PhysicsICSE Class 10 · Conservation of Energy

Energy of a Freely Falling Body: Proof that U + K Is Constant, with Numericals (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

For a body dropped from rest at height h with no air resistance, U = mgh and K = 0 at the top. After it falls x, U = mg(h − x) and K = mgx. Just before the ground, U = 0 and K = mgh. U + K = mgh at every point, so mechanical energy is conserved.

Going down from A to C, the blue U bar shrinks while the orange K bar grows, but the whole bar stays the same length: U + K = mgh.

Illustration drawn for QuestionBuddy.

Understand Energy of a Freely Falling Body (U + K Constant)

A freely falling body moves under gravity alone; air resistance is neglected. As it falls, it loses potential energy $U$ and gains exactly as much kinetic energy $K$ . The syllabus asks for the theoretical verification that $U + K$ stays constant.

Set-up. A body of mass $m$ is released from rest at point A, at height $h$ above the ground. Take the ground as the reference level ( $U = 0$ there). Picture a vertical line: A at the top, B a distance $x$ below A, and C just above the ground.

At A (height $h$ , at rest):

$U_{A} = m g h$ , $K_{A} = 0$ , so $U_{A} + K_{A} = m g h$ .

At B (fallen $x$ , height $h - x$ ):

$U_{B} = m g (h - x)$ .
From $v^{2} = u^{2} + 2 g x$ with $u = 0$ : $v_{B}^{2} = 2 g x$ , so $K_{B} = \frac{1}{2} m (2 g x) = m g x$ .
$U_{B} + K_{B} = m g h - m g x + m g x = m g h$ .

At C (fallen $h$ , height 0):

$U_{C} = 0$ ; $v_{C}^{2} = 2 g h$ , so $K_{C} = \frac{1}{2} m (2 g h) = m g h$ .
$U_{C} + K_{C} = m g h$ .

The sum $U + K = m g h$ at A, B and C: the mechanical energy is conserved. The loss of PE between any two points equals the gain in KE.

Results worth knowing.

**Speed after falling $h$ **: $m g h = \frac{1}{2} m v^{2}$ , so $v = 2 g h$ . The mass cancels, so all bodies dropped from the same height reach the ground with the same speed (without air resistance).
Halfway down: $U = K = \frac{1}{2} m g h$ .
**Ratio at height $y$ **: $\frac{K}{U} = \frac{h - y}{y}$ . For example, at one quarter of the height, $K = 3 U$ .
Thrown down or up: start from the total $\frac{1}{2} m u^{2} + m g h$ and share it between $U$ and $K$ at any other point.

Graphs. Against the distance fallen $x$ , $U$ falls in a straight line from $m g h$ to 0 while $K$ rises in a straight line from 0 to $m g h$ ; the two lines cross at $x = \frac{h}{2}$ , and their sum is a horizontal line at $m g h$ .

With air resistance. Part of the energy becomes heat and sound in the air, so $U + K$ decreases as the body falls:

$m g h = \frac{1}{2} m v^{2} + W_{air} .$

The total energy (mechanical + heat + sound) is still conserved.

Units. Use SI throughout: $m$ in kg, $h$ in m, $g$ in m s⁻², energy in J. A weight given as $W$ kgf means a mass of $W$ kg.

Energy of a Freely Falling Body (U + K Constant): formulas and key facts

Mechanical energy of the falling body: $U + K = m g (h - x) + m g x = m g h$; Constant at every point, with no air resistance.
Speed after falling h: $v = 2 g h$; Independent of mass.
KE after falling x: $K = m g x$; Equals the PE lost.
Ratio at height y: $\frac{K}{U} = \frac{h - y}{y}$; Body dropped from rest at height h.
With air resistance: $m g h = \frac{1}{2} m v^{2} + W_{air}$; W_air = energy turned into heat and sound.

How to solve Energy of a Freely Falling Body (U + K Constant) questions

Treat the total energy at the start as a fixed budget, then share it between U and K at the point asked about.

1Fix the reference level (usually the ground) and write the total energy at the start: $m g h$ (plus $\frac{1}{2} m u^{2}$ if it is thrown).
2At the point asked, find $U = m g \times$ (height above the reference).
3Then $K = total - U$ (minus any energy lost to air).
4For speed, use $v = \frac{2 K}{m}$ , or $v = 2 g x$ for a body dropped from rest.
5Check that $U + K$ at every point equals the starting total.

