Carrying a Heavy Bag Across a Flat Room Does No Work at All
Calculate work from force, displacement and the angle between them, see why a horizontal walk with a load on your head does zero work, connect work to the change in energy, and convert between erg, calorie, electron volt, kilowatt hour and horsepower.
Why does physics say no work is done when you carry a load across a room?
A coolie lifts a heavy trunk onto his head, walks fifty metres along a level platform, and sets it down. Ask him whether he has done work and the answer is obvious. Ask physics and the answer is zero.
Both answers are correct, because they are answers to different questions. Everyday "work" means effort and tiredness. In physics, work is done only when a force actually moves something along its own direction.
While the coolie walks on a level platform:
- The force he applies on the trunk is vertically upward, holding it against gravity
- The displacement of the trunk is horizontal, along the platform
- The two are at right angles, so the force has no component along the displacement
So the work done on the trunk by that upward force is zero. He tires because his muscles keep contracting and relaxing, which consumes energy inside his body — but none of that energy is transferred to the trunk as work.
That single example forces a precise definition, and the whole of this part of the chapter follows from it:
- Work is the product of the force and the displacement in the direction of the force
- The work-energy theorem says the work done on a body equals the change in its energy, which is why work and energy share the same unit
- Power is the rate of doing work, and it is what separates a labourer from a crane doing the same job
And then there is the untidy business of units. Work and energy come in the joule, the erg, the calorie, the electron volt and the kilowatt hour; power comes in the watt, the kilowatt, the megawatt, the gigawatt and the horsepower. Every one of them appears in the syllabus, and converting between them accurately is worth as many marks as any formula.
Unless a question says otherwise, **take m s** for the numericals here.
This page covers the second part of the ICSE Class 10 Physics chapter on force, work, power and energy: work and its special cases, the work-energy theorem, power, and the units of all three.
Both answers are correct, because they are answers to different questions. Everyday "work" means effort and tiredness. In physics, work is done only when a force actually moves something along its own direction.
While the coolie walks on a level platform:
- The force he applies on the trunk is vertically upward, holding it against gravity
- The displacement of the trunk is horizontal, along the platform
- The two are at right angles, so the force has no component along the displacement
So the work done on the trunk by that upward force is zero. He tires because his muscles keep contracting and relaxing, which consumes energy inside his body — but none of that energy is transferred to the trunk as work.
That single example forces a precise definition, and the whole of this part of the chapter follows from it:
- Work is the product of the force and the displacement in the direction of the force
- The work-energy theorem says the work done on a body equals the change in its energy, which is why work and energy share the same unit
- Power is the rate of doing work, and it is what separates a labourer from a crane doing the same job
And then there is the untidy business of units. Work and energy come in the joule, the erg, the calorie, the electron volt and the kilowatt hour; power comes in the watt, the kilowatt, the megawatt, the gigawatt and the horsepower. Every one of them appears in the syllabus, and converting between them accurately is worth as many marks as any formula.
Unless a question says otherwise, **take m s** for the numericals here.
This page covers the second part of the ICSE Class 10 Physics chapter on force, work, power and energy: work and its special cases, the work-energy theorem, power, and the units of all three.
Formula
How do you calculate the work done by a force at an angle?
Multiply the force by the displacement and by the cosine of the angle between them.
where is the force, the displacement, and the angle between the direction of the force and the direction of the displacement. The SI unit of work is the joule (J), and one joule is the work done when a force of one newton moves a body one metre along its own direction.
Work is a scalar quantity, even though force and displacement are both vectors — which is why the answer is a plain number with a sign and no direction.
The three cases that matter, and each is a standard question:
- ****, force and displacement in the same direction. Then and , the maximum positive work
- ****, force perpendicular to displacement. Then and , no work at all
- ****, force opposite to displacement. Then and , negative work
Worked example 1 — a general angle. A force of N pulls a box through m, the rope making an angle of with the floor. Find the work done.
Notice that only half the force does useful work here. The other component lifts the box slightly against the floor and contributes nothing to the horizontal motion.
Worked example 2 — force along the displacement. A force of N pushes a trolley m along the direction of the force.
