PhysicsICSE Class 10 · Chapter 3

Work, Energy and Power ICSE Class 10: Formulas, Units and Numericals

Updated 14 October 2026 · Fact-checked

Work is done when a force moves its point of application: W = Fs cos θ, so it is zero when the force is perpendicular to the motion. Energy is the capacity to do work, and power is the rate of doing work, P = W/t. ICSE Class 10 also needs units such as erg, calorie, kWh, eV and horsepower.

What Work, Energy and Power is about

In physics, work has a precise meaning: a force does work only when its point of application moves, and only the part of the motion along the force counts. Energy is the capacity to do work, and power tells you how fast the work is done.

The CISCE 2027 syllabus covers: the definition of work; $W = F s cos θ$ with the special cases $θ = 0^{\circ}$ and $9 0^{\circ}$ ; $W = m g h$ ; energy as the capacity to do work; units of work and energy and their relation to the joule (erg, calorie, kW h, eV); power $P = \frac{W}{t}$ with SI and cgs units, kW, MW, GW and horsepower ( $1$ hp $= 746$ W); and simple numerical problems.

The CISCE examiners' comments for 2024 show two recurring slips: choosing a displacement that is not along the force (for example the height instead of the slant length), and writing a comparison of powers as an unsimplified fraction, in the wrong order, or by multiplying instead of dividing. In numericals, take $g = 10$ m s⁻² unless told otherwise, and convert minutes and hours to seconds.

Weightage: CISCE publishes no chapter or unit weightage for ICSE Physics. Work, Energy and Power is in the 1. Force, Work, Power and Energy (iii) part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Physics (Science Paper 1) syllabus, examination year 2027

Key concepts

Work and W = Fs cos θ

Work is done when a force acting on a body produces a displacement. It is measured by the product of the force and the displacement in the direction of the force:
$W = F s cos θ$ ,
where $θ$ is the angle between the force and the displacement. Work is a scalar. SI unit: joule (1 J = 1 N × 1 m). cgs unit: erg (1 dyne × 1 cm); $1 J = 1 0^{7}$ erg.

$θ = 0^{\circ}$ : $W = F s$ (maximum).
$θ = 9 0^{\circ}$ : $W = 0$ .
$θ = 18 0^{\circ}$ : $W = - F s$ (negative work).

Worked example

A force of 10 N pulls a box 5 m along the floor. The force acts at $6 0^{\circ}$ to the floor. Find the work done.

Try it first, then show the solutionSolution

$W = F s cos θ = 10 \times 5 \times cos 6 0^{\circ}$ .
$= 50 \times 0.5 = 25$ J.

Answer: 25 J

Positive, negative and zero work

Positive: force and displacement in the same direction (you lifting a bag; gravity on a falling ball).
Negative: force opposite to displacement (friction on a sliding box; gravity on a ball thrown upwards; brakes on a car).
Zero: no displacement (pushing a wall), or force perpendicular to displacement (a coolie walking on level ground with a load on his head does no work against gravity; the centripetal force in circular motion).

Feeling tired does not mean physical work is done.

Worked example

State whether the work done is positive, negative or zero: (a) by gravity on a ball thrown up while it rises, (b) by a person holding a heavy suitcase still, (c) by friction on a sliding book.

Try it first, then show the solutionSolution

(a) Gravity acts down, motion is up: negative.
(b) No displacement: zero.
(c) Friction opposes motion: negative.

Answer: (a) negative (b) zero (c) negative

Work against gravity: W = mgh

To lift a body of mass $m$ through a vertical height $h$ at steady speed, the upward force equals its weight $m g$ , so
$W = m g h$ .

Only the vertical height counts, whatever the path (stairs, ramp or ladder), because gravity is vertical. On a frictionless ramp of length $l$ and height $h$ , the force along the ramp $F$ satisfies $F \times l = m g h$ .

Worked example

A girl of mass 50 kg climbs a staircase to a floor 4 m higher. How much work does she do against gravity? ( $g = 10$ m s⁻²)

Try it first, then show the solutionSolution

$W = m g h = 50 \times 10 \times 4$ .
$= 2000$ J.

Answer: 2000 J

Energy as the capacity to do work

A body has energy if it can do work. The energy of a body is measured by the amount of work it can do, so energy and work have the same units.

When work is done on a body, its energy increases by that amount; when a body does work, it loses that much energy. (The forms of energy, U = mgh and K = ½mv², are in the next chapter.)

