Friction Does Negative Work, and That Is Not a Figure of Speech
Learn to calculate work with the cosine factor when a force acts at an angle, recognise the three cases of zero work, tell positive work from negative, and use the work-energy theorem.
How can work done be a negative number?
When the force acts against the motion, it takes energy out of the object instead of putting it in.
Push a box m along the floor with a force of N. You do
of work on it. Meanwhile friction of N drags backwards along the same m, so friction does
The net work on the box is J, and that is the energy the box actually gained.
So the sign is not a bookkeeping convention. Positive work adds energy to a body; negative work removes it. Friction, brakes and a fielder's hands all do negative work, which is precisely how they slow things down.
This page covers the first part of the CBSE Class 9 Science chapter on work, energy and simple machines.
Push a box m along the floor with a force of N. You do
of work on it. Meanwhile friction of N drags backwards along the same m, so friction does
The net work on the box is J, and that is the energy the box actually gained.
So the sign is not a bookkeeping convention. Positive work adds energy to a body; negative work removes it. Friction, brakes and a fielder's hands all do negative work, which is precisely how they slow things down.
This page covers the first part of the CBSE Class 9 Science chapter on work, energy and simple machines.
Formula
What is the formula for work when the force acts at an angle?
Only the part of the force along the displacement does work, so
where is the force, the displacement, and the angle between them.
The unit is the joule (J), and — the work done when a force of N moves a body m in the direction of the force. Work is a scalar: it has a sign but no direction.
**The values of to know:**
Worked example 1 — force along the motion. A force of N moves a body m in the direction of the force.
Worked example 2 — force at an angle. The same N force, the same m, but applied at to the displacement.
Half the work, from the same force over the same distance. Only the component of N lay along the motion; the other component pulled uselessly upward.
Worked example 3 — pulling a trolley. A boy pulls a trolley m with a force of N applied along a handle at to the ground.
Worked example 4 — lifting. A kg bag is lifted m vertically. The force needed equals the weight, and it acts along the displacement, so :
The angle is between the force and the displacement, not between the force and the ground or the force and the horizontal. In worked example 3 the handle happened to make with the ground and the trolley moved along the ground, so the two angles coincided. When they do not, using the wrong one is the standard error — so mark the displacement arrow on your diagram before measuring anything.
Work needs both a force and a displacement. If either is zero, the work is zero however large the other is — which is the subject of the next section.
where is the force, the displacement, and the angle between them.
The unit is the joule (J), and — the work done when a force of N moves a body m in the direction of the force. Work is a scalar: it has a sign but no direction.
**The values of to know:**
Worked example 1 — force along the motion. A force of N moves a body m in the direction of the force.
Worked example 2 — force at an angle. The same N force, the same m, but applied at to the displacement.
Half the work, from the same force over the same distance. Only the component of N lay along the motion; the other component pulled uselessly upward.
Worked example 3 — pulling a trolley. A boy pulls a trolley m with a force of N applied along a handle at to the ground.
Worked example 4 — lifting. A kg bag is lifted m vertically. The force needed equals the weight, and it acts along the displacement, so :
The angle is between the force and the displacement, not between the force and the ground or the force and the horizontal. In worked example 3 the handle happened to make with the ground and the trolley moved along the ground, so the two angles coincided. When they do not, using the wrong one is the standard error — so mark the displacement arrow on your diagram before measuring anything.
Work needs both a force and a displacement. If either is zero, the work is zero however large the other is — which is the subject of the next section.
When is the work done exactly zero?
In three distinct cases, and a question asking for zero work is always asking which of the three applies.
**Case 1 — no displacement ().** Hold a heavy bag perfectly still at arm's length. However long you hold it and however tired you become,
**Case 2 — no force (). A body moving on a perfectly smooth horizontal surface with nothing pushing it. No force, so no work is done on it, and by the first law it keeps moving anyway.
Case 3 — force perpendicular to the displacement ().** Since , the work is zero however large the force and the displacement.
Worked example — carrying a load on level ground. A man carries a kg load on his head and walks m along level ground.
The force supporting the load is vertical, of size N. The displacement is horizontal. So and
No work at all is done on the load by the supporting force, over fifty metres.
