PhysicsICSE Class 10 · Different Types of Energy

Kinetic Energy: Derivation of K = ½mv², Work-Energy Relation and Solved Numericals (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

Kinetic energy is the energy a body has because of its motion. A body of mass m moving at speed v has K = ½mv², equal to the work done to bring it from rest to that speed. Doubling the speed makes the kinetic energy four times as large. Its forms are translational, rotational and vibrational.

Energy depends on v squared: doubling the speed makes the kinetic energy four times as large, so K = ½mv² grows fast.

Illustration drawn for QuestionBuddy.

Understand Kinetic Energy (K = ½mv²)

Kinetic energy (KE, $K$ ) is the energy a body possesses because of its motion. It is measured by the work the body can do before coming to rest, or equally by the work needed to bring it from rest to its speed.

**Derivation of $K = \frac{1}{2} m v^{2}$ (named in the syllabus).** A body of mass $m$ is at rest on a smooth surface. A constant force $F$ acts on it and gives it a uniform acceleration $a$ , so $F = ma$ . After a displacement $s$ in the direction of the force, its speed is $v$ . From $v^{2} = u^{2} + 2 a s$ with $u = 0$ :

$s = \frac{v ^{2}}{2 a} .$

Work done by the force:

$W = F \times s = ma \times \frac{v ^{2}}{2 a} = \frac{1}{2} m v^{2} .$

This work appears as the kinetic energy of the body, so $K = \frac{1}{2} m v^{2}$ .

Work-energy relation. If the speed changes from $u$ to $v$ , the net work done on the body equals the change in its kinetic energy:

$W = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} .$

A braking force does negative work and removes KE; this gives the stopping force or distance.

What KE depends on. $K \propto m$ and $K \propto v^{2}$ . Double the mass: double the KE. Double the speed: four times the KE. Since stopping distance under the same braking force is proportional to KE, a car at twice the speed needs about four times the distance to stop. KE is never negative.

Forms of kinetic energy (only simple examples are needed):

Translational: the whole body moves from one place to another (a moving bus, a falling stone, a flying bullet).
Rotational: the body spins about an axis (a ceiling fan, a spinning top, the Earth spinning on its axis).
Vibrational: the body moves to and fro about a mean position (a plucked guitar string, the prongs of a struck tuning fork).

A rolling wheel has both translational and rotational KE. ICSE numericals on KE are for translational motion only.

Units. Joule. Use $m$ in kg and $v$ in m/s; convert km/h by multiplying by $\frac{5}{18}$ (dividing by 3.6) before squaring.

Kinetic Energy (K = ½mv²): formulas and key facts

Kinetic energy: $K = \frac{1}{2} m v^{2}$; v in m/s, m in kg.
Speed from KE: $v = \frac{2 K}{m}$; Rearranged.
Work-energy relation: $W = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$; Net work = change in KE.
Stopping force: $F \times s = \frac{1}{2} m v^{2}$; Average force that stops the body in distance s.
Speed conversion: $1 km/h = \frac{5}{18} m/s$; Convert before squaring.

How to solve Kinetic Energy (K = ½mv²) questions

Convert units first, because the speed is squared.

1Convert mass to kg and speed to m/s.
2For KE, substitute in $K = \frac{1}{2} m v^{2}$ ; remember to square $v$ .
3For a change of speed, find the KE before and after; the work done is the difference.
4For a stopping force, set $F \times s$ equal to the KE lost.
5For 'what if' questions, use proportion: $K \propto m$ and $K \propto v^{2}$ .
6Give the answer in joules, with a sign for work if asked.

Kinetic Energy (K = ½mv²) in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Mass versus speed

Suppose a 70 kg sprinter runs at 10 m/s.

Find the sprinter's KE. How fast would a 7 kg bowling ball have to move to have the same KE?

Try it first, then show the solutionSolution

$K = \frac{1}{2} \times 70 \times 1 0^{2} = 3500$ J.
$v = \frac{2 \times 3500}{7} = 1000 \approx 31.6$ m/s.

Answer: 3500 J; about 31.6 m/s.

Memory hook: One tenth of the mass needs only about three times the speed, because speed is squared.

Why speed limits matter

Imagine a 1000 kg car on a highway at 90 km/h, and the same car at 60 km/h.

Compare its kinetic energy at the two speeds.

Try it first, then show the solutionSolution

90 km/h $= 25$ m/s: $K = \frac{1}{2} \times 1000 \times 2 5^{2} = 312500$ J.
60 km/h $\approx 16.7$ m/s: $K \approx 138900$ J; ratio $= (\frac{90}{60})^{2} = 2.25$ .

Answer: At 90 km/h the car has 2.25 times the KE it has at 60 km/h.

Memory hook: 1.5 times the speed, 2.25 times the energy to stop.

Tricks to remember Kinetic Energy (K = ½mv²)

Short memory aids. Say each one aloud once, then use it on a question the same day.