Energy of a Freely Falling Body (U + K Constant) 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.

A falling mango

Suppose a mango of mass 0.25 kg drops from a branch 2.45 m above the ground. ( $g = 10 m s^{- 2}$ , no air resistance)

Find its speed and kinetic energy just before it lands.

Try it first, then show the solutionSolution

$v = 2 \times 10 \times 2.45 = 49 = 7$ m/s.
$K = m g h = 0.25 \times 10 \times 2.45 = 6.125$ J (check: $\frac{1}{2} \times 0.25 \times 49 = 6.125$ J).

Answer: 7 m/s; about 6.1 J.

Memory hook: PE at the top becomes KE at the bottom, joule for joule.

A platform diver

Imagine a diver of mass 50 kg steps off a 10 m platform. ( $g = 10 m s^{- 2}$ , no air resistance)

Find her KE and PE halfway down, and her speed as she enters the water.

Try it first, then show the solutionSolution

Total $= 50 \times 10 \times 10 = 5000$ J.
Halfway (5 m up): $U = 2500$ J, $K = 2500$ J.
At the water: $v = 2 \times 10 \times 10 = 200 \approx 14.1$ m/s.

Answer: 2500 J each halfway; about 14.1 m/s at the water.

Memory hook: Halfway down, the energy is split half and half.

Tricks to remember Energy of a Freely Falling Body (U + K Constant)

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

A-B-C table
Draw three rows A, B, C and three columns U, K, Total. The Total column is the same number all the way down.
Use it when: The three-point proof.
Lost here, found there
Every joule of PE lost appears as a joule of KE gained: ΔK = mgx.
Use it when: Finding K quickly at any point.
Root two-g-h
'Two G H, then square root': v = √(2gh).
Use it when: Speed after a fall.
Quarter height, three to one
At a quarter of the height, three quarters of the energy is kinetic: K = 3U.
Use it when: Ratio questions.

Energy of a Freely Falling Body (U + K Constant): solved examples

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

Example 1: Energy at half height · 1 mark · MCQBoard level

A body falls freely from a height h. When it is at height $\frac{h}{2}$ :
(a) U = K
(b) U = 2K
(c) K = 2U
(d) K = 0

Try it first, then show the solutionSolution

At $\frac{h}{2}$ , $U = \frac{m g h}{2}$ , and the PE lost, $\frac{m g h}{2}$ , has become KE.

Answer: (a) U = K

Example 2: Speed and KE at the ground · 2 marks · Short answerBoard level

A 4 kg ball falls from rest through 20 m to the ground. Find its speed and kinetic energy just before it hits the ground. ( $g = 10 m s^{- 2}$ )

Try it first, then show the solutionSolution

$v = 2 g h = 2 \times 10 \times 20 = 20$ m/s. (1)
$K = m g h = 4 \times 10 \times 20 = 800$ J. (1)

Answer: 20 m/s; 800 J

Example 3: Verifying U + K with numbers · 3 marks · Short answerHarder board level

A 3 kg stone is released from rest at A, 40 m above the ground. B is 15 m below A, and C is just above the ground. Find U, K and U + K at A, B and C. ( $g = 10 m s^{- 2}$ )

Try it first, then show the solutionSolution

A: $U = 3 \times 10 \times 40 = 1200$ J, $K = 0$ ; total 1200 J. (1)
B (25 m high): $U = 3 \times 10 \times 25 = 750$ J; $v^{2} = 2 \times 10 \times 15 = 300$ , $K = \frac{1}{2} \times 3 \times 300 = 450$ J; total 1200 J. (1)
C: $U = 0$ ; $v^{2} = 2 \times 10 \times 40 = 800$ , $K = \frac{1}{2} \times 3 \times 800 = 1200$ J; total 1200 J. (1)

Answer: U + K = 1200 J at A, B and C

Example 4: Where is K three times U? · 3 marks · Short answerHOTS / competency

A ball is dropped from a height of 80 m. (a) At what height above the ground is its kinetic energy three times its potential energy? (b) What is its speed there? ( $g = 10 m s^{- 2}$ )

Try it first, then show the solutionSolution

(a) Total $= m g \times 80$ . With $K = 3 U$ , total $= 4 U$ , so $4 m g y = 80 m g$ and $y = 20$ m. (1½)
(b) It has fallen $80 - 20 = 60$ m: $v = 2 \times 10 \times 60 = 1200 \approx 34.6$ m/s. (1½)