Worked example

A worker does 600 J of work lifting a box onto a shelf. How much energy does the box gain, and in what form?

Try it first, then show the solutionSolution

Work done on the box is stored in it.
It gains 600 J of gravitational potential energy.

Answer: 600 J of potential energy

Units of work and energy

erg: $1 erg = 1 0^{- 7}$ J.
calorie (heat): $1 cal \approx 4.2$ J.
kilowatt hour (commercial unit of electrical energy): energy used by a 1 kW device in 1 hour; $1 kW h = 1000 \times 3600 = 3.6 \times 1 0^{6}$ J.
electron volt (atomic scale): $1 eV = 1.6 \times 1 0^{- 19}$ J.

Largest to smallest: kW h > calorie > joule > erg > eV.

Worked example

Convert (a) 5 kW h into joules, (b) $2.5 \times 1 0^{8}$ erg into joules, (c) 840 J into calories.

Try it first, then show the solutionSolution

(a) $5 \times 3.6 \times 1 0^{6} = 1.8 \times 1 0^{7}$ J.
(b) $2.5 \times 1 0^{8} \times 1 0^{- 7} = 25$ J.
(c) $840 \div 4.2 = 200$ cal.

Answer: (a) 1.8 × 10⁷ J (b) 25 J (c) 200 cal

Power and its units

Power is the rate of doing work (or of transferring energy):
$P = \frac{W}{t}$ .
Power is a scalar. SI unit: watt (1 W = 1 J s⁻¹). cgs unit: erg s⁻¹ ( $1 W = 1 0^{7}$ erg s⁻¹).

$1$ kW $= 1 0^{3}$ W, $1$ MW $= 1 0^{6}$ W, $1$ GW $= 1 0^{9}$ W.
$1$ horsepower (hp) $= 746$ W.

When a force keeps a body moving at steady speed $v$ , $P = \frac{F s}{t} = F v$ .

Note: kW is a unit of power; kW h is a unit of energy.

Worked example

A motor does 3000 J of work in 1 minute. Find its power in watts and in horsepower.

Try it first, then show the solutionSolution

$t = 60$ s; $P = \frac{3000}{60} = 50$ W.
In hp: $\frac{50}{746} \approx 0.067$ hp.

Answer: 50 W ≈ 0.067 hp

Comparing powers

To compare two powers, find each power (convert times to the same unit), then write the ratio in the order asked and simplify to whole numbers, e.g. $4 : 1$ .

If the same work is done in half the time, the power doubles; if twice the work is done in the same time, the power doubles.

Worked example

Rita does 900 J of work in 5 minutes and Sam does 600 J in 10 minutes. Find the ratio of Rita's power to Sam's.

Try it first, then show the solutionSolution

$P_{R} = \frac{900}{300 s} = 3$ W; $P_{S} = \frac{600}{600 s} = 1$ W.
$P_{R} : P_{S} = 3 : 1$ .

Answer: 3 : 1

Physics in real life: Work, Energy and Power

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.

Power of a person climbing stairs

Suppose a student of mass 50 kg runs up 30 steps, each 20 cm high, in 12 s.

Find the work done against gravity and the power developed. ( $g = 10$ m s⁻²)

Try it first, then show the solutionSolution

Height $= 30 \times 0.2 = 6$ m.
$W = 50 \times 10 \times 6 = 3000$ J.
$P = \frac{3000}{12} = 250$ W.

Answer: 3000 J; 250 W

Memory hook: Height counts, not the number of steps.

No work against gravity on level ground

Imagine a porter walks 100 m along a level platform carrying a 20 kg suitcase on his head.

How much work does he do against gravity on the suitcase?

Try it first, then show the solutionSolution

The force he applies on the suitcase is upward; the displacement is horizontal.
$θ = 9 0^{\circ}$ , so $W = F s cos 9 0^{\circ} = 0$ .

Answer: Zero

Memory hook: Up force, sideways walk: zero work against gravity.

Power from a waterfall

Suppose 1000 kg of water falls through 50 m every second at a small hydro plant.

Find the power available, and how many 100 W bulbs could be lit if 60% of it is turned into electrical power.

Try it first, then show the solutionSolution

Work done by gravity per second $= m g h = 1000 \times 10 \times 50 = 5 \times 1 0^{5}$ J.
Power $= 5 \times 1 0^{5}$ W $= 0.5$ MW.
Useful power $= 3 \times 1 0^{5}$ W, enough for $3000$ bulbs.

