ScienceCBSE Class 9 · Work, Energy, and Simple Machines

Work, Kinetic Energy, Potential Energy and Conservation of Energy (Class 9)

Updated 17 October 2026 · Fact-checked

Work done by a constant force is W = F × s, where s is the displacement along the force; the unit is the joule (1 J = 1 N m). No work is done if there is no force, no displacement, or the force is perpendicular to the displacement. The work-energy theorem says work done on an object equals the change in its energy; kinetic energy is K = ½mv², gravitational potential energy is U = mgh, and when only gravity acts the total K + U is conserved.

As the ball falls, potential energy U shrinks and kinetic energy K grows by the same amount, so U + K stays constant when only gravity acts.

Illustration drawn for QuestionBuddy.

Understand Work, Kinetic Energy, Potential Energy and Conservation

Work. Lifting three bags, or one bag three times as high, takes three times as much work, so work grows with force and with displacement along the force: W = F × s. The SI unit is the joule (J); 1 J is the work done when a 1 N force moves an object 1 m along the force, and 1 J = 1 N m = 1 kg m² s⁻². Always say which force does work on which object. On a force-displacement graph, work is the area under the graph. Work has no direction but can be positive or negative. Zero work is done when there is no force, no displacement (pushing a wall, holding a barbell still), or the force is perpendicular to the displacement (carrying a box along level ground; the upward force does no work). Positive work: force and displacement in the same direction (pushing a wheelchair, lifting a dumbbell). Negative work: force opposite to the displacement (a goalkeeper stopping a ball, friction, brakes). Both objects in an interaction pair do work on each other: the goalkeeper does negative work on the ball and the ball does positive work on her hands.

Energy and the work-energy theorem. An object that can do work has energy. Work done on an object equals the change in its energy: positive work increases energy, negative work decreases it. The unit is the joule, after James Prescott Joule. Forms of energy: mechanical, thermal, light, sound, electrical, chemical, nuclear. Conversions: bulb (electrical to light and heat), muscles (chemical to mechanical), solar panel (light to electrical), photosynthesis (light to chemical). Energy can also be transferred as heat, radiation, in circuits and in sound waves.

Kinetic energy. Energy due to motion. A constant force F changes the velocity from u to v over displacement s: v² = u² + 2as, so s = (v² − u²) ÷ 2a, and W = F × s = ma × (v² − u²) ÷ 2a = ½m(v² − u²). Starting from rest, K = ½mv². Doubling the speed makes the kinetic energy four times; doubling the mass doubles it. A cricket ball of about 0.2 kg bowled at 43 m s⁻¹ has about 185 J.

Potential energy. Energy stored because of deformed shape (a stretched slingshot band, a compressed spring) or relative positions (magnetic, electric, gravitational). To lift a mass m slowly through height h the force is mg, so the work mgh is stored: U = mgh, with U = 0 at the ground, valid near the Earth's surface. It depends only on the height, not on the path. Work done against friction is not stored as potential energy.

Conservation of mechanical energy. Mechanical energy = K + U. For a body falling from rest at height h: at the top U = mgh and K = 0; after time t, v = gt, K = ½mg²t² and U = mgh − ½mg²t²; just before landing K = mgh and U = 0. The total mgh is constant: potential energy lost equals kinetic energy gained. For a slide or free fall from height h, ½mv² = mgh gives v = √(2gh), independent of mass and of the shape of the slide (ignoring friction). A pendulum has only potential energy at the extremes and only kinetic energy at the lowest point; it stops slowly because of friction and air resistance. In an isolated system total energy is always conserved, though mechanical energy may change into heat and sound; a perpetual motion machine is impossible.

Work, Kinetic Energy, Potential Energy and Conservation: formulas and key facts

Work: $W = F \times s (s a l o n g t h e f or ce) ∣1 J = 1 N m$
Work-energy theorem: $W or k d o n eo nan o bj ec t = c han g e ini t se n er g y ∣ W = ½ m (v^{2} - u^{2})$
Kinetic energy: $K = ½ m v^{2}$
Gravitational potential energy: $U = m g h (g = 9.8 m s^{- 2})$
Conservation of mechanical energy: $K + U = co n s t an tw h e n o n l y g r a v i t y a c t s ∣ v = \sqrt (2 g h) a f t er f a l l in g h$
Zero work: $F = 0∣ s = 0∣ F p er p e n d i c u l a r t os$

How to solve Work, Kinetic Energy, Potential Energy and Conservation questions

Use this method for work and energy numericals.

1List m, F, s, u, v, h with SI units (kg, N, m, m s⁻¹) and use g = 9.8 m s⁻² unless told otherwise.
2Decide whether work is positive, negative or zero by comparing the direction of force and displacement.
3For a change of speed use work = change in kinetic energy; for a change of height use work = change in potential energy.
4For free fall or a smooth slide use K + U = constant between two points.
5With friction, write: work by push + work by friction = change in K (friction work is negative).
6State the answer in joules with the sign if the sense of work matters.