**More cases of , all of them standard:
- The centripetal force on a body in circular motion is always directed to the centre while the motion is along the tangent — so it does no work, which is exactly why the speed in uniform circular motion never changes
- The Moon in a circular orbit: gravity is perpendicular to its motion, so gravity does no work on it
- The normal reaction from a floor on a box being pushed horizontally
Zero work does not mean zero effort. The man carrying the load is genuinely tired, and his muscles are consuming energy — they must keep contracting to hold the load up, and each contraction costs chemical energy. But no energy is transferred to the load**, and work in physics means exactly that transfer.
That link to circular motion is worth keeping. The third part of the motion chapter established that uniform circular motion is accelerated because the direction changes. This page adds the other half: the force responsible does no work, so the speed stays constant while the direction changes. The two facts fit together, and neither makes sense without the other.
**Case 1 — no displacement ().** Hold a heavy bag perfectly still at arm's length. However long you hold it and however tired you become,
**Case 2 — no force (). A body moving on a perfectly smooth horizontal surface with nothing pushing it. No force, so no work is done on it, and by the first law it keeps moving anyway.
Case 3 — force perpendicular to the displacement ().** Since , the work is zero however large the force and the displacement.
Worked example — carrying a load on level ground. A man carries a kg load on his head and walks m along level ground.
The force supporting the load is vertical, of size N. The displacement is horizontal. So and
No work at all is done on the load by the supporting force, over fifty metres.
**More cases of , all of them standard:
- The centripetal force on a body in circular motion is always directed to the centre while the motion is along the tangent — so it does no work, which is exactly why the speed in uniform circular motion never changes
- The Moon in a circular orbit: gravity is perpendicular to its motion, so gravity does no work on it
- The normal reaction from a floor on a box being pushed horizontally
Zero work does not mean zero effort. The man carrying the load is genuinely tired, and his muscles are consuming energy — they must keep contracting to hold the load up, and each contraction costs chemical energy. But no energy is transferred to the load**, and work in physics means exactly that transfer.
That link to circular motion is worth keeping. The third part of the motion chapter established that uniform circular motion is accelerated because the direction changes. This page adds the other half: the force responsible does no work, so the speed stays constant while the direction changes. The two facts fit together, and neither makes sense without the other.
How do you tell positive work from negative work?
By the angle. If the force has any component along the motion, the work is positive; if it has a component against the motion, the work is negative.
- : is positive, so positive work, and energy is transferred to the body
- : zero work
- : is negative, so negative work, and energy is taken from the body
Worked example 1 — the pushed box, both forces. A box is pushed m with N while friction of N opposes it.
Worked example 2 — lifting and lowering. A kg object is lifted m, then lowered m. Taking , its weight is N.
While lifting: the hand's force is upward and the motion upward, so the hand does J. Gravity acts downward while the motion is upward, so gravity does J.
While lowering: the motion is now downward. Gravity does J, and the hand does J.
The same force does positive work in one case and negative in the other. Gravity has not changed; the direction of the displacement has. So gravity always does negative work is false — it is negative on a rising body and positive on a falling one.
Worked example 3 — a ball thrown upward. On the way up, gravity opposes the motion and does negative work, which is why the ball slows down. At the top the velocity is zero. On the way down, gravity acts along the motion and does positive work, which is why the ball speeds up. One force, two signs, and the whole flight explained.
Worked example 4 — braking. A car's brakes apply a force opposite to its motion, so they do negative work, removing kinetic energy from the car. The energy does not vanish; it appears as heat in the brake pads and tyres.
Negative work is how anything is ever slowed down. Friction, brakes, air resistance, a fielder's hands, a crash barrier — every one of them works by doing negative work on a moving object. So a question asking what does the work of stopping this object? is asking which force is doing the negative work, and by how much.
- : is positive, so positive work, and energy is transferred to the body
- : zero work
- : is negative, so negative work, and energy is taken from the body
Worked example 1 — the pushed box, both forces. A box is pushed m with N while friction of N opposes it.
Worked example 2 — lifting and lowering. A kg object is lifted m, then lowered m. Taking , its weight is N.