Answer: 0.5 MW; 3000 bulbs

Memory hook: Power is energy per second.

Topics in this chapter

Work, Energy and Power formula sheet

Work: $W = F s cos θ$; θ = angle between force and displacement.
Work against gravity: $W = m g h$; h = vertical height.
Power: $P = \frac{W}{t} = F v$; Fv for steady speed v along the force.
Energy units: $1 J = 1 0^{7} erg, 1 cal \approx 4.2 J$; erg = dyne × cm.
Commercial and atomic units: $1 kW h = 3.6 \times 1 0^{6} J, 1 eV = 1.6 \times 1 0^{- 19} J$; kW h is energy, kW is power.
Power units: $1 hp = 746 W$; kW = 10³ W, MW = 10⁶ W, GW = 10⁹ W.

How the board asks Work, Energy and Power

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
MCQ: units, conversions, special cases	Section A, Question 1: kW h in joules, which displacement to use, when work is zero.	Know the conversions; displacement along the force.
Short numerical (2 marks)	Work by an inclined force; work against gravity; power of a person or pump.	W = Fs cos θ or mgh; P = W ÷ t with t in seconds.
Compare two powers	Two people or machines doing work in different times; find the ratio.	Same units; ratio in the order asked; simplify.
Definitions and reasons	Define work, power, kW h; explain zero or negative work.	Precise definitions; give the angle or condition.
Multi-step numerical (3-4 marks)	Pumps, cranes, lifts, engines: work, power in W and hp, energy in kW h.	Find W, then P, then convert units step by step.

Tricks to remember Work, Energy and Power

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

Cos for the component
Only the part of the force along the motion does work: F cos θ.
Use it when: Inclined forces.
0, 90, 180
0° full work, 90° no work, 180° negative work.
Use it when: Special cases.
kW h = 3.6 million
1000 W × 3600 s = 3.6 × 10⁶ J.
Use it when: Commercial unit conversions.
Horse = 746
One horsepower is 746 watts.
Use it when: hp conversions.

Work, Energy and Power: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: The kilowatt hour · 1 mark · MCQBoard level

1 kW h is equal to:
(a) $3.6 \times 1 0^{3}$ J
(b) $3.6 \times 1 0^{5}$ J
(c) $3.6 \times 1 0^{6}$ J
(d) $3.6 \times 1 0^{7}$ J

Try it first, then show the solutionSolution

$1000 W \times 3600 s = 3.6 \times 1 0^{6}$ J.

Answer: (c)

Example 2: Zero work by a porter · 1 mark · Assertion-reasonBoard level

Assertion (A): A porter walking on a level road with a load on his head does no work against gravity on the load.
Reason (R): The force he applies on the load is perpendicular to its displacement.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

Try it first, then show the solutionSolution

$cos 9 0^{\circ} = 0$ .

Answer: (c)

Example 3: Work by an inclined force · 2 marks · Short answerBoard level

A boy pulls a 15 kg trolley 8 m along a level floor with a rope inclined at $6 0^{\circ}$ to the horizontal. The tension in the rope is 40 N. Calculate (a) the work done by the tension, (b) the work done by gravity on the trolley.

Try it first, then show the solutionSolution

(a) $W = F s cos θ = 40 \times 8 \times cos 6 0^{\circ} = 160$ J.
(b) Gravity acts vertically and the displacement is horizontal ( $θ = 9 0^{\circ}$ ), so the work done by gravity is zero.

Answer: (a) 160 J (b) 0

Example 4: Power of a pump · 3 marks · Short answerHarder board level

A pump lifts 500 kg of water to a tank 20 m high in 50 s. Find (a) the work done, (b) the power in watts and in horsepower. ( $g = 10$ m s⁻²)

Try it first, then show the solutionSolution

(a) $W = m g h = 500 \times 10 \times 20 = 1 \times 1 0^{5}$ J.
(b) $P = \frac{1 0 ^{5}}{50} = 2000$ W.
In hp: $\frac{2000}{746} \approx 2.68$ hp.

