PhysicsICSE Class 10 · Work, Energy and Power

Work Done by a Force: W = Fs cos θ, W = mgh and Solved Numericals (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

Work is force multiplied by displacement in the direction of the force: W = Fs cos θ, where θ is the angle between them. Its SI unit is the joule. Work is greatest at θ = 0°, zero at θ = 90° and negative at θ = 180°; lifting a mass m through height h needs W = mgh.

Work depends on the angle between force and displacement: maximum at 0°, zero at 90°, negative at 180°.

Illustration drawn for QuestionBuddy.

Understand Work Done by a Force (W = Fs cos θ)

Work is done when a force acting on a body produces a displacement of the body. The amount of work is the product of the force and the component of the displacement in the direction of the force:

$W = F s cos θ,$

where $θ$ is the angle between the force $F$ and the displacement $s$ . Work is a scalar. Its SI unit is the joule: $1 J = 1 N \times 1 m$ , the work done when a force of 1 N moves its point of application by 1 m in its own direction. The cgs unit is the erg ( $1$ dyne $\times$ $1$ cm), and $1 J = 1 0^{7}$ erg.

Special cases (named in the syllabus).

$θ = 0°$ : force and displacement in the same direction, $cos 0° = 1$ , so $W = F s$ , the largest possible. Example: lifting a bag straight up.
$θ = 90°$ : force perpendicular to displacement, $cos 90° = 0$ , so $W = 0$ . Example: a porter walking on level ground with a load on his head does no work against gravity on it; the centripetal force does no work.
$θ = 180°$ : force opposite to displacement, $cos 180° = - 1$ , so $W = - F s$ (negative work). Example: friction on a sliding box; gravity on a ball going up.

For other angles use $cos 60° = 0.5$ and $cos 30° \approx 0.866$ .

Which displacement? Always the displacement measured along the force. In the 2024 MCQ, a force acting along a slope moved a load up the slope, so the displacement to use was the slant length, 5 m, not the vertical height of 3 m.

Work against gravity. To lift a body of mass $m$ at steady speed through a vertical height $h$ , you apply an upward force equal to its weight $m g$ , along the vertical displacement, so

$W = m g h .$

Only the vertical height counts, whatever path is taken (stairs, a ramp, a winding hill road), because gravity acts vertically. On a smooth ramp of length $l$ rising through $h$ , a force $F$ along the ramp does $F l = m g h$ : a longer ramp needs a smaller force but the same work. Gravity itself does $- m g h$ on a rising body and $+ m g h$ on a falling one.

No displacement, no work. Holding a heavy bag still, or pushing a wall that does not move, does no work in the physics sense, however tiring it feels.

Energy link. Work done on a body is stored as its energy, so energy is measured by the work a body can do, in the same unit, the joule.

Work Done by a Force (W = Fs cos θ): formulas and key facts

Work by a constant force: $W = F s cos θ$; θ = angle between force and displacement.
Special cases: $θ = 0^{\circ} : W = F s; θ = 9 0^{\circ} : W = 0; θ = 18 0^{\circ} : W = - F s$; Maximum, zero and negative work.
Work against gravity: $W = m g h$; h = vertical height, whatever the path.
Smooth ramp: $F \times l = m g h$; l = length of ramp along the force.
Units: $1 J = 1 N m = 1 0^{7} erg$; 1 erg = 1 dyne cm.

How to solve Work Done by a Force (W = Fs cos θ) questions

Find the force, the displacement and the angle between them before you multiply.

1Identify which force's work is asked: your pull, gravity, friction or a normal force.
2Find the displacement and the angle $θ$ between that force and the displacement.
3If the force is along the motion, $W = F s$ ; if perpendicular, $W = 0$ ; if opposite, $W = - F s$ ; otherwise $W = F s cos θ$ .
4For lifting, use $W = m g h$ with the vertical height only; state the value of $g$ used.
5Convert units: mass in kg, distance in m, so that $W$ comes out in joules.
6State whether the work is positive, negative or zero when asked.

Work Done by a Force (W = Fs cos θ) 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.