Answer: (a) 20 m (b) About 34.6 m/s

Example 5: Energy lost to air · 3 marks · Short answerHarder board level

A body of mass 1.5 kg falls from a height of 60 m and reaches the ground at 30 m/s. (a) Find the PE at the start and the KE at the ground. (b) How much energy was lost to air resistance, and where did it go? ( $g = 10 m s^{- 2}$ )

Try it first, then show the solutionSolution

(a) $U = 1.5 \times 10 \times 60 = 900$ J; $K = \frac{1}{2} \times 1.5 \times 3 0^{2} = 675$ J. (2)
(b) $900 - 675 = 225$ J, changed into heat (and a little sound) in the air and the body. Total energy is still conserved. (1)

Answer: (a) 900 J, 675 J (b) 225 J, as heat and sound

Example 6: Case study: a falling coconut · 4 marks · Case-basedHOTS / competency

A coconut of mass 1.2 kg breaks off a tree 12.8 m above the ground. ( $g = 10 m s^{- 2}$ , no air resistance)
(a) Find its PE at the start.
(b) Find its PE and KE when it is 5 m above the ground.
(c) Find its speed just before it lands.
(d) A 0.3 kg dry husk falls from the same branch at the same time. Would it land at a different speed? Explain using energy.

Try it first, then show the solutionSolution

(a) $U = 1.2 \times 10 \times 12.8 = 153.6$ J. (1)
(b) $U = 1.2 \times 10 \times 5 = 60$ J; $K = 153.6 - 60 = 93.6$ J. (1)
(c) $v = 2 \times 10 \times 12.8 = 256 = 16$ m/s. (1)
(d) No (without air resistance): $m g h = \frac{1}{2} m v^{2}$ gives $v = 2 g h$ , in which $m$ cancels, so it also lands at 16 m/s. (1)

Answer: (a) 153.6 J (b) 60 J, 93.6 J (c) 16 m/s (d) Same speed; mass cancels.

Where students lose marks in Energy of a Freely Falling Body (U + K Constant)

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

Mistake: Writing U at B as mgx instead of mg(h − x)
PE depends on the height above the reference, not on the distance fallen.
Wrong working
3 kg dropped from 20 m, after falling 5 m: $U = 3 \times 10 \times 5 = 150$ J.
Right working
Height is 15 m: $U = 3 \times 10 \times 15 = 450$ J, and $K = 150$ J.
Do this: Write 'height of B above ground = h − x' as a separate line.
Usually costs: Usually the whole B row of the proof.
Mistake: Using v = gh
The speed after falling h is $2 g h$ .
Wrong working
$h = 20$ m: $v = 10 \times 20 = 200$ m/s.
Right working
$v = 2 \times 10 \times 20 = 20$ m/s.
Do this: Derive it from $m g h = \frac{1}{2} m v^{2}$ if unsure.
Usually costs: The answer and often the following part.
Mistake: Converting kgf wrongly
A weight of W kgf is the weight of a W kg mass.
Wrong working
'Weight 50 kgf, so m = 500 kg' or 'm = 50 g'.
Right working
$m = 50$ kg; weight $= 50 \times 10 = 500$ N.
Do this: Write m = … kg before any energy formula.
Usually costs: CISCE examiners reported this in 2025 Q6(iii), with the error carried into later parts.
Mistake: Mixing up mechanical energy and total energy
Mechanical energy is conserved only with no friction or air resistance; total energy is always conserved.
Wrong working
'With air resistance, energy is not conserved.'
Right working
'With air resistance, mechanical energy decreases, but total energy is conserved: the loss appears as heat and sound.'
Do this: Say which energy you mean every time.
Usually costs: Reported for 2024 Q6(ii), where candidates could not separate the two.

Exam tips

The 'show that U + K is constant' question (3 marks) needs all three points with working: U, K and total at A, B and C, and the kinematic step $v^{2} = 2 g x$ at B.
State the assumption: no air resistance. Without it, the proof is incomplete.
In 2025 Q6(iii) the expected first lines were $K = \frac{1}{2} m v^{2}$ with $m$ in kg, then $m g h = \frac{1}{2} m v^{2}$ for the height; convert kgf to kg first.
Give speeds in m/s and energies in J; keep $g$ as the question gives it (10 m s⁻² or 9.8 m s⁻²).