Worked example 3 — negative work. A frictional force of N acts on a block that slides m.
Friction always opposes the motion, so :
The minus sign means energy has been taken out of the block rather than put into it, which is exactly what friction does.
Work done against gravity. Lifting a body of mass vertically through a height needs an upward force equal to its weight , and the displacement is along that force, so
Worked example 4. Find the work done in lifting a kg bucket through a height of m.
Worked example 5 — a larger one. Find the work done in lifting kg of water from a well m deep.
The three standard zero-work situations, which an examiner asks for by name:
- A coolie walking on a level platform with a load on his head — the force is vertical, the displacement horizontal
- A body moving in a circle at a steady speed — the force needed to keep it on the circle points to the centre, always perpendicular to the motion. This is why the moon does no work on its orbit and why the earth's pull does no work on the moon over a complete revolution
- A body pushed against an immovable wall — there is a force but no displacement, so and
One subtlety worth stating. The formula needs the displacement, not the distance travelled. A body that goes out and comes back has zero displacement, so the work done by a constant force over a closed path is zero — which is another way of saying that gravity gives back on the way down exactly what it took on the way up.
where is the force, the displacement, and the angle between the direction of the force and the direction of the displacement. The SI unit of work is the joule (J), and one joule is the work done when a force of one newton moves a body one metre along its own direction.
Work is a scalar quantity, even though force and displacement are both vectors — which is why the answer is a plain number with a sign and no direction.
The three cases that matter, and each is a standard question:
- ****, force and displacement in the same direction. Then and , the maximum positive work
- ****, force perpendicular to displacement. Then and , no work at all
- ****, force opposite to displacement. Then and , negative work
Worked example 1 — a general angle. A force of N pulls a box through m, the rope making an angle of with the floor. Find the work done.
Notice that only half the force does useful work here. The other component lifts the box slightly against the floor and contributes nothing to the horizontal motion.
Worked example 2 — force along the displacement. A force of N pushes a trolley m along the direction of the force.
Worked example 3 — negative work. A frictional force of N acts on a block that slides m.
Friction always opposes the motion, so :
The minus sign means energy has been taken out of the block rather than put into it, which is exactly what friction does.
Work done against gravity. Lifting a body of mass vertically through a height needs an upward force equal to its weight , and the displacement is along that force, so
Worked example 4. Find the work done in lifting a kg bucket through a height of m.
Worked example 5 — a larger one. Find the work done in lifting kg of water from a well m deep.
The three standard zero-work situations, which an examiner asks for by name:
- A coolie walking on a level platform with a load on his head — the force is vertical, the displacement horizontal
- A body moving in a circle at a steady speed — the force needed to keep it on the circle points to the centre, always perpendicular to the motion. This is why the moon does no work on its orbit and why the earth's pull does no work on the moon over a complete revolution
- A body pushed against an immovable wall — there is a force but no displacement, so and
One subtlety worth stating. The formula needs the displacement, not the distance travelled. A body that goes out and comes back has zero displacement, so the work done by a constant force over a closed path is zero — which is another way of saying that gravity gives back on the way down exactly what it took on the way up.
What does the work-energy theorem actually tell you?
The work done on a body equals the change in its energy. Do positive work on it and its energy rises by exactly that amount; do negative work and its energy falls by that amount.
Which is why work and energy are measured in the same unit. They are not two different quantities that happen to share the joule — work is energy being transferred, and the joule measures the amount transferred.
The theorem explains where work goes. When you do work on a body, that energy has to end up somewhere:
- Lifting it stores the energy as potential energy, recoverable when it comes down
- Speeding it up stores the energy as kinetic energy
- Dragging it against friction converts the energy into heat, which is why the surfaces warm up
Worked example 1 — work turning into speed. A force of N acts on a body of mass kg, initially at rest, over a distance of m along a smooth surface. Find its final speed.
The work done is
By the theorem, all of that becomes kinetic energy:
Check with the laws of motion. The acceleration is m s, and from with ,
The same value by a completely different route. That agreement is not a coincidence — the work-energy theorem is the laws of motion rewritten in terms of energy, and either method may be used.