Work, Kinetic Energy, Potential Energy and Conservation 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.

Is the student doing work?

Suppose a student pushes a 20 kg crate 5 m along a floor with a steady horizontal force of 20 N, then carries a 5 kg bag horizontally for 10 m, and finally pushes against a wall for a minute.

Find the work done on the crate and decide whether work is done on the bag and on the wall.

Try it first, then show the solutionSolution

Crate: W = F × s = 20 × 5 = 100 J (positive: force and displacement are in the same direction).
Bag: the upward force supporting the bag is perpendicular to the horizontal displacement, so no work is done on the bag by that force.
Wall: the wall does not move (s = 0), so no work is done on it, though the student's muscles tire by using internal energy.

Answer: 100 J on the crate; zero on the bag; zero on the wall.

Memory hook: No displacement, no work.

Speed at the bottom of a slide

Suppose a child of any mass slides from rest down a smooth slide whose top is 2.5 m above the ground.

Find her speed at the bottom (g = 9.8 m s⁻²) and say whether it depends on her mass.

Try it first, then show the solutionSolution

mgh = ½mv², so v = √(2gh) = √(2 × 9.8 × 2.5) = √49 = 7 m s⁻¹.
The mass cancels, so the speed does not depend on her mass (nor on the shape of a smooth slide).

Answer: 7 m s⁻¹, independent of mass.

Memory hook: Height decides the speed.

Tricks to remember Work, Kinetic Energy, Potential Energy and Conservation

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

Force along the move
Work needs a force and a displacement in that force's direction. Perpendicular: zero.
Use it when: Deciding if work is done.
Square the speed
K = ½mv²: double the speed, four times the energy.
Use it when: Kinetic energy comparisons.
mgh up, ½mv² down
Potential energy lost in a fall appears as kinetic energy: mgh = ½mv².
Use it when: Free fall and slide questions.
Friction takes, push gives
Net work = push work − friction work = change in K.
Use it when: Work-energy problems with friction.

Work, Kinetic Energy, Potential Energy and Conservation: solved examples

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

Example 1: Positive, negative and zero work · 3 marks · Short answerBoard level

(a) A force of 20 N moves a crate 5 m in the direction of the force. Find the work done. (1)
(b) Friction of 10 N acts on a block as it slides 4 m. Find the work done by friction. (1)
(c) A person carries a bag horizontally along a level road. Is any work done on the bag? Give a reason. (1)

Try it first, then show the solutionSolution

(a) W = F × s = 20 × 5 = 100 J, positive. (1)
(b) Friction opposes the displacement, so W = −10 × 4 = −40 J (negative work). (1)
(c) No work by the supporting force, because the force is vertical and the displacement is horizontal (perpendicular). (1)

Answer: (a) 100 J; (b) −40 J; (c) none, the force is perpendicular to the displacement.

Example 2: Kinetic and potential energy · 3 marks · Short answerBoard level

(Take g = 9.8 m s⁻².) (a) Find the kinetic energy of a 2 kg ball moving at 6 m s⁻¹. (1)
(b) What is it if the speed doubles? (1)
(c) Find the potential energy of a 3 kg object at a height of 5 m. (1)

Try it first, then show the solutionSolution

(a) K = ½mv² = ½ × 2 × 6² = 36 J. (1)
(b) At 12 m s⁻¹: K = ½ × 2 × 144 = 144 J, which is four times as large. (1)
(c) U = mgh = 3 × 9.8 × 5 = 147 J. (1)

Answer: (a) 36 J; (b) 144 J; (c) 147 J.

Example 3: A coconut falling from a tree · 5 marks · Long answerHarder board level

(Take g = 9.8 m s⁻².) A 1.5 kg coconut falls from rest from a height of 9.8 m.
(a) Find its potential energy at the top. (1)
(b) After 1 s find its speed, height, K and U, and verify that K + U is constant. (2)
(c) Find its speed just before it lands. (1)
(d) At what height is its kinetic energy equal to its potential energy? (1)

Try it first, then show the solutionSolution

(a) U = mgh = 1.5 × 9.8 × 9.8 = 144.06 J. (1)
(b) v = gt = 9.8 m s⁻¹; fall = ½ × 9.8 × 1² = 4.9 m, so height = 9.8 − 4.9 = 4.9 m. K = ½ × 1.5 × 9.8² = 72.03 J; U = 1.5 × 9.8 × 4.9 = 72.03 J; K + U = 144.06 J, constant. (2)
(c) mgh = ½mv² gives v = √(2gh) = √(2 × 9.8 × 9.8) = √192.08 = 13.9 m s⁻¹. (1)
(d) K = U means each is half of 144.06 J, so mgh' = 72.03 J, giving h' = 4.9 m (half the starting height). (1)

Answer: (a) 144.06 J; (b) 9.8 m s⁻¹, 4.9 m, 72.03 J each, total 144.06 J; (c) 13.9 m s⁻¹; (d) 4.9 m.

