PhysicsICSE Class 10 · Conservation of Energy

Principle of Conservation of Energy: Statement, Friction and Energy-Balance Numericals (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The principle of conservation of energy states that energy can neither be created nor destroyed; it can only change from one form to another, so the total energy of an isolated system stays constant. Mechanical energy (U + K) is conserved only when there is no friction or air resistance.

On the rough track, kinetic energy is smaller, but the lost part appears as heat, so the End bar is as tall as the Start bar.

Illustration drawn for QuestionBuddy.

Understand Principle of Conservation of Energy

Statement. *Energy can neither be created nor destroyed. It can only be transformed from one form into another, and the total energy of an isolated system remains constant.*

An isolated system is one that exchanges no energy with its surroundings. When energy seems to vanish, it has always gone somewhere: heat, sound, light, or the energy of another body.

Two kinds of 'conservation'.

Statement	When is it true?
Total energy is conserved	Always
Mechanical energy ( $U + K$ ) is conserved	Only if no friction or air resistance acts

Friction and air resistance turn some mechanical energy into heat (and a little sound). Then $U + K$ decreases, but the heat produced is exactly the mechanical energy lost, so the total is unchanged.

The energy-balance equation. Between a starting point 1 and a later point 2:

$U_{1} + K_{1} = U_{2} + K_{2} + W_{f},$

where $W_{f}$ is the energy turned into heat against friction (zero on a smooth path). This one line solves almost every numerical in the topic.

Consequences on a smooth (frictionless) track.

A body released from rest at height $h$ reaches the bottom with $v = 2 g h$ , whatever the shape of the track. Only the drop in height matters.
It can rise again to its starting height, never higher. A body released from rest cannot reach a point higher than its start.
A body thrown upward at speed $u$ rises to $h = \frac{u ^{2}}{2 g}$ .

On a rough track. The body reaches the bottom more slowly, and it cannot climb back to its starting height. To reach the next hill it needs enough energy for both the height and the friction loss on the way.

Applying it. In every problem:
1. Fix the reference level (usually the ground).
2. Write the total energy at the start.
3. Account for every place it goes: PE, KE, heat.

Everyday examples.

Bicycle brakes: KE of the bicycle becomes heat in the brake pads and rims.
A ball bouncing lower each time: some KE becomes heat and sound at each impact.
A hydroelectric station: PE of stored water becomes KE of falling water, then electrical energy, with some heat in the turbines.

In none of these is energy destroyed; it changes into less useful forms, mostly heat spread through the surroundings.

Principle of Conservation of Energy: formulas and key facts

Energy balance: $U_{1} + K_{1} = U_{2} + K_{2} + W_{f}$; W_f = 0 on a smooth path.
Speed at the bottom of a smooth track: $v = 2 g h$; Any shape of track; released from rest.
Height reached when thrown up at u: $h = \frac{u ^{2}}{2 g}$; No air resistance.
Energy lost to friction: $W_{f} = m g h - \frac{1}{2} m v^{2}$; For a body starting from rest at height h.

How to solve Principle of Conservation of Energy questions

Write one energy-balance line between the two points asked about, then solve for the unknown.

1Convert units: mass in kg (a weight of W kgf means W kg), heights in m, speeds in m/s.
2Choose the reference level and find $U_{1}$ , $K_{1}$ at the start.
3Find $U_{2}$ from the height of the second point.
4Include $W_{f}$ if the path is rough or a loss is given.
5Solve $U_{1} + K_{1} = U_{2} + K_{2} + W_{f}$ for $K_{2}$ , $v$ , $h$ or $W_{f}$ .
6Check: on a smooth path the body never ends higher than its start with more speed.

Principle of Conservation of Energy 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.

Freewheeling down a slope

Suppose a cyclist stops pedalling at the top of a slope 5 m high while moving at 2 m/s, and freewheels to the bottom. Ignore friction and air resistance. ( $g = 10 m s^{- 2}$ )

Find her speed at the bottom.

Try it first, then show the solutionSolution

Per kilogram: $\frac{1}{2} v^{2} = \frac{1}{2} (2)^{2} + 10 \times 5 = 2 + 50 = 52$ .
$v = 104 \approx 10.2$ m/s.

Answer: About 10.2 m/s, whatever her mass.