Worked example 2 — negative work bringing a body to rest. A body of mass kg moving at m s is brought to rest by a retarding force. Find the work done by that force.
**The work is negative joule**, and the magnitude J is the energy that the body lost. **If the force acted over a distance of m, its magnitude was N.
Worked example 3 — the energy route being the easier one.** A stone of mass kg is thrown vertically upward with a speed of m s. How high does it rise?
All the kinetic energy becomes potential energy at the top, where the stone momentarily stops:
The mass cancelled, which is why a heavy stone and a light one thrown at the same speed rise to the same height. Solving this with the equations of motion would have needed the same two lines, but the energy method makes the cancellation of the mass obvious rather than accidental.
The link back to the previous section. If a coolie does zero work on the trunk, the theorem says the trunk's energy does not change — and it does not. Its height is the same, its speed is the same, so there is nothing for the work to have altered. The theorem and the agree, as they must.
Which is why work and energy are measured in the same unit. They are not two different quantities that happen to share the joule — work is energy being transferred, and the joule measures the amount transferred.
The theorem explains where work goes. When you do work on a body, that energy has to end up somewhere:
- Lifting it stores the energy as potential energy, recoverable when it comes down
- Speeding it up stores the energy as kinetic energy
- Dragging it against friction converts the energy into heat, which is why the surfaces warm up
Worked example 1 — work turning into speed. A force of N acts on a body of mass kg, initially at rest, over a distance of m along a smooth surface. Find its final speed.
The work done is
By the theorem, all of that becomes kinetic energy:
Check with the laws of motion. The acceleration is m s, and from with ,
The same value by a completely different route. That agreement is not a coincidence — the work-energy theorem is the laws of motion rewritten in terms of energy, and either method may be used.
Worked example 2 — negative work bringing a body to rest. A body of mass kg moving at m s is brought to rest by a retarding force. Find the work done by that force.
**The work is negative joule**, and the magnitude J is the energy that the body lost. **If the force acted over a distance of m, its magnitude was N.
Worked example 3 — the energy route being the easier one.** A stone of mass kg is thrown vertically upward with a speed of m s. How high does it rise?
All the kinetic energy becomes potential energy at the top, where the stone momentarily stops:
The mass cancelled, which is why a heavy stone and a light one thrown at the same speed rise to the same height. Solving this with the equations of motion would have needed the same two lines, but the energy method makes the cancellation of the mass obvious rather than accidental.
The link back to the previous section. If a coolie does zero work on the trunk, the theorem says the trunk's energy does not change — and it does not. Its height is the same, its speed is the same, so there is nothing for the work to have altered. The theorem and the agree, as they must.
How do you calculate power and convert between all its units?
Power is the rate of doing work — the work done divided by the time taken.
The SI unit is the watt (W), and one watt is one joule per second. Power measures how fast, not how much: a labourer and a crane lifting the same load through the same height do the same work but have very different powers, because the crane does it in a fraction of the time.
Worked example 1 — a man climbing stairs. A man of mass kg climbs steps, each m high, in s. Find the work done and his power.
The total height climbed:
Worked example 2 — a water pump. A pump raises kg of water through a height of m in s. Find its power in watt and in horsepower.
The units of work and energy, all of which must be recognised:
- The joule (J) — the SI unit
- The erg — the CGS unit, and J erg, since N dyne and m cm
- The calorie — the heat unit, with calorie J
- The kilowatt hour (kWh) — the commercial unit, the energy used by a kW device in one hour, so kWh J
- The electron volt (eV) — the atomic unit, with eV J
The units of power:
- The watt (W) — one joule per second
- The kilowatt (kW) W, the megawatt (MW) W, the gigawatt (GW) W
- The horsepower (hp), with hp W
Worked example 3 — the commercial unit. A W bulb is used for hours. Find the energy consumed in kilowatt hour and in joule.
One unit on the electricity bill is one kilowatt hour, and the enormous number of joules it represents is why the bill uses kWh at all.