Only the vertical height counts

Suppose you carry 8 kg of books up three floors of a building, each floor 3.5 m high, once by the stairs and once by a long sloping ramp.

How much work do you do against gravity on the books each time? ( $g = 10 m s^{- 2}$ )

Try it first, then show the solutionSolution

Vertical height $= 3 \times 3.5 = 10.5$ m.
$W = m g h = 8 \times 10 \times 10.5 = 840$ J, by either route.

Answer: 840 J both times; the route does not matter.

Memory hook: Gravity only counts the climb, not the walk.

Angle of pull

Imagine pulling a wheeled suitcase 50 m along a level platform, with a steady pull of 40 N along its handle, which makes $60°$ with the ground.

Find the work done by your pull. How much would the same 40 N pull do over 50 m if you lowered the handle to $30°$ ?

Try it first, then show the solutionSolution

$W = F s cos 60° = 40 \times 50 \times 0.5 = 1000$ J.
At $30°$ : $W = 40 \times 50 \times 0.866 \approx 1732$ J.

Answer: 1000 J at 60°; about 1732 J at 30°.

Memory hook: The flatter the pull, the more of it moves the load.

Tricks to remember Work Done by a Force (W = Fs cos θ)

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

Zero, ninety, one-eighty
0° → full Fs; 90° → nothing; 180° → minus Fs. Learn these three before any other angle.
Use it when: The special cases named in the syllabus.
Along the force
Displacement for work is always measured along the force. Ask: 'which way does the force point?'
Use it when: Slopes and pulled loads, as in 2024 Q1(iii).
mgh: the lift tax
Lifting costs mgh, however you climb. Ramps reduce the force, never the work.
Use it when: Stairs, ramps and hill roads.
Tired is not work
No displacement, no work: holding, pushing a wall, standing with a load.
Use it when: Zero-work reasoning questions.

Work Done by a Force (W = Fs cos θ): solved examples

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

Example 1: Work at an angle · 1 mark · MCQBoard level

A force of 20 N moves a body through 3 m along a line at $60°$ to the force. The work done is:
(a) 60 J
(b) 30 J
(c) 52 J
(d) 0 J

Try it first, then show the solutionSolution

$W = F s cos θ = 20 \times 3 \times cos 60°$ .
$= 60 \times 0.5 = 30$ J.

Answer: (b) 30 J

Example 2: Lifting, then carrying · 2 marks · Short answerBoard level

A 12 kg suitcase is lifted from the floor onto a shelf 1.5 m high, and later carried 5 m along a level corridor at that height. Find the work done against gravity in each part. ( $g = 10 m s^{- 2}$ )

Try it first, then show the solutionSolution

Lifting: $W = m g h = 12 \times 10 \times 1.5 = 180$ J. (1)
Carrying at the same height: the upward force is at $90°$ to the horizontal displacement, so $W = 0$ . (1)

Answer: 180 J; 0 J

Example 3: A smooth ramp · 3 marks · Short answerHarder board level

A 25 kg box is pushed at steady speed up a smooth ramp 10 m long that rises through 4 m, by a force along the ramp. ( $g = 10 m s^{- 2}$ )
(a) Find the work done against gravity.
(b) Find the force needed.
(c) What force would lift it straight up, and does the ramp save work?

Try it first, then show the solutionSolution

(a) $W = m g h = 25 \times 10 \times 4 = 1000$ J. (1)
(b) $F \times 10 = 1000$ , so $F = 100$ N. (1)
(c) Lifting straight up needs $m g = 250$ N. The ramp needs less force over a longer distance, but the work (1000 J) is the same: it saves force, not work. (1)

Answer: (a) 1000 J (b) 100 N (c) 250 N; no saving of work.