While lifting: the hand's force is upward and the motion upward, so the hand does J. Gravity acts downward while the motion is upward, so gravity does J.
While lowering: the motion is now downward. Gravity does J, and the hand does J.
The same force does positive work in one case and negative in the other. Gravity has not changed; the direction of the displacement has. So gravity always does negative work is false — it is negative on a rising body and positive on a falling one.
Worked example 3 — a ball thrown upward. On the way up, gravity opposes the motion and does negative work, which is why the ball slows down. At the top the velocity is zero. On the way down, gravity acts along the motion and does positive work, which is why the ball speeds up. One force, two signs, and the whole flight explained.
Worked example 4 — braking. A car's brakes apply a force opposite to its motion, so they do negative work, removing kinetic energy from the car. The energy does not vanish; it appears as heat in the brake pads and tyres.
Negative work is how anything is ever slowed down. Friction, brakes, air resistance, a fielder's hands, a crash barrier — every one of them works by doing negative work on a moving object. So a question asking what does the work of stopping this object? is asking which force is doing the negative work, and by how much.
What does the work-energy theorem let you calculate?
It says the net work done on a body equals the change in its kinetic energy, which lets you find a speed without ever finding the time or the acceleration.
Where it comes from. Take the third equation of motion, , and multiply throughout by :
Rearranged, . The theorem is the third equation of motion, rewritten in terms of energy — nothing new has been assumed.
Worked example 1 — finding a final speed. A force of N acts on a kg body at rest over a distance of m.
Check with the equations of motion: , and , so m/s. The two agree, as they must.
Worked example 2 — finding a braking force. A car of mass kg travelling at m/s is brought to rest in m.
So the brakes do J of work over m, and the force is
opposing the motion. Notice how little the theorem needed — no time, no acceleration, just the energy change and the distance.
Worked example 3 — a speed change both ways. A kg body speeds up from m/s to m/s.
So the net work done on it was J.
Worked example 4 — catching a ball. A kg ball arriving at m/s is brought to rest.
The hands do J of work, whether the catch takes a long time or a short one. Compare that with the previous chapter, where the force depended on the time but the momentum change did not. Here the work depends on the distance the hands move back, and the energy change does not. Momentum and time go together; energy and distance go together — and that pairing is worth learning once and for all.
The theorem uses the net work. Worked example 1 involved a single force and nothing else. When friction also acts, the net work is the push's positive work plus friction's negative work, and using only the applied force overstates the speed gained.
Where it comes from. Take the third equation of motion, , and multiply throughout by :
Rearranged, . The theorem is the third equation of motion, rewritten in terms of energy — nothing new has been assumed.
Worked example 1 — finding a final speed. A force of N acts on a kg body at rest over a distance of m.
Check with the equations of motion: , and , so m/s. The two agree, as they must.
Worked example 2 — finding a braking force. A car of mass kg travelling at m/s is brought to rest in m.
So the brakes do J of work over m, and the force is
opposing the motion. Notice how little the theorem needed — no time, no acceleration, just the energy change and the distance.
Worked example 3 — a speed change both ways. A kg body speeds up from m/s to m/s.
So the net work done on it was J.
Worked example 4 — catching a ball. A kg ball arriving at m/s is brought to rest.
The hands do J of work, whether the catch takes a long time or a short one. Compare that with the previous chapter, where the force depended on the time but the momentum change did not. Here the work depends on the distance the hands move back, and the energy change does not. Momentum and time go together; energy and distance go together — and that pairing is worth learning once and for all.
The theorem uses the net work. Worked example 1 involved a single force and nothing else. When friction also acts, the net work is the push's positive work plus friction's negative work, and using only the applied force overstates the speed gained.
Exam tip
Exam tip: mark the displacement, then measure the angle to it
**, with between the force and the displacement** — not between the force and the ground. Draw the displacement arrow first.
Learn the four values: , , , .
Three zero-work cases: (holding a bag still), , and (carrying a load on level ground, or any centripetal force).
Zero work is not zero effort — your muscles still spend energy, but none reaches the load.
**Positive work below , negative above it. Friction and brakes always do negative work, which is how they slow things.
The same force can do work of either sign.** Gravity does J on a rising kg object and J on it falling.