Worked example 4 — a machine rated in horsepower. A motor is rated hp. Express its power in watt and find the work it does in one minute.
Worked example 5 — converting to erg. Express J in erg.
Worked example 6 — energy in electron volt. Express J in electron volt.
Why so many units exist at all. Each one is sized for the job it is used in. The joule is far too large for an electron and far too small for a household bill, so the electron volt and the kilowatt hour exist for those two extremes. The calorie belongs to heat and food, the erg to the CGS system, and the horsepower to engines.
One warning about power and energy. They are not interchangeable, and the bill charges for energy. **A W appliance used for one hour and a W appliance used for two hours consume exactly the same energy** — one kilowatt hour each — even though their powers differ by a factor of two.
The SI unit is the watt (W), and one watt is one joule per second. Power measures how fast, not how much: a labourer and a crane lifting the same load through the same height do the same work but have very different powers, because the crane does it in a fraction of the time.
Worked example 1 — a man climbing stairs. A man of mass kg climbs steps, each m high, in s. Find the work done and his power.
The total height climbed:
Worked example 2 — a water pump. A pump raises kg of water through a height of m in s. Find its power in watt and in horsepower.
The units of work and energy, all of which must be recognised:
- The joule (J) — the SI unit
- The erg — the CGS unit, and J erg, since N dyne and m cm
- The calorie — the heat unit, with calorie J
- The kilowatt hour (kWh) — the commercial unit, the energy used by a kW device in one hour, so kWh J
- The electron volt (eV) — the atomic unit, with eV J
The units of power:
- The watt (W) — one joule per second
- The kilowatt (kW) W, the megawatt (MW) W, the gigawatt (GW) W
- The horsepower (hp), with hp W
Worked example 3 — the commercial unit. A W bulb is used for hours. Find the energy consumed in kilowatt hour and in joule.
One unit on the electricity bill is one kilowatt hour, and the enormous number of joules it represents is why the bill uses kWh at all.
Worked example 4 — a machine rated in horsepower. A motor is rated hp. Express its power in watt and find the work it does in one minute.
Worked example 5 — converting to erg. Express J in erg.
Worked example 6 — energy in electron volt. Express J in electron volt.
Why so many units exist at all. Each one is sized for the job it is used in. The joule is far too large for an electron and far too small for a household bill, so the electron volt and the kilowatt hour exist for those two extremes. The calorie belongs to heat and food, the erg to the CGS system, and the horsepower to engines.
One warning about power and energy. They are not interchangeable, and the bill charges for energy. **A W appliance used for one hour and a W appliance used for two hours consume exactly the same energy** — one kilowatt hour each — even though their powers differ by a factor of two.
How do you set out a numerical on work, power and energy?
Write down what is given with its units, convert everything to SI, choose the formula from what is asked, and state the unit in the answer. These questions are short, and almost all the lost marks are unit errors rather than method errors.
Worked example 1 — force from work and distance. A body is moved through m and the work done is J, the force acting along the motion. Find the force.
Worked example 2 — height from work. A man does J of work in lifting a kg load. Through what height did he lift it?
Worked example 3 — time from power. A machine of power W does J of work. How long does it take?
Worked example 4 — a pump filling a tank. A pump lifts kg of water to a tank m above the ground in s. Find the power of the pump in watt and in kilowatt.
Worked example 5 — an angled force with two parts. A boy pulls a toy car through m with a force of N inclined at to the ground, while a friction force of N opposes the motion. Find the net work done on the car.
Work done by the pull:
Work done by friction, which acts opposite to the displacement:
The net work:
**And by the work-energy theorem, the car gains J of kinetic energy** — not J, because friction took part of it away as heat. Adding the two works with their signs is the whole of this question, and treating both as positive is the error it is built to catch.
Worked example 6 — a body on a smooth incline. A block of mass kg is raised through a vertical height of m along a smooth inclined plane. Find the work done against gravity.
The length of the incline is irrelevant. Work against gravity depends only on the vertical height gained, because gravity acts vertically and only the vertical component of the displacement is along it. A longer, gentler slope needs a smaller force over a greater distance for exactly the same work — which is the whole principle of the inclined plane as a machine.