Example 4: Positive and negative work by gravity · 3 marks · Short answerHarder board level

A ball of mass 0.5 kg is thrown vertically upwards and rises 8 m before falling back to the hand. Find the work done by gravity (a) on the way up, (b) on the way down, (c) for the whole trip. ( $g = 10 m s^{- 2}$ )

Try it first, then show the solutionSolution

(a) Up: gravity (5 N) acts down, displacement up, $θ = 180°$ : $W = - 5 \times 8 = - 40$ J. (1)
(b) Down: same direction, $θ = 0°$ : $W = + 40$ J. (1)
(c) Total: $- 40 + 40 = 0$ J; the ball returns to the same height. (1)

Answer: (a) −40 J (b) +40 J (c) 0 J

Example 5: Pushing a lawn mower · 3 marks · Short answerHOTS / competency

A gardener pushes a lawn mower 20 m across a level lawn with a force of 100 N along the handle, which is inclined at $30°$ below the horizontal.
(a) Find the work done by the push.
(b) Find the work done if the handle is held at $60°$ below the horizontal with the same push.
(c) Which angle is better, and why?

Try it first, then show the solutionSolution

(a) $W = 100 \times 20 \times cos 30° \approx 1732$ J. (1)
(b) $W = 100 \times 20 \times cos 60° = 1000$ J. (1)
(c) $30°$ : a larger part of the push acts along the motion; at $60°$ more of it only presses the mower into the ground, doing no work. (1)

Answer: (a) About 1732 J (b) 1000 J (c) 30°, since more of the force acts along the displacement.

Example 6: Case study: a delivery trolley · 4 marks · Case-basedHOTS / competency

A worker pushes a 30 kg trolley up a smooth ramp 5 m long that rises 3 m, with a force along the ramp, at steady speed. At the top, he pushes it 4 m along a level corridor at steady speed against a friction force of 50 N. ( $g = 10 m s^{- 2}$ )
(a) For the work on the ramp, which displacement do you use with the worker's force: 3 m, 4 m or 5 m?
(b) Find the worker's force on the ramp and the work it does.
(c) Find the work done by gravity on the ramp.
(d) Find the total work done by the worker.

Try it first, then show the solutionSolution

(a) 5 m: the force acts along the ramp, so the displacement along it is the slant length. (1)
(b) $F \times 5 = m g h = 30 \times 10 \times 3 = 900$ J, so $F = 180$ N and $W = 900$ J. (1)
(c) Gravity acts down while the trolley rises 3 m: $W = - 900$ J. (1)
(d) Corridor: $50 \times 4 = 200$ J; total $= 900 + 200 = 1100$ J. (1)

Answer: (a) 5 m (b) 180 N, 900 J (c) −900 J (d) 1100 J

Where students lose marks in Work Done by a Force (W = Fs cos θ)

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

Mistake: Using a displacement not along the force
In $W = F s cos θ$ the displacement is taken along the force; a force acting along a slope uses the slant length.
Wrong working
Force along a 3-4-5 slope: displacement taken as the height, 3 m.
Right working
Displacement along the force: 5 m.
Do this: Draw the force arrow and measure the displacement in that direction.
Usually costs: Reported for 2024 Q1(iii), where candidates chose 3 m instead of 5 m.
Mistake: Using the path length for work against gravity
Gravity is vertical; only the vertical height matters.
Wrong working
$W = 45 \times 10 \times 300$ for a 300 m path up a 12 m hill.
Right working
$W = 45 \times 10 \times 12 = 5400$ J.
Do this: Write 'h = vertical height' next to $m g h$ .
Usually costs: The answer mark.
Mistake: Missing the minus sign for opposing forces
When the force is opposite to the displacement, $cos 180° = - 1$ .
Wrong working
Work by friction of 10 N over 2 m is 20 J.
Right working
It is −20 J.
Do this: Decide the angle first: 0°, 90° or 180°.
Usually costs: Sign marks in 'positive or negative work' questions.
Mistake: Mass in grams or distance in cm with joules
The joule needs newtons and metres.
Wrong working
$W = 500 \times 10 \times 2$ for a 500 g mass lifted 2 m.
Right working
$W = 0.5 \times 10 \times 2 = 10$ J.
Do this: Convert to kg and m before substituting.
Usually costs: The answer mark.