Memory hook: Starting KE plus PE lost gives the final KE.

Where braking energy goes

Imagine a car of mass 1000 kg moving at 20 m/s brakes to a stop on a level road.

How much kinetic energy is lost, and has it been destroyed?

Try it first, then show the solutionSolution

$K = \frac{1}{2} \times 1000 \times 2 0^{2} = 2 \times 1 0^{5}$ J.
It becomes heat in the brakes and tyres (and some sound); none is destroyed.

Answer: $2 \times 1 0^{5}$ J, changed into heat and sound.

Memory hook: Hot brakes are the missing kinetic energy.

Tricks to remember Principle of Conservation of Energy

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

No new energy, no lost energy
Energy is like money moved between accounts: the total never changes, it just changes account.
Use it when: Stating and applying the principle.
Mechanical needs smooth
'Mechanical energy is conserved' is true only on a Smooth path; 'Energy is conserved' is true on Every path.
Use it when: Two-path questions.
Start = End + Heat
U₁ + K₁ = U₂ + K₂ + heat. Say it as 'start equals end plus heat'.
Use it when: Every friction numerical.
Shape doesn't matter
On a smooth track only the height drop decides the speed, not the curve.
Use it when: Curved-track and slide questions.

Principle of Conservation of Energy: solved examples

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

Example 1: Where does the KE go? · 1 mark · MCQBoard level

A ball rolling on a rough floor slows down and stops. Its kinetic energy has been:
(a) destroyed
(b) changed mainly into heat
(c) changed into potential energy
(d) stored in the ball as extra mass

Try it first, then show the solutionSolution

The floor is level, so the PE does not change. Friction turns the KE into heat (and a little sound).

Answer: (b) changed mainly into heat

Example 2: Stating and illustrating the principle · 3 marks · Short answerBoard level

(a) State the principle of conservation of energy. (b) A cyclist applies the brakes and stops. Is mechanical energy conserved? Is energy conserved? Explain.

Try it first, then show the solutionSolution

(a) Energy can neither be created nor destroyed; it only changes from one form to another, and the total energy of an isolated system stays constant. (1)
(b) Mechanical energy is not conserved: the KE of the cycle falls to zero. (1)
Total energy is conserved: the KE has become heat in the brakes and rims, and some sound. (1)

Answer: (b) Mechanical energy: no. Total energy: yes; KE becomes heat and sound.

Example 3: A smooth curved track · 3 marks · Short answerHarder board level

A 2 kg trolley starts from rest 1.8 m above the lowest point of a smooth curved track. (a) Find its speed at the lowest point. (b) The track then rises again. How high can the trolley climb? (c) Would a steeper or longer track change your answers? ( $g = 10 m s^{- 2}$ )

Try it first, then show the solutionSolution

(a) $m g h = \frac{1}{2} m v^{2}$ : $v = 2 \times 10 \times 1.8 = 6$ m/s. (1)
(b) All the KE becomes PE again: it climbs to 1.8 m, its starting height. (1)
(c) No: on a smooth track only the change in height matters. (1)

Answer: (a) 6 m/s (b) 1.8 m (c) No

Example 4: Smooth path versus rough path · 4 marks · Long answerHarder board level

Two identical toy cars, each of mass 0.5 kg, are released from rest at the top of a hill 2 m high. Car P runs down a smooth path; car Q runs down a rough path and loses 3 J to friction on the way down. ( $g = 10 m s^{- 2}$ )
(a) Find each car's speed at the bottom.
(b) The next hill is 1.5 m high. Which car can certainly reach its top? Explain using energy.

Try it first, then show the solutionSolution

Starting PE $= 0.5 \times 10 \times 2 = 10$ J for each. (½)
(a) P: $K = 10$ J, $v = \frac{2 \times 10}{0.5} = 40 \approx 6.3$ m/s. (1)
Q: $K = 10 - 3 = 7$ J, $v = 28 \approx 5.3$ m/s. (1)
(b) PE needed at the top $= 0.5 \times 10 \times 1.5 = 7.5$ J. P has 10 J and loses nothing, so it reaches the top. Q has only 7 J at the bottom and still meets friction, so it cannot. (1½)

Answer: (a) P about 6.3 m/s, Q about 5.3 m/s (b) P only

Example 5: A skater on a rough track · 4 marks · Long answerHOTS / competency

A skater of mass 45 kg starts from rest at P, 5 m above the ground, and reaches Q at ground level at 9 m/s. From Q the track rises to R, 2 m above the ground, and a further 300 J is lost to friction between Q and R. ( $g = 10 m s^{- 2}$ )
(a) Find the KE at Q.
(b) Find the energy lost between P and Q.
(c) Find the KE and speed at R.