Half m v squared
Say it as 'half, m, v, squared': four words, and the last one is the one students forget.
Use it when: Every KE numerical.
Double speed, four times
v × 2 → K × 4; v × 3 → K × 9.
Use it when: Ratio and MCQ questions.
Derivation chain
F = ma, s = v²/2a, W = Fs = ½mv².
Use it when: The 'derive K = ½mv²' question.
Move, spin, shake
Translational (move), rotational (spin), vibrational (shake).
Use it when: Forms of kinetic energy.

Kinetic Energy (K = ½mv²): solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: KE of a moving body · 1 mark · MCQBoard level

A 4 kg body moves at 3 m/s. Its kinetic energy is:
(a) 12 J
(b) 18 J
(c) 36 J
(d) 6 J

Try it first, then show the solutionSolution

$K = \frac{1}{2} \times 4 \times 3^{2}$ .
$= 2 \times 9 = 18$ J.

Answer: (b) 18 J

Example 2: Speed from KE · 2 marks · Short answerBoard level

A ball of mass 0.2 kg has a kinetic energy of 10 J. Find its speed.

Try it first, then show the solutionSolution

$v = \frac{2 K}{m} = \frac{2 \times 10}{0.2}$ . (1)
$= 100 = 10$ m/s. (1)

Answer: 10 m/s

Example 3: Work done to speed up · 3 marks · Short answerHarder board level

A scooter and its rider have a total mass of 120 kg. The scooter speeds up from 18 km/h to 54 km/h. Find (a) the initial and final KE, (b) the work done on it.

Try it first, then show the solutionSolution

18 km/h $= 5$ m/s; 54 km/h $= 15$ m/s. (½)
(a) $K_{1} = \frac{1}{2} \times 120 \times 5^{2} = 1500$ J; $K_{2} = \frac{1}{2} \times 120 \times 1 5^{2} = 13500$ J. (1½)
(b) $W = 13500 - 1500 = 12000$ J. (1)

Answer: (a) 1500 J, 13 500 J (b) 12 000 J

Example 4: Braking force · 3 marks · Short answerHarder board level

A car of mass 800 kg moving at 72 km/h is brought to rest in 40 m by its brakes. Find (a) its KE before braking, (b) the average braking force.

Try it first, then show the solutionSolution

72 km/h $= 20$ m/s; $K = \frac{1}{2} \times 800 \times 2 0^{2} = 1.6 \times 1 0^{5}$ J. (1½)
Work done by the brakes removes all the KE: $F \times 40 = 1.6 \times 1 0^{5}$ , so $F = 4000$ N. (1½)

Answer: (a) $1.6 \times 1 0^{5}$ J (b) 4000 N

Example 5: Stopping distance at double speed · 3 marks · Short answerHOTS / competency

With the same braking force, a car stops in 12 m from 36 km/h. Without using the braking force's value, find its stopping distance from 72 km/h. Explain.

Try it first, then show the solutionSolution

The brakes do work $F \times s$ equal to the KE lost, so $s \propto K \propto v^{2}$ . (1)
Doubling the speed makes the KE four times as much. (1)
$s = 4 \times 12 = 48$ m. (1)

Answer: 48 m: the KE, and so the stopping distance, is four times as large.

Example 6: Case study: a bicycle ride · 4 marks · Case-basedHOTS / competency

A cyclist and her bicycle (total mass 80 kg) ride at 6 m/s. She rings the bell, then brakes to a stop.
(a) Name the form of KE of (i) the bicycle frame, (ii) the spinning wheels, (iii) the bell when rung.
(b) Find the translational KE before braking.
(c) How much work do the brakes (friction) do in stopping her?

Try it first, then show the solutionSolution

(a) (i) Translational (ii) Rotational, as well as translational (iii) Vibrational. (2)
(b) $K = \frac{1}{2} \times 80 \times 6^{2} = 1440$ J. (1)
(c) The brakes remove all of it: $W = - 1440$ J (negative work). (1)

Answer: (a) Translational; rotational (and translational); vibrational (b) 1440 J (c) −1440 J

Where students lose marks in Kinetic Energy (K = ½mv²)

The slips examiners see most often, and what to write instead.

