed.
Give me a hint
Rearrange Eₖ = ½mv² to v = √(2Eₖ/m); convert kilojoules to joules.
Compare my reasoning
- Eₖ = 250 000 J.
- v = √(2 × 250 000 / 1000) = √500.
- Take the positive speed magnitude.
v ≈ 22.4 m s⁻¹.
Look for these in your work
- Energy converted to joules
- Equation rearranged
- Positive speed and unit stated
Common mistakes
- Forgetting to square speed
- Using velocity direction as a negative kinetic energy
- Confusing work done by the resultant force with energy stored at one instant
What can you explain now?
Which equation gives kinetic energy for mass m and speed v?
Compare with the explanation
Eₖ = ½mv²
Kinetic energy is half mass times speed squared.
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Make it stick
How does the work–motion derivation lead from net work to a change in kinetic energy?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.