Answer: (a) 10⁵ J (b) 2000 W ≈ 2.68 hp

Example 5: Comparing two powers · 3 marks · Short answerHarder board level

A lifts a 40 kg load through 6 m in 20 s. B lifts a 60 kg load through 5 m in 30 s. Compare their powers. ( $g = 10$ m s⁻²)

Try it first, then show the solutionSolution

$P_{A} = \frac{40 \times 10 \times 6}{20} = 120$ W.
$P_{B} = \frac{60 \times 10 \times 5}{30} = 100$ W.
$P_{A} : P_{B} = 120 : 100 = 6 : 5$ .

Answer: 6 : 5

Example 6: Unit conversions · 2 marks · Long answerBoard level

Express (a) 5 kW h in joules, (b) 1 J in erg.

Try it first, then show the solutionSolution

(a) $5 \times 3.6 \times 1 0^{6} = 1.8 \times 1 0^{7}$ J.
(b) $1 J = 1 N \times 1 m = 1 0^{5} dyne \times 1 0^{2} cm = 1 0^{7}$ erg.

Answer: (a) 1.8 × 10⁷ J (b) 10⁷ erg

Example 7: Up a ramp · 3 marks · Long answerHarder board level

A box of mass 20 kg is pulled up a smooth ramp 5 m long, rising through a vertical height of 3 m, by a force along the ramp. (a) Find the work done against gravity. (b) Find the force along the ramp. (c) Which length did you use as the displacement for gravity, and why? ( $g = 10$ m s⁻²)

Try it first, then show the solutionSolution

(a) $W = m g h = 20 \times 10 \times 3 = 600$ J.
(b) $F \times 5 = 600$ , so $F = 120$ N.
(c) The vertical 3 m, because gravity acts vertically; for the pulling force the displacement along it is the 5 m slant length.

Answer: (a) 600 J (b) 120 N

Example 8: Power of a car engine · 4 marks · Long answerHOTS / competency

A car moves at a steady 20 m/s against a total resistance of 600 N. (a) Find the power of the engine in kW and hp. (b) Find the work done in one minute. (c) If the speed and the resistance both doubled, what power would be needed?

Try it first, then show the solutionSolution

(a) Steady speed, so engine force $= 600$ N. $P = F v = 600 \times 20 = 12000$ W $= 12$ kW $\approx 16.09$ hp.
(b) $W = P t = 12000 \times 60 = 7.2 \times 1 0^{5}$ J.
(c) $P = 1200 \times 40 = 48000$ W $= 48$ kW, four times as much.

Answer: (a) 12 kW ≈ 16.09 hp (b) 7.2 × 10⁵ J (c) 48 kW

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

A fitness test on the stairs

Suppose a student of mass 50 kg runs up a flight of 30 steps, each 20 cm high, in 12 s. ( $g = 10$ m s⁻²)

(i) · 3 marks
Find the work done against gravity.
Answer
Answer: 3000 J
Vertical height $= 30 \times 0.2 = 6$ m. $W = 50 \times 10 \times 6 = 3000$ J.
(ii) · 3 marks
Find the power developed, in watts and in horsepower.
Answer
Answer: 250 W ≈ 0.335 hp
$P = \frac{3000}{12} = 250$ W $= \frac{250}{746} \approx 0.335$ hp.
(iii) · 4 marks
Express the work done in calories. If the student walks up slowly in 30 s instead, what changes and what stays the same?
Answer
Answer: About 714 cal; work same, power less
$\frac{3000}{4.2} \approx 714$ cal. The work done is the same (same weight, same height), but the power falls to $\frac{3000}{30} = 100$ W because it takes longer.

A lift in a building

Imagine a lift cabin of mass 500 kg carries six people of 60 kg each up 30 m in 20 s at steady speed. ( $g = 10$ m s⁻²)

(i) · 3 marks
Find the work done by the motor in one trip.
Answer
Answer: 2.58 × 10⁵ J
Total mass $= 500 + 360 = 860$ kg. $W = 860 \times 10 \times 30 = 258000$ J.
(ii) · 3 marks
Find the power of the motor in kW and hp.
Answer
Answer: 12.9 kW ≈ 17.29 hp
$P = \frac{258 000}{20} = 12900$ W $= 12.9$ kW $= \frac{12 900}{746} \approx 17.29$ hp.
(iii) · 4 marks
Find the energy used for 50 such trips, in kW h (to 2 decimal places), assuming no losses.
Answer
Answer: 3.58 kW h
$50 \times 258000 = 1.29 \times 1 0^{7}$ J. $\frac{1.29 \times 1 0 ^{7}}{3.6 \times 1 0 ^{6}} \approx 3.58$ kW h.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