Exam tips

Write the formula, substitution with units, and the answer with its unit: $W = F s cos θ = \dots = \dots$ J.
Learn the definition: 'work is done when a force produces a displacement; it is measured by the product of force and displacement in the direction of the force'.
For zero-work reasons, say either 'no displacement' or 'force perpendicular to displacement', whichever applies.
State the value of $g$ you use; ICSE questions usually give $g = 10$ or $9.8 m s^{- 2}$ .

Practice questions on Work Done by a Force (W = Fs cos θ)

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 work done by a force is negative when the angle between the force and the displacement is:
(a) $0°$
(b) $60°$
(c) $90°$
(d) $180°$
Hint
Sign of $cos θ$ .
Answer
(d) $180°$
Q2 · 2 marks · Short answerBoard level
A force of 50 N moves a body 4 m in its own direction. Find the work done in joules and in erg.
Hint
$1 J = 1 0^{7}$ erg.
Answer
$W = 50 \times 4 = 200$ J $= 2 \times 1 0^{9}$ erg.
Q3 · 2 marks · Short answerHarder board level
A girl of mass 45 kg walks up a hill 12 m high along a winding path 300 m long. How much work does she do against gravity? ( $g = 10 m s^{- 2}$ )
Hint
Does the path length matter?
Answer
$W = 45 \times 10 \times 12 = 5400$ J; only the vertical height counts.
Q4 · 2 marks · Short answerHarder board level
State whether the work done is positive, negative or zero: (a) friction on a book sliding across a table, (b) gravity on an apple falling from a tree, (c) the Sun's pull on a planet in a circular orbit, (d) a man pushing hard against a wall that does not move.
Hint
Compare the directions of force and displacement.
Answer
(a) Negative (b) Positive (c) Zero (force perpendicular to motion) (d) Zero (no displacement)
Q5 · 2 marks · Short answerHarder board level
A 5 kg bucket of water is pulled up at steady speed from a well 15 m deep and then carried 10 m along level ground. Find the work done against gravity in each part. ( $g = 9.8 m s^{- 2}$ )
Hint
Only the vertical part counts against gravity.
Answer
Raising: $5 \times 9.8 \times 15 = 735$ J. Carrying level: 0 J.
Q6 · 2 marks · Short answerHarder board level
A box moves 6 m along a level floor while being pulled by a force of 80 N at $60°$ to the horizontal. Find the work done by the force. How much work would the same force do over 6 m if it were horizontal?
Hint
$cos 60° = 0.5$ .
Answer
$80 \times 6 \times 0.5 = 240$ J; horizontal: $80 \times 6 = 480$ J.
Q7 · 3 marks · Short answerHOTS / competency
A 2 kg block slides down a smooth slope 10 m long from a height of 5 m. (a) Find the work done by gravity. (b) Why is it not $m g \times 10$ ? (c) How much work does the normal force of the slope do? ( $g = 10 m s^{- 2}$ )
Hint
Gravity is vertical; the normal force is perpendicular to the slope.
Answer
(a) $m g h = 2 \times 10 \times 5 = 100$ J. (b) Only the displacement along gravity (vertical, 5 m) counts, not the 10 m along the slope. (c) Zero: it is perpendicular to the motion.

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.Work done by 10 N over 2 m in its own direction?Show answer
20 J.
2.Work against gravity lifting 5 kg through 2 m ( $g = 10$ )?Show answer
100 J.
3.Value of $W$ when $θ = 90°$ ?Show answer
Zero.
4.How many erg in 1 J?Show answer
$1 0^{7}$ .
5.Give one example of negative work.Show answer
Friction on a sliding box (or gravity on a ball moving up).

Work Done by a Force (W = Fs cos θ): frequently asked questions

What is the formula for work done when the force is at an angle?

W = Fs cos θ, where θ is the angle between the force and the displacement. Only the part of the displacement along the force counts.

Why is no work done when you carry a bag on level ground?

Against gravity, your upward force on the bag is at 90° to its horizontal displacement, so W = Fs cos 90° = 0.

Does a ramp reduce the work needed to lift a load?

No. A smooth ramp reduces the force needed but increases the distance; the work against gravity is still mgh.