Worked example 7 — cost from energy. A kW heater runs for hours a day for days. Find the energy consumed and the cost at ₹ per unit.
The three-step habit worth forming. Convert to SI, compute, then convert to whatever unit the question wants. Doing the conversion in the middle of the calculation is where errors creep in, and keeping it at the two ends makes the working easy to check.
Worked example 1 — force from work and distance. A body is moved through m and the work done is J, the force acting along the motion. Find the force.
Worked example 2 — height from work. A man does J of work in lifting a kg load. Through what height did he lift it?
Worked example 3 — time from power. A machine of power W does J of work. How long does it take?
Worked example 4 — a pump filling a tank. A pump lifts kg of water to a tank m above the ground in s. Find the power of the pump in watt and in kilowatt.
Worked example 5 — an angled force with two parts. A boy pulls a toy car through m with a force of N inclined at to the ground, while a friction force of N opposes the motion. Find the net work done on the car.
Work done by the pull:
Work done by friction, which acts opposite to the displacement:
The net work:
**And by the work-energy theorem, the car gains J of kinetic energy** — not J, because friction took part of it away as heat. Adding the two works with their signs is the whole of this question, and treating both as positive is the error it is built to catch.
Worked example 6 — a body on a smooth incline. A block of mass kg is raised through a vertical height of m along a smooth inclined plane. Find the work done against gravity.
The length of the incline is irrelevant. Work against gravity depends only on the vertical height gained, because gravity acts vertically and only the vertical component of the displacement is along it. A longer, gentler slope needs a smaller force over a greater distance for exactly the same work — which is the whole principle of the inclined plane as a machine.
Worked example 7 — cost from energy. A kW heater runs for hours a day for days. Find the energy consumed and the cost at ₹ per unit.
The three-step habit worth forming. Convert to SI, compute, then convert to whatever unit the question wants. Doing the conversion in the middle of the calculation is where errors creep in, and keeping it at the two ends makes the working easy to check.
Exam tip
Which steps protect the marks in a work and power question?
List the given quantities with units, convert to SI, and state the unit of every answer. An answer of "" with no unit earns nothing, however correct the number.
- **Use **, and read as the angle between the force and the displacement, not between the force and the horizontal unless they happen to be the same
- Remember the three cases: , ,
- Give the zero-work examples by name — a coolie on a level platform, a body in circular motion, a push on an immovable wall
- **For work against gravity use ** with the vertical height only; the length of an incline does not enter
- **Take m s unless the question gives another value, and say which value you used
- Add works with their signs when more than one force acts
- Convert minutes and hours to seconds** before using in watt
- Learn the conversions: J erg, calorie J, kWh J, eV J, hp W
- Keep power and energy apart — the bill charges for energy in kWh, not for power in kW
- State the answer in a sentence with its unit
The misconception to name. Feeling tired is not the same as doing work. A man holding a heavy suitcase without moving does no work on it at all, however exhausted he becomes, because there is no displacement. His muscles are using chemical energy to stay contracted, and that energy goes to heat inside his body rather than to the suitcase. Physics measures the transfer to the body, not the effort of the person.
A second trap. Using the length of a slope instead of the vertical height in . **A kg block raised m vertically along any smooth path needs J**, whether the slope is m long or m long. Substituting the slope length inflates the answer, and it is the commonest error in incline questions.
- **Use **, and read as the angle between the force and the displacement, not between the force and the horizontal unless they happen to be the same
- Remember the three cases: , ,
- Give the zero-work examples by name — a coolie on a level platform, a body in circular motion, a push on an immovable wall
- **For work against gravity use ** with the vertical height only; the length of an incline does not enter
- **Take m s unless the question gives another value, and say which value you used
- Add works with their signs when more than one force acts
- Convert minutes and hours to seconds** before using in watt
- Learn the conversions: J erg, calorie J, kWh J, eV J, hp W
- Keep power and energy apart — the bill charges for energy in kWh, not for power in kW
- State the answer in a sentence with its unit
The misconception to name. Feeling tired is not the same as doing work. A man holding a heavy suitcase without moving does no work on it at all, however exhausted he becomes, because there is no displacement. His muscles are using chemical energy to stay contracted, and that energy goes to heat inside his body rather than to the suitcase. Physics measures the transfer to the body, not the effort of the person.