A force of 8 N moves a body 6 m in its own direction. Find the work done. (2 marks)

Step	Marks
$W = F s cos 0^{\circ} = 8 \times 6$	1
$= 48$ J	1

A pump lifts 300 kg of water to a height of 12 m in 1 minute. Find its power. ( $g = 10$ m s⁻²) (3 marks)

Step	Marks
$W = m g h = 300 \times 10 \times 12 = 36000$ J	1
$t = 60$ s	1
$P = \frac{36 000}{60} = 600$ W	1

A boy of mass 60 kg runs up stairs 5 m high in 10 s; his sister of mass 50 kg climbs the same stairs in 25 s. Find the ratio of their powers. ( $g = 10$ m s⁻²) (4 marks)

Step	Marks
Boy: $W = 60 \times 10 \times 5 = 3000$ J; $P = 300$ W	1
Sister: $W = 50 \times 10 \times 5 = 2500$ J	1
$P = \frac{2500}{25} = 100$ W	1
Ratio $300 : 100 = 3 : 1$	1

Where students lose marks in Work, Energy and Power

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

Mistake: Wrong displacement for the force
CISCE examiners' comments, 2024: in an MCQ where a force moved a load along a slope, candidates ignored the direction of the force and chose the height (3 m) instead of the slant length (5 m).
Do this: Use the displacement in the direction of the force: along the slope for a pull along the slope; vertical height for work against gravity.
Usually costs: 1 mark.
Mistake: Ratio of powers in the wrong form
CISCE examiners' comments, 2024: candidates did not simplify the ratio, left it as a fraction instead of 4 : 1, multiplied instead of dividing, or gave 1 : 4 instead of 4 : 1.
Do this: Compute each power, then write the ratio in the order named and simplify to whole numbers.
Usually costs: 1 mark in a 2-mark part.
Mistake: Not converting time to seconds
Power in watts needs time in seconds; leaving minutes gives an answer 60 times too large.
Wrong working
P = 600 J ÷ 10 = 60 W (10 minutes)
Right working
P = 600 J ÷ 600 s = 1 W
Do this: Write t in seconds before dividing.
Usually costs: 1 mark.
Mistake: kW and kW h mixed up
kW is a unit of power; kW h is a unit of energy. Interchanging them is a common unit error in definitions and conversions.
Do this: kW h = kW × h: power multiplied by time gives energy.
Usually costs: 1 mark.