Mistake: Forgetting to square the speed
KE uses $v^{2}$ .
Wrong working
$K = \frac{1}{2} \times 4 \times 3 = 6$ J.
Right working
$K = \frac{1}{2} \times 4 \times 9 = 18$ J.
Do this: Write $v^{2}$ in the formula line, then substitute.
Usually costs: The answer mark.
Mistake: Using km/h in the formula
Squaring a speed in km/h gives a wrong number and a wrong unit.
Wrong working
$K = \frac{1}{2} \times 800 \times 7 2^{2}$ .
Right working
72 km/h $= 20$ m/s; $K = \frac{1}{2} \times 800 \times 2 0^{2} = 1.6 \times 1 0^{5}$ J.
Do this: Multiply km/h by $\frac{5}{18}$ first.
Usually costs: The answer mark.
Mistake: Doubling KE when speed doubles
KE is proportional to $v^{2}$ , not $v$ .
Wrong working
Speed doubled, so KE doubled.
Right working
Speed doubled, so KE becomes four times.
Do this: Square the factor of change in speed.
Usually costs: MCQ and ratio marks.
Mistake: Skipping steps in the derivation
The marks are for $F = ma$ , $v^{2} = 2 a s$ , and $W = F s$ combined.
Wrong working
$K = \frac{1}{2} m v^{2}$ because KE depends on mass and speed.
Right working
$W = F s = ma \cdot \frac{v ^{2}}{2 a} = \frac{1}{2} m v^{2}$ , stored as KE.
Do this: Learn the three-line chain.
Usually costs: Derivation marks.

Exam tips

The derivation of $K = \frac{1}{2} m v^{2}$ is named in the 2027 syllabus ('derivation included'); start from a body at rest acted on by a constant force.
KE numericals are limited to translational motion; for rotational and vibrational KE only simple examples are asked.
In ratio questions, write $\frac{K _{1}}{K _{2}} = \frac{m _{1} v _{1}^{2}}{m _{2} v _{2}^{2}}$ and simplify.
When a body is brought to rest, the work done by the stopping force equals the KE it had (with a minus sign).

Practice questions on Kinetic Energy (K = ½mv²)

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The kinetic energy of a body depends on:
(a) its mass only
(b) its speed only
(c) both its mass and its speed
(d) its height above the ground
Hint
Look at $\frac{1}{2} m v^{2}$ .
Answer
(c)
Q2 · 1 mark · Very short answerBoard level
Find the KE of a car of mass 1500 kg moving at 10 m/s.
Hint
Square the speed.
Answer
$\frac{1}{2} \times 1500 \times 100 = 75000$ J.
Q3 · 2 marks · Short answerHarder board level
What happens to the KE of a body if (a) its speed is tripled, (b) its mass is doubled and its speed halved?
Hint
$K \propto m v^{2}$ .
Answer
(a) It becomes 9 times. (b) $2 \times \frac{1}{4} = \frac{1}{2}$ : it halves.
Q4 · 3 marks · Short answerHarder board level
A bullet of mass 20 g moving at 300 m/s enters a wooden block and stops after travelling 6 cm into it. Find (a) the bullet's KE, (b) the average resistive force of the wood.
Hint
Convert g to kg and cm to m.
Answer
(a) $\frac{1}{2} \times 0.02 \times 30 0^{2} = 900$ J. (b) $F = \frac{900}{0.06} = 1.5 \times 1 0^{4}$ N.
Q5 · 2 marks · Short answerHarder board level
Name the form of kinetic energy in: (a) a spinning top, (b) the prongs of a struck tuning fork, (c) a falling raindrop, (d) the Earth turning on its axis.
Hint
Move, spin or shake?
Answer
(a) Rotational (b) Vibrational (c) Translational (d) Rotational
Q6 · 3 marks · Short answerHarder board level
A 5 kg trolley at rest on a smooth floor is pushed by a horizontal force of 30 N through 4 m. Find (a) its KE, (b) its speed.
Hint
Work done = KE gained.
Answer
(a) $K = 30 \times 4 = 120$ J. (b) $v = \frac{2 \times 120}{5} = 48 \approx 6.93$ m/s.
Q7 · 3 marks · Short answerHOTS / competency
A 2 kg ball moves at 10 m/s. (a) Find its KE. (b) At what speed would its KE be doubled? (c) What mass moving at 10 m/s has the same KE as a 1 kg body moving at 20 m/s?
Hint
Doubling KE multiplies v by $2$ .
Answer
(a) 100 J. (b) $102 \approx 14.1$ m/s. (c) The 1 kg body has 200 J; $m = \frac{2 \times 200}{1 0 ^{2}} = 4$ kg.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.KE of a 2 kg body at 5 m/s?Show answer
25 J.
2.Speed doubled: KE becomes?Show answer
Four times.
3.36 km/h in m/s?Show answer
10 m/s.
4.Form of KE of a vibrating guitar string?Show answer
Vibrational.
5.Work needed to stop a body with KE 500 J?Show answer
500 J (done against it, i.e. −500 J by the stopping force).

Kinetic Energy (K = ½mv²): frequently asked questions

How is K = ½mv² derived?

A constant force F = ma accelerates a body from rest to speed v over a distance s = v²/2a. The work done, Fs = ma × v²/2a = ½mv², is stored as kinetic energy.

Why does kinetic energy become four times when speed doubles?

Because kinetic energy depends on the square of the speed: (2v)² = 4v².

What are the forms of kinetic energy?

Translational (a body moving from place to place), rotational (a body spinning, like a fan) and vibrational (a body moving to and fro, like a guitar string).