A second trap. Using the length of a slope instead of the vertical height in . **A kg block raised m vertically along any smooth path needs J**, whether the slope is m long or m long. Substituting the slope length inflates the answer, and it is the commonest error in incline questions.
Did you know
Why is a unit of power named after a horse?
The horsepower exists because engines had to be sold to people who already owned horses. The only useful way to describe a new engine was to say how many horses it could replace, so the rate at which a working horse could raise a load became a unit.
And the number is oddly specific: ** hp W. It is not a round figure because it was never defined from the watt — it was measured from the work a horse could do and then converted afterwards. Every awkward conversion factor in this chapter has the same origin: two units defined independently, then found to be related by whatever number the measurement gave.
That is why the list looks arbitrary until you know where each unit came from.
- The erg** came from the CGS system, so J erg is a pure consequence of using centimetres and grams instead of metres and kilograms
- The calorie was defined from heating water, before anyone knew heat was a form of energy at all. **The factor is the measured exchange rate between mechanical work and heat
- The kilowatt hour was invented by electricity suppliers, who needed a unit big enough to bill in
- The electron volt was invented by atomic physicists, who needed one small enough to be useful for a single electron
The kilowatt hour is worth a second look, because its name gives away its definition. A kilowatt is a power and an hour is a time, so kilowatt hour is a power times a time, which is an energy. It is not a unit of power, despite containing the word "watt" — and reading the name as a multiplication is the quickest way never to confuse the two again.
A comparison that puts the sizes in perspective.** One kilowatt hour is million joules; one electron volt is joule. The ratio between them is about twenty-five powers of ten, which is why no single unit could ever have served both purposes.
And the horsepower survives for the same reason the others do. Engine ratings, pump ratings and motor ratings were quoted in horsepower long before the watt became standard, and a unit in daily use is very hard to displace. So a modern water pump is sold as "one horsepower" and a modern electric kettle as "two kilowatt", and the same physical quantity ends up with two names depending on which trade is selling it.
One last observation about the zero-work idea. The coolie's zero work is not a quirk of the definition — it is the reason a bicycle is efficient. A cyclist on level ground does no work against gravity at all, only against friction and air resistance, which is why cycling is so much easier than climbing. And it is why a pump's power depends on the height it raises water to and not on how far along the pipe the water travels.
And the number is oddly specific: ** hp W. It is not a round figure because it was never defined from the watt — it was measured from the work a horse could do and then converted afterwards. Every awkward conversion factor in this chapter has the same origin: two units defined independently, then found to be related by whatever number the measurement gave.
That is why the list looks arbitrary until you know where each unit came from.
- The erg** came from the CGS system, so J erg is a pure consequence of using centimetres and grams instead of metres and kilograms
- The calorie was defined from heating water, before anyone knew heat was a form of energy at all. **The factor is the measured exchange rate between mechanical work and heat
- The kilowatt hour was invented by electricity suppliers, who needed a unit big enough to bill in
- The electron volt was invented by atomic physicists, who needed one small enough to be useful for a single electron
The kilowatt hour is worth a second look, because its name gives away its definition. A kilowatt is a power and an hour is a time, so kilowatt hour is a power times a time, which is an energy. It is not a unit of power, despite containing the word "watt" — and reading the name as a multiplication is the quickest way never to confuse the two again.
A comparison that puts the sizes in perspective.** One kilowatt hour is million joules; one electron volt is joule. The ratio between them is about twenty-five powers of ten, which is why no single unit could ever have served both purposes.
And the horsepower survives for the same reason the others do. Engine ratings, pump ratings and motor ratings were quoted in horsepower long before the watt became standard, and a unit in daily use is very hard to displace. So a modern water pump is sold as "one horsepower" and a modern electric kettle as "two kilowatt", and the same physical quantity ends up with two names depending on which trade is selling it.