Example 4: Work-energy theorem with friction · 5 marks · Long answerHarder board level

A 4 kg box is pushed from rest along a rough floor by a horizontal force of 30 N over 5 m. The friction is 10 N throughout.
(a) Find the work done by the push and the work done by friction. (1)
(b) Find the box's kinetic energy after 5 m. (1)
(c) Find its speed. (1)
(d) The push stops. How far does the box slide before it stops? (2)

Try it first, then show the solutionSolution

(a) Push: 30 × 5 = 150 J. Friction: −10 × 5 = −50 J. (1)
(b) Net work = 150 − 50 = 100 J, which is the kinetic energy gained from rest. (1)
(c) ½ × 4 × v² = 100, so v² = 50 and v = 7.07 m s⁻¹. (1)
(d) Friction alone must remove 100 J: 10 × d = 100, so d = 10 m. Check: a = 10 ÷ 4 = 2.5 m s⁻², d = v² ÷ 2a = 50 ÷ 5 = 10 m. (2)

Answer: (a) 150 J and −50 J; (b) 100 J; (c) 7.07 m s⁻¹; (d) 10 m.

Example 5: Slide with and without friction · 4 marks · Case-basedHOTS / competency

(Take g = 9.8 m s⁻².) A 20 kg child slides from rest from a height of 2.5 m.
(a) Find her speed at the bottom if friction is negligible. (1)
(b) Find her kinetic energy there, and that of a 30 kg child on the same slide. (1)
(c) If friction removes 20% of the 20 kg child's initial potential energy, find her speed at the bottom. (1)
(d) Where does the missing energy go? (1)

Try it first, then show the solutionSolution

(a) v = √(2gh) = √(2 × 9.8 × 2.5) = √49 = 7 m s⁻¹. (1)
(b) 20 kg: K = ½ × 20 × 49 = 490 J (equal to mgh = 20 × 9.8 × 2.5 = 490 J). 30 kg: K = ½ × 30 × 49 = 735 J. The speeds are the same, but the energy is greater for the larger mass. (1)
(c) Initial U = 490 J; 20% lost = 98 J; K = 392 J; ½ × 20 × v² = 392 gives v² = 39.2 and v = 6.26 m s⁻¹. (1)
(d) It changes into heat (and sound) through friction between the child and the slide; total energy is conserved though mechanical energy is not. (1)

Answer: (a) 7 m s⁻¹; (b) 490 J and 735 J; (c) 6.26 m s⁻¹; (d) heat and sound.

Where students lose marks in Work, Kinetic Energy, Potential Energy and Conservation

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

Mistake: Using the total distance instead of the displacement along the force.
A round trip returns to the start, so the net work is zero.
Wrong working
A force of 10 N moves a block 3 m forward and 3 m back: W = 10 × 6 = 60 J
Right working
Forward 30 J and back −30 J; net displacement zero, net work 0 for this force.
Do this: Use displacement along the force with its sign.
Usually costs: 1 to 2 marks.
Mistake: Forgetting that kinetic energy depends on v².
Speed is multiplied instead of squared.
Wrong working
Doubling the speed doubles K.
Right working
Doubling the speed makes K four times.
Do this: Write ½mv² with the speed squared first.
Usually costs: 1 to 2 marks.
Mistake: Using g = 10 in one line and 9.8 in another.
Values get mixed during copying.
Wrong working
U = 2 × 10 × 20, K = ½ × 2 × (√(2 × 9.8 × 20))²
Right working
Use the same g throughout: 9.8 m s⁻² here.
Do this: State g once and use it everywhere.
Usually costs: 1 mark.
Mistake: Saying work is done by a person holding a heavy load still.
The person feels tired.
Wrong working
Holding a barbell above the head does work.
Right working
No displacement means no work on the barbell; the muscles use internal energy.
Do this: Check for displacement.
Usually costs: 1 mark.
Mistake: Ignoring the negative sign of friction work.
Only magnitudes are used.
Wrong working
Net work = 150 + 50 = 200 J
Right working
Net work = 150 − 50 = 100 J
Do this: Friction opposes motion: negative.
Usually costs: 2 marks.

Exam tips

Write the formula, then substitute with units; the unit joule earns the last mark.
For any problem with a change in speed, think of the work-energy theorem first.
In energy-conservation questions write the energy at the start, then at the end, and equate.
State the sign of work and give the reason (force and displacement same or opposite).
Use one value of g throughout and state it.