Work is a scalar in joules, with . Give the sign but never a direction.
Work-energy theorem: . Use it when the question gives a distance and wants a speed or a force, and skips the time.
Use the net work, adding the negative contributions.
And remember which pairs go together: force with time gives momentum, force with distance gives energy.
Learn the four values: , , , .
Three zero-work cases: (holding a bag still), , and (carrying a load on level ground, or any centripetal force).
Zero work is not zero effort — your muscles still spend energy, but none reaches the load.
**Positive work below , negative above it. Friction and brakes always do negative work, which is how they slow things.
The same force can do work of either sign.** Gravity does J on a rising kg object and J on it falling.
Work is a scalar in joules, with . Give the sign but never a direction.
Work-energy theorem: . Use it when the question gives a distance and wants a speed or a force, and skips the time.
Use the net work, adding the negative contributions.
And remember which pairs go together: force with time gives momentum, force with distance gives energy.
Did you know
Why a longer braking distance saves you and a longer time does too
Two pieces of vehicle design look as though they do the same thing, and they rest on two different equations.
A crumple zone works on time. It folds slowly, stretching the collision out, and because the change in momentum is fixed, a longer time means a smaller force. That is the momentum form of the second law, and it is the fielder's soft hands applied to a car.
A long braking distance works on distance. The kinetic energy that has to be removed is fixed, and spreading that removal over a greater distance means a smaller force. That is the work-energy theorem.
The two are genuinely different levers on the same problem, and the arithmetic shows how differently they behave. Kinetic energy depends on the square of the speed, so doubling a car's speed quadruples the energy the brakes must remove. With the same braking force, the stopping distance quadruples too.
Work it through. At m/s a kg car carries J, and a N braking force needs m to remove it. At m/s the same car carries J, and the same braking force now needs
Twice the speed, four times the distance. Not twice — four times, and that is the single most useful number in road safety.
Momentum, by contrast, depends on speed to the first power only, so doubling the speed merely doubles the momentum and doubles the stopping time.
So the same collision, described by momentum and by energy, gives two different scaling rules — and knowing which quantity a question is about is what tells you whether the answer doubles or quadruples.
A crumple zone works on time. It folds slowly, stretching the collision out, and because the change in momentum is fixed, a longer time means a smaller force. That is the momentum form of the second law, and it is the fielder's soft hands applied to a car.
A long braking distance works on distance. The kinetic energy that has to be removed is fixed, and spreading that removal over a greater distance means a smaller force. That is the work-energy theorem.
The two are genuinely different levers on the same problem, and the arithmetic shows how differently they behave. Kinetic energy depends on the square of the speed, so doubling a car's speed quadruples the energy the brakes must remove. With the same braking force, the stopping distance quadruples too.
Work it through. At m/s a kg car carries J, and a N braking force needs m to remove it. At m/s the same car carries J, and the same braking force now needs
Twice the speed, four times the distance. Not twice — four times, and that is the single most useful number in road safety.
Momentum, by contrast, depends on speed to the first power only, so doubling the speed merely doubles the momentum and doubles the stopping time.
So the same collision, described by momentum and by energy, gives two different scaling rules — and knowing which quantity a question is about is what tells you whether the answer doubles or quadruples.
Exam relevance
How does the work-energy theorem feed into JEE Main and NEET?
This page is the foundation for the Class 11 Physics chapter Work, Energy and Power, examined in JEE Main and in NEET Physics, and the formula on this page is the one that chapter opens with.
The cosine factor is treated there as the scalar product of two vectors, written , which is the same statement with vector notation. Once forces vary, the product becomes an integral and the work becomes the area under a force-displacement graph — an extension that reuses the slope-and-area reasoning from the motion chapter.
The zero-work cases become standing facts in that chapter and beyond. That the centripetal force does no work is the reason speed is constant in uniform circular motion, and it reappears in Class 11 Motion in a Plane, in Gravitation for circular orbits, and in Class 12 Moving Charges and Magnetism, where the magnetic force on a moving charge is always perpendicular to its velocity and therefore does no work either. That last case is a recurring JEE Main item, and it is this page's rule in a new setting.