One last observation about the zero-work idea. The coolie's zero work is not a quirk of the definition — it is the reason a bicycle is efficient. A cyclist on level ground does no work against gravity at all, only against friction and air resistance, which is why cycling is so much easier than climbing. And it is why a pump's power depends on the height it raises water to and not on how far along the pipe the water travels.
Exam relevance
How do work, power and energy feed into JEE and NEET?
This is foundation work for Class 11 Work, Energy and Power, examined in JEE Main, JEE Advanced and NEET Physics.
**Where leads. Class 11 writes it as a dot product** of the force and displacement vectors, and then generalises it to a varying force as an integral, . **The is exactly what the dot product encodes**, and the zero-work case at becomes the statement that perpendicular vectors have zero dot product. JEE Main sets questions where a force varies with position and the work is the area under a force-displacement graph — the same idea with the constant force removed.
Where the work-energy theorem leads. Class 11 proves it rather than stating it, and uses it as the standard alternative to the equations of motion for any problem involving speeds and distances but not times. A JEE problem on a block sliding down a rough incline is solved in three lines by the theorem and in three-quarters of a page by kinematics, and recognising which questions suit it is a real exam skill.
Where the circular-motion zero-work case leads. It becomes the key fact that a centripetal force does no work, so the speed of a body in uniform circular motion never changes. That single sentence answers a whole family of assertion-reason questions in both JEE and NEET.
Where power leads. Class 11 adds instantaneous power, , which is the source of a standard JEE Main numerical: a vehicle of given engine power climbing a slope at a steady speed. **The relation "power equals force times velocity" is with both sides divided by the time, so it is not a new formula.
Where the units lead. The electron volt becomes the working unit of energy in Class 12 Atoms, Nuclei and Dual Nature of Radiation, where photon energies and ionisation energies are always quoted in eV. NEET uses it in the photoelectric-effect questions, and a candidate who cannot convert eV to joule cannot finish them.
Question types to expect.** At this level: work at an angle, zero-work cases, calculations, power of a pump or a man climbing stairs, and unit conversions. In competitive papers: work by a variable force, work-energy theorem on inclines with friction, instantaneous power, and energy conversions in eV.
The single trap that costs marks. Using the distance along a slope instead of the vertical height in . Gravity does work only through the vertical displacement, and the same error at JEE level appears as taking the wrong component of a displacement in a dot product.
A second trap. Adding the magnitudes of several works instead of their signed values. Friction contributes negative work, so in the toy-car example the net work is J and not J — and the theorem then gives the right gain in kinetic energy. In competitive problems the same slip appears whenever a retarding force is present.
Board versus competitive emphasis. The ICSE paper marks the formula, the substitution, the unit, the named zero-work example and the conversion factor; a competitive paper marks a single value, often reached through the theorem or through . The transferable habit is asking which component of the force lies along the motion before writing anything down — it decides the sign, it decides whether the work is zero, and at every later level it decides the dot product.
**Where leads. Class 11 writes it as a dot product** of the force and displacement vectors, and then generalises it to a varying force as an integral, . **The is exactly what the dot product encodes**, and the zero-work case at becomes the statement that perpendicular vectors have zero dot product. JEE Main sets questions where a force varies with position and the work is the area under a force-displacement graph — the same idea with the constant force removed.
Where the work-energy theorem leads. Class 11 proves it rather than stating it, and uses it as the standard alternative to the equations of motion for any problem involving speeds and distances but not times. A JEE problem on a block sliding down a rough incline is solved in three lines by the theorem and in three-quarters of a page by kinematics, and recognising which questions suit it is a real exam skill.
Where the circular-motion zero-work case leads. It becomes the key fact that a centripetal force does no work, so the speed of a body in uniform circular motion never changes. That single sentence answers a whole family of assertion-reason questions in both JEE and NEET.
Where power leads. Class 11 adds instantaneous power, , which is the source of a standard JEE Main numerical: a vehicle of given engine power climbing a slope at a steady speed. **The relation "power equals force times velocity" is with both sides divided by the time, so it is not a new formula.