Practice questions on Work, Kinetic Energy, Potential Energy and Conservation

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

Q1 · 2 marks · Very short answerBoard level
State three situations in which no work is done although a force is applied or an effort is made, with one example each.
Hint
Think of F = 0, s = 0, and a perpendicular force.
Answer
No displacement (pushing a wall or holding a barbell still); force perpendicular to displacement (carrying a bag along level ground); no force (a body moving at constant velocity in space with no force acting) (2 marks: any three with examples earn 2; two earn 1).
Q2 · 3 marks · Short answerHarder board level
A car of mass 1000 kg moves at 54 km h⁻¹. Find (a) its kinetic energy, (b) its kinetic energy at double the speed, and (c) the work needed to raise its speed from 54 km h⁻¹ to 108 km h⁻¹.
Hint
Convert 54 km h⁻¹ to m s⁻¹ first.
Answer
54 km h⁻¹ = 15 m s⁻¹. (a) K = ½ × 1000 × 15² = 112 500 J (1 mark). (b) At 30 m s⁻¹: K = ½ × 1000 × 900 = 450 000 J, four times as large (1 mark). (c) Work = 450 000 − 112 500 = 337 500 J (1 mark).
Q3 · 2 marks · Assertion-reasonBoard level
Assertion (A): A person carrying a box along a level road does no work on the box. Reason (R): The force supporting the box is perpendicular to the displacement of the box. Choose: (a) A and R both true, R explains A; (b) A and R both true, R does not explain A; (c) A true, R false; (d) A false, R true.
Hint
Compare the directions of force and displacement.
Answer
(a). The supporting force is vertical and the displacement is horizontal, so W = F s cos 90° = 0, meaning no work is done by that force.
Q4 · 3 marks · Short answerHarder board level
A roller coaster car of mass 400 kg starts from rest at the top of a 25 m hill on a smooth track (g = 9.8 m s⁻²). (a) Find its speed when it is 5 m above the ground. (b) Find its kinetic energy there. (c) Can it climb a second hill 26 m high? Give a reason.
Hint
The height lost is 25 − 5 = 20 m.
Answer
(a) v = √(2g × 20) = √392 = 19.8 m s⁻¹ (1 mark). (b) K = mg × 20 = 400 × 9.8 × 20 = 78 400 J, which equals ½ × 400 × 392 (1 mark). (c) No: without friction the car cannot rise above its starting height of 25 m, because all its energy would then be potential and 26 m needs more energy than it has (1 mark).
Q5 · 4 marks · Case-basedHOTS / competency
A 0.5 kg ball is thrown vertically upwards at 14 m s⁻¹ (g = 9.8 m s⁻², ignore air resistance). (a) Find its initial kinetic energy. (b) Find the maximum height. (c) At a height of 5 m find its kinetic and potential energies and its speed. (d) Show that the total energy is the same at all these points.
Hint
At the top all the energy is potential.
Answer
(a) K = ½ × 0.5 × 14² = 49 J (1 mark). (b) mgh = 49 gives h = 49 ÷ (0.5 × 9.8) = 10 m (1 mark). (c) At 5 m: U = 0.5 × 9.8 × 5 = 24.5 J, so K = 49 − 24.5 = 24.5 J; ½ × 0.5 × v² = 24.5 gives v² = 98 and v = 9.9 m s⁻¹ (1 mark). (d) Start: 49 + 0 = 49 J; at 5 m: 24.5 + 24.5 = 49 J; at the top: 0 + 49 = 49 J (1 mark).

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.What is 1 joule in terms of newton and metre?Show answer
1 J = 1 N m.
2.State the work-energy theorem.Show answer
The work done on an object equals the change in its energy.
3.What is the formula for kinetic energy?Show answer
K = ½mv².
4.What is the formula for gravitational potential energy?Show answer
U = mgh.
5.When is the work done by a force zero?Show answer
When F = 0, when s = 0, or when the force is perpendicular to the displacement.
6.What speed does a body get falling from height h?Show answer
v = √(2gh).

Work, Kinetic Energy, Potential Energy and Conservation: frequently asked questions

Is work a vector?

No. Work has no direction, but it can be positive or negative.

Why does a heavier child not reach the bottom of a slide faster?

mgh = ½mv² and the mass cancels, so the speed depends only on the height (ignoring friction).

Where does mechanical energy go when a pendulum stops?

It changes into heat (and a little sound) through friction and air resistance, so total energy is conserved.

Does potential energy depend on the path taken?

No, only on the height. Climbing by stairs or by lift gives the same gain.

Is the work done by gravity on a falling ball positive?

Yes. Gravity acts downward and so does the displacement, so the work is positive and equals the gain in kinetic energy.