Practice questions on Energy of a Freely Falling Body (U + K Constant)

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
Just before a freely falling body hits the ground, it has:
(a) maximum PE and zero KE
(b) zero PE and maximum KE
(c) equal PE and KE
(d) zero PE and zero KE
Hint
Ground is the reference level.
Answer
(b) zero PE and maximum KE
Q2 · 2 marks · Short answerBoard level
A 0.6 kg ball slips off a shelf 7.2 m above the floor. Find its speed and kinetic energy just before it hits the ground. ( $g = 10 m s^{- 2}$ )
Hint
$v = 2 g h$ ; $K = m g h$ .
Answer
$v = 144 = 12$ m/s; $K = 0.6 \times 10 \times 7.2 = 43.2$ J.
Q3 · 3 marks · Short answerHarder board level
A 1 kg body is dropped from 30 m. Copy and complete a table of U, K and U + K at heights of 30 m, 20 m, 10 m and 0 m. ( $g = 10 m s^{- 2}$ )
Hint
Total = 300 J at every height.
Answer
30 m: U 300 J, K 0; 20 m: U 200 J, K 100 J; 10 m: U 100 J, K 200 J; 0 m: U 0, K 300 J. U + K = 300 J throughout.
Q4 · 2 marks · Short answerHarder board level
A body falls from rest from a height of 50 m. At what height above the ground is its speed 20 m/s? ( $g = 10 m s^{- 2}$ )
Hint
Distance fallen $= \frac{v ^{2}}{2 g}$ .
Answer
Fallen $x = \frac{2 0 ^{2}}{2 \times 10} = 20$ m, so its height is $50 - 20 = 30$ m.
Q5 · 2 marks · Short answerHarder board level
A body dropped from 60 m is now 15 m above the ground. Find the ratio of its kinetic energy to its potential energy.
Hint
K is mg × distance fallen; U is mg × height.
Answer
$\frac{K}{U} = \frac{60 - 15}{15} = 3$ , i.e. $3 : 1$ .
Q6 · 3 marks · Short answerHarder board level
A 2 kg body falls from a height of 20 m. Air resistance turns 40 J of its energy into heat on the way down. Calculate how fast it is moving as it reaches the ground. ( $g = 10 m s^{- 2}$ )
Hint
$m g h = \frac{1}{2} m v^{2} + W_{air}$ .
Answer
$m g h = 400$ J, so $K = 400 - 40 = 360$ J; $v = \frac{2 \times 360}{2} = 360 \approx 19.0$ m/s.
Q7 · 4 marks · Long answerHOTS / competency
From a balcony 10 m high, a 0.4 kg ball is thrown straight down at 5 m/s. (a) Find its total mechanical energy at the start (ground as reference). (b) How fast is it moving as it lands? (c) How much faster is this than if it had simply been dropped? ( $g = 10 m s^{- 2}$ , no air resistance)
Hint
Start with K + U, not just U.
Answer
(a) $\frac{1}{2} \times 0.4 \times 25 + 0.4 \times 10 \times 10 = 5 + 40 = 45$ J. (b) $v = \frac{2 \times 45}{0.4} = 225 = 15$ m/s. (c) Dropped: $200 \approx 14.1$ m/s, so about 0.9 m/s faster.

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.Speed of a body after falling 5 m from rest ( $g = 10$ m s⁻²)?Show answer
10 m/s.
2.For a body dropped from h, what is U + K at every point?Show answer
mgh.
3.Ratio U : K at half the height?Show answer
1 : 1.
4.Does the landing speed depend on the mass (no air resistance)?Show answer
No: $v = 2 g h$ .
5.KE of a 2 kg body just before landing after a 10 m fall ( $g = 10$ m s⁻²)?Show answer
200 J.

Energy of a Freely Falling Body (U + K Constant): frequently asked questions

How do you show that the mechanical energy of a freely falling body is conserved?

Find U and K at the top (mgh, 0), after falling x (mg(h − x), mgx using v² = 2gx) and at the bottom (0, mgh). The sum is mgh each time, so U + K is constant.

Is energy conserved when air resistance acts on a falling body?

The total energy is conserved, but mechanical energy is not: part of it becomes heat and sound in the air, so U + K falls as the body descends.

Why do a heavy and a light ball dropped together land at the same speed?

From mgh = ½mv², the mass cancels and v = √(2gh). Without air resistance, the landing speed depends only on the height.