Practice questions on Work, Energy and Power

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 zero when the angle between the force and the displacement is:
(a) $0^{\circ}$
(b) $4 5^{\circ}$
(c) $9 0^{\circ}$
(d) $18 0^{\circ}$
Hint
cos 90° = 0.
Answer
(c)
Q2 · 1 mark · MCQBoard level
One horsepower is equal to:
(a) 100 W
(b) 500 W
(c) 746 W
(d) 1000 W
Hint
Standard value.
Answer
(c)
Q3 · 1 mark · MCQHarder board level
Which of these is the largest unit of energy?
(a) erg
(b) calorie
(c) electron volt
(d) kilowatt hour
Hint
Compare each with the joule.
Answer
(d)
Q4 · 1 mark · Assertion-reasonBoard level
Assertion (A): The SI unit of power is the watt.
Reason (R): 1 watt is the power of a source that does 1 joule of work per second.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Definition of the watt.
Answer
(c)
Q5 · 2 marks · Short answerHarder board level
A man of mass 60 kg climbs a staircase of vertical height 10 m in 15 s. Find the work done and his power. ( $g = 10$ m s⁻²)
Hint
mgh, then ÷ t.
Answer
$W = 6000$ J; $P = 400$ W.
Q6 · 2 marks · Short answerHarder board level
Express (a) 2 kW h in MJ, (b) 3600 J in kW h.
Hint
1 kW h = 3.6 MJ.
Answer
(a) 7.2 MJ (b) 0.001 kW h
Q7 · 3 marks · Long answerBoard level
(a) Define work and state its SI unit. (b) Define power and state its SI unit. (c) When is the work done by a force negative? Give one example.
Hint
Definitions and the angle.
Answer
(a) Work is done when a force produces displacement; W = Fs cos θ; joule. (b) Rate of doing work, P = W/t; watt. (c) When the force is opposite to the displacement (θ = 180°), e.g. friction on a sliding box.
Q8 · 3 marks · Long answerHarder board level
A crane lifts a 2 tonne load through 15 m in 30 s. Find the work done and the power in kW and hp. ( $g = 10$ m s⁻², 1 tonne = 1000 kg)
Hint
m = 2000 kg.
Answer
$W = 2000 \times 10 \times 15 = 3 \times 1 0^{5}$ J; $P = 1 0^{4}$ W $= 10$ kW $\approx 13.40$ hp.
Q9 · 3 marks · Long answerHOTS / competency
Rahul holds a 20 kg bag still above the ground for 10 minutes. Priya carries the same bag up a 2 m high flight of stairs. (a) Who does more work on the bag? (b) Why does Rahul still feel tired? (c) How much work does Priya do against gravity? ( $g = 10$ m s⁻²)
Hint
Work needs displacement.
Answer
(a) Priya: Rahul's bag has no displacement, so he does no work on it. (b) His muscles use chemical energy to keep exerting a force, though no mechanical work is done on the bag. (c) $20 \times 10 \times 2 = 400$ J.
Q10 · 3 marks · Long answerHarder board level
A sledge is pulled 10 m along level snow by a rope with tension 50 N at $3 0^{\circ}$ to the horizontal. Find the work done by the tension (to the nearest joule), and the work if the rope were horizontal.
Hint
cos 30° ≈ 0.866.
Answer
$50 \times 10 \times 0.866 \approx 433$ J; horizontal: $500$ J.
Q11 · 1 mark · MCQHOTS / competency
If the same work is done in half the time, the power:
(a) halves
(b) stays the same
(c) doubles
(d) becomes four times
Hint
P = W/t.
Answer
(c)
Q12 · 4 marks · Long answerHOTS / competency
At a small hydroelectric plant, 1000 kg of water falls through 50 m every second. (a) Find the power available in MW. (b) If 60% is converted to electrical power, how many 100 W lamps can it run? (c) How much electrical energy, in kW h, does it supply in 1 hour? ( $g = 10$ m s⁻²)
Hint
Power = mgh per second.
Answer
(a) $1000 \times 10 \times 50 = 5 \times 1 0^{5}$ W $= 0.5$ MW. (b) $0.6 \times 5 \times 1 0^{5} = 3 \times 1 0^{5}$ W, enough for 3000 lamps. (c) $300$ kW × 1 h $= 300$ kW h.

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.Formula for work?Show answer
W = Fs cos θ
2.Work done when force ⟂ displacement?Show answer
Zero.
3.1 J in erg?Show answer
10⁷ erg
4.1 kW h in joules?Show answer
3.6 × 10⁶ J
5.1 eV in joules?Show answer
1.6 × 10⁻¹⁹ J
6.1 hp in watts?Show answer
746 W
7.Define power.Show answer
Rate of doing work, P = W/t; SI unit watt.

Best order to study it

Definition of work and W = Fs cos θ.
Positive, negative and zero work.
Work against gravity, W = mgh.
Energy and its units.
Power, its units and P = Fv.
Numericals: comparisons, pumps, lifts and engines.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you use the right angle θ?
Do you use vertical height for gravity?
Can you convert kW h, erg, cal and eV to joules?
Do you convert time to seconds?
Can you write a ratio of powers in the order asked?
Can you convert watts to horsepower?

Work, Energy and Power: frequently asked questions

Is work a vector or a scalar?

A scalar. It has magnitude only, though it can be positive or negative.

Why is no work done when I push a wall?

The wall does not move, so the displacement is zero and W = Fs cos θ = 0, even though you get tired.

Which value of 1 calorie should I use?

About 4.2 J, unless the question gives a different value.

What is the difference between kW and kW h?

kW measures power (how fast energy is used); kW h measures energy (power × time).

Is P = Fv in the syllabus?

The syllabus links power to force; P = Fv follows directly from P = W/t when a force moves a body at steady speed, and is useful in engine problems.

Based on: CISCE ICSE 2027 Physics (Science Paper 1) syllabus; CISCE ICSE 2027 Physics (Science Paper 1) specimen question paper and marking scheme; CISCE Analysis of Pupil Performance, ICSE Physics (Science Paper 1), 2025; CISCE Analysis of Pupil Performance, ICSE Physics (Science Paper 1), 2024