Where the units lead. The electron volt becomes the working unit of energy in Class 12 Atoms, Nuclei and Dual Nature of Radiation, where photon energies and ionisation energies are always quoted in eV. NEET uses it in the photoelectric-effect questions, and a candidate who cannot convert eV to joule cannot finish them.
Question types to expect.** At this level: work at an angle, zero-work cases, calculations, power of a pump or a man climbing stairs, and unit conversions. In competitive papers: work by a variable force, work-energy theorem on inclines with friction, instantaneous power, and energy conversions in eV.
The single trap that costs marks. Using the distance along a slope instead of the vertical height in . Gravity does work only through the vertical displacement, and the same error at JEE level appears as taking the wrong component of a displacement in a dot product.
A second trap. Adding the magnitudes of several works instead of their signed values. Friction contributes negative work, so in the toy-car example the net work is J and not J — and the theorem then gives the right gain in kinetic energy. In competitive problems the same slip appears whenever a retarding force is present.
Board versus competitive emphasis. The ICSE paper marks the formula, the substitution, the unit, the named zero-work example and the conversion factor; a competitive paper marks a single value, often reached through the theorem or through . The transferable habit is asking which component of the force lies along the motion before writing anything down — it decides the sign, it decides whether the work is zero, and at every later level it decides the dot product.
Key takeaways
What must you be able to do from this part?
One formula with a cosine, one theorem and a list of conversions.
- ****, with the angle between the force and the displacement; work is a scalar, measured in joule
- ** gives ; gives ; gives
- A force of N through m at does J**; a friction force of N over m does J
- **Work against gravity is , using the vertical** height only — kg through m gives J, and kg through m gives kJ
- Zero work is done by a coolie walking on a level platform, by the centripetal force in circular motion, and when a body does not move at all
- The work-energy theorem: work done equals the change in energy, which is why both use the joule
- ** N over m on a kg body at rest** gives m s, confirmed by
- **Stopping a kg body from m s** requires J of work
- **A stone thrown up at m s rises m, independent of its mass
- , in watt, one joule per second — power measures how fast, not how much
- A kg man climbing steps of m in s** does J at W
- **A pump raising kg through m in s** gives W, about hp
- Energy units: J erg, calorie J, kWh J, eV J
- Power units: kW W, MW W, GW W, hp W
- **A W bulb for hours uses kWh**, and a hp motor is W
- Add works with their signs: a N pull at over m against N of friction gives J
- **A kW heater for hours a day for days** uses kWh, costing ₹ at ₹ a unit
- Power and energy are different — the bill charges for energy
The quickest self-test is one you can do on a staircase. Time yourself running up a flight, measure its height, estimate your mass, and work out your power in watt and in horsepower — then ask whether a one-horsepower water pump would beat you.
- ****, with the angle between the force and the displacement; work is a scalar, measured in joule
- ** gives ; gives ; gives
- A force of N through m at does J**; a friction force of N over m does J
- **Work against gravity is , using the vertical** height only — kg through m gives J, and kg through m gives kJ
- Zero work is done by a coolie walking on a level platform, by the centripetal force in circular motion, and when a body does not move at all
- The work-energy theorem: work done equals the change in energy, which is why both use the joule
- ** N over m on a kg body at rest** gives m s, confirmed by
- **Stopping a kg body from m s** requires J of work
- **A stone thrown up at m s rises m, independent of its mass
- , in watt, one joule per second — power measures how fast, not how much
- A kg man climbing steps of m in s** does J at W
- **A pump raising kg through m in s** gives W, about hp
- Energy units: J erg, calorie J, kWh J, eV J
- Power units: kW W, MW W, GW W, hp W
- **A W bulb for hours uses kWh**, and a hp motor is W
- Add works with their signs: a N pull at over m against N of friction gives J
- **A kW heater for hours a day for days** uses kWh, costing ₹ at ₹ a unit
- Power and energy are different — the bill charges for energy
The quickest self-test is one you can do on a staircase. Time yourself running up a flight, measure its height, estimate your mass, and work out your power in watt and in horsepower — then ask whether a one-horsepower water pump would beat you.