Try it first, then show the solutionSolution

(a) $K_{Q} = \frac{1}{2} \times 45 \times 9^{2} = 1822.5$ J. (1)
(b) $U_{P} = 45 \times 10 \times 5 = 2250$ J, so lost $= 2250 - 1822.5 = 427.5$ J. (1)
(c) $K_{R} = K_{Q} - U_{R} - W_{f} = 1822.5 - 45 \times 10 \times 2 - 300 = 622.5$ J. (1)
$v_{R} = \frac{2 \times 622.5}{45} \approx 5.26$ m/s. (1)

Answer: (a) 1822.5 J (b) 427.5 J (c) 622.5 J, about 5.26 m/s

Example 6: Case study: a throw-up catch · 4 marks · Case-basedHOTS / competency

A fielder throws a 0.16 kg cricket ball straight up at 15 m/s. ( $g = 10 m s^{- 2}$ )
(a) Find its KE at launch.
(b) How high would it rise with no air resistance?
(c) It actually rises to 10 m. How much energy was lost on the way up?
(d) If the same amount is lost on the way down, at what speed does it return to the launch height?

Try it first, then show the solutionSolution

(a) $K = \frac{1}{2} \times 0.16 \times 1 5^{2} = 18$ J. (1)
(b) $h = \frac{u ^{2}}{2 g} = \frac{225}{20} = 11.25$ m. (1)
(c) PE at 10 m $= 0.16 \times 10 \times 10 = 16$ J, so lost $= 18 - 16 = 2$ J, as heat in the air. (1)
(d) $K = 16 - 2 = 14$ J, so $v = \frac{2 \times 14}{0.16} = 175 \approx 13.2$ m/s: slower than it was thrown, though total energy is conserved. (1)

Answer: (a) 18 J (b) 11.25 m (c) 2 J (d) About 13.2 m/s

Where students lose marks in Principle of Conservation of Energy

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

Mistake: Not separating mechanical energy from total energy
Mechanical energy is conserved only without friction; total energy always.
Wrong working
'On the rough path the law of conservation of energy is not obeyed.'
Right working
'On the rough path mechanical energy is not conserved, but the law of conservation of energy is obeyed on both paths.'
Do this: Write two separate sentences, one for each kind.
Usually costs: Reported by CISCE examiners for 2024 Q6(ii).
Mistake: Answering in the wrong terms
If the question names paths or points, the answer must use those names.
Wrong working
'The left one' or 'A' when asked 'along which path?'
Right working
'Path 1.'
Do this: Reuse the exact labels of the question.
Usually costs: Also reported for 2024 Q6(ii).
Mistake: Forgetting the friction loss at a later point
Energy lost before the point asked about must be subtracted too.
Wrong working
30 kg, $K_{Q} = 1500$ J, R 3 m high, 200 J lost from Q to R: $K_{R} = 1500 - 900 = 600$ J.
Right working
$K_{R} = 1500 - 900 - 200 = 400$ J.
Do this: Write the full balance $U_{1} + K_{1} = U_{2} + K_{2} + W_{f}$ .
Usually costs: Reported for 2025 Q6(iii), where some candidates also set U equal to K at R.
Mistake: Saying energy is 'used up' or 'destroyed'
Energy is only converted, usually into heat.
Wrong working
'The KE is destroyed by friction.'
Right working
'Friction converts the KE into heat.'
Do this: Always name the form the energy changes into.
Usually costs: Reasoning marks.

Exam tips

In 2024 Q6(ii) the expected answers were: the marble on the frictionless path surely reaches the next peak; mechanical energy is conserved on path 1 only; the law of conservation of energy holds on both paths.
Quote the statement word for word: 'Energy can neither be created nor destroyed…', adding 'total energy of an isolated system remains constant'.
For friction numericals, write the energy-balance line first; method marks are given for it even if arithmetic slips.
In 2025 Q6(iii), the KE at R was expected as total minus PE at R minus the given friction loss; set out each term.