The work-energy theorem is used throughout Class 11 mechanics as a shortcut wherever the time is not wanted — for a block sliding down a rough incline, for a spring being compressed, for a body on a vertical circular track.
Negative work by friction becomes the standard way of computing heat generated, and it leads into the distinction between conservative and non-conservative forces in Class 11.
What the questions look like. Numericals are the main form, and the two standard shapes are find the speed after a force acts over a distance and find the force from a stopping distance. Assertion-reason items favour the zero-work statements, especially the centripetal one and the carried-load one. Conceptual questions give a situation and ask for the sign of the work done by a named force, which is where the lifting-and-lowering example above pays off.
How board and competitive emphasis differ. A board paper asks you to define work, state the three zero-work cases, and compute once. A competitive paper gives a situation with two or three forces and asks for the net work, or asks for the work done by one named force among several — so identifying which force and which displacement go together is the skill being tested rather than the formula.
The single trap that costs the most marks. Measuring from the wrong line. The angle is between the force and the displacement, and in problems on an inclined plane the displacement is along the slope while the weight is vertical — so the angle is neither the slope angle nor its complement without thinking. Drawing the displacement arrow before measuring is what prevents it.
A second trap worth naming. Assuming that if a body is tired-making to hold, work is being done on it. Zero displacement means zero work, and a question describing someone holding a load or a coolie walking on level ground is asking for zero — an answer many students refuse to give because it feels wrong.
The cosine factor is treated there as the scalar product of two vectors, written , which is the same statement with vector notation. Once forces vary, the product becomes an integral and the work becomes the area under a force-displacement graph — an extension that reuses the slope-and-area reasoning from the motion chapter.
The zero-work cases become standing facts in that chapter and beyond. That the centripetal force does no work is the reason speed is constant in uniform circular motion, and it reappears in Class 11 Motion in a Plane, in Gravitation for circular orbits, and in Class 12 Moving Charges and Magnetism, where the magnetic force on a moving charge is always perpendicular to its velocity and therefore does no work either. That last case is a recurring JEE Main item, and it is this page's rule in a new setting.
The work-energy theorem is used throughout Class 11 mechanics as a shortcut wherever the time is not wanted — for a block sliding down a rough incline, for a spring being compressed, for a body on a vertical circular track.
Negative work by friction becomes the standard way of computing heat generated, and it leads into the distinction between conservative and non-conservative forces in Class 11.
What the questions look like. Numericals are the main form, and the two standard shapes are find the speed after a force acts over a distance and find the force from a stopping distance. Assertion-reason items favour the zero-work statements, especially the centripetal one and the carried-load one. Conceptual questions give a situation and ask for the sign of the work done by a named force, which is where the lifting-and-lowering example above pays off.
How board and competitive emphasis differ. A board paper asks you to define work, state the three zero-work cases, and compute once. A competitive paper gives a situation with two or three forces and asks for the net work, or asks for the work done by one named force among several — so identifying which force and which displacement go together is the skill being tested rather than the formula.
The single trap that costs the most marks. Measuring from the wrong line. The angle is between the force and the displacement, and in problems on an inclined plane the displacement is along the slope while the weight is vertical — so the angle is neither the slope angle nor its complement without thinking. Drawing the displacement arrow before measuring is what prevents it.
A second trap worth naming. Assuming that if a body is tired-making to hold, work is being done on it. Zero displacement means zero work, and a question describing someone holding a load or a coolie walking on level ground is asking for zero — an answer many students refuse to give because it feels wrong.
Key takeaways
Work, its sign, and the work-energy theorem: quick revision
- ****, with between the force and the displacement. Work is a scalar in joules, and .
- Values to know: , , , .
- N over m along the motion gives J; the same at gives J.
- An N pull at over m gives J; lifting kg through m gives J.
- Three zero-work cases: , , and .
- Carrying a kg load m on level ground gives J.
- The centripetal force does no work, which is why speed is constant in uniform circular motion. The normal reaction on a horizontally pushed box does none either.
- Zero work is not zero effort — muscles spend energy, but none is transferred to the load.
- Positive work for adds energy; negative work for removes