Practice questions on Principle of Conservation of Energy

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
Which statement is always true?
(a) Mechanical energy is conserved.
(b) The total energy of an isolated system is conserved.
(c) Kinetic energy is conserved.
(d) Potential energy is conserved.
Hint
Friction can spoil all but one.
Answer
(b) The total energy of an isolated system is conserved.
Q2 · 2 marks · Short answerBoard level
A 5 kg box slides from rest down a smooth ramp of vertical height 1.25 m. Find its speed at the bottom. ( $g = 10 m s^{- 2}$ )
Hint
Smooth means $m g h = \frac{1}{2} m v^{2}$ .
Answer
$v = 2 \times 10 \times 1.25 = 5$ m/s.
Q3 · 2 marks · Short answerHarder board level
A 0.3 kg ball is thrown vertically upward at 9 m/s. Using conservation of energy, work out how high it rises and the PE it has at the top. ( $g = 10 m s^{- 2}$ , no air resistance)
Hint
All the KE becomes PE at the top.
Answer
$h = \frac{81}{20} = 4.05$ m; $U = 0.3 \times 10 \times 4.05 = 12.15$ J (equal to the starting KE).
Q4 · 2 marks · Short answerHarder board level
A cyclist and bicycle of total mass 60 kg freewheel from rest down a hill 8 m high and are moving at 10 m/s at the foot of the hill. What amount of energy became heat and sound? ( $g = 10 m s^{- 2}$ )
Hint
Loss = PE at the top − KE at the bottom.
Answer
$4800 - \frac{1}{2} \times 60 \times 100 = 4800 - 3000 = 1800$ J.
Q5 · 3 marks · Short answerHarder board level
A 0.4 kg ball is released from rest 5 m up a rough slope and reaches the bottom at 8 m/s. What percentage of its starting energy was lost to friction? ( $g = 10 m s^{- 2}$ )
Hint
Compare the loss with $m g h$ .
Answer
$U = 20$ J; $K = \frac{1}{2} \times 0.4 \times 64 = 12.8$ J; lost $= 7.2$ J $= 36%$ .
Q6 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A ball released from rest on a smooth curved track can never rise above its starting height.
Reason (R): On a smooth track the total mechanical energy remains constant, and it is all potential energy at the start.
(a) Both A and R are true and R explains A.
(b) Both are true but R does not explain A.
(c) A is true, R is false.
(d) A is false, R is true.
Hint
Rising higher would need more PE than it started with.
Answer
(a) Both A and R are true and R explains A.
Q7 · 3 marks · Short answerHOTS / competency
A 2 kg block slides from rest down a smooth slope of height h, then moves along a rough horizontal floor where a friction force of 5 N acts. It stops after travelling 12 m on the floor. Find h. ( $g = 10 m s^{- 2}$ )
Hint
All the PE ends as work done against friction on the floor.
Answer
Work against friction $= 5 \times 12 = 60$ J $= m g h = 2 \times 10 \times h$ , so $h = 3$ m.

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.State the principle of conservation of energy in one line.Show answer
Energy can neither be created nor destroyed, only changed from one form to another.
2.What does friction convert mechanical energy into?Show answer
Mainly heat (and some sound).
3.Speed at the bottom of a smooth slide 5 m high, starting from rest ( $g = 10$ m s⁻²)?Show answer
10 m/s.
4.Height reached by a ball thrown up at 10 m/s ( $g = 10$ m s⁻², no air resistance)?Show answer
5 m.
5.When is mechanical energy conserved?Show answer
When no friction or air resistance acts.

Principle of Conservation of Energy: frequently asked questions

If friction 'wastes' energy, how is energy still conserved?

The wasted mechanical energy becomes heat and sound. Adding these to the remaining mechanical energy gives the same total as before, so total energy is conserved.

Why can a ball on a smooth track not rise above its starting point?

Its mechanical energy stays equal to the PE it started with. To rise higher it would need more PE than that, which would mean energy appearing from nowhere.

Does the shape of a smooth slide change the speed at the bottom?

No. With no friction, the speed depends only on the drop in height: v = √(2gh). A steeper slide gets you there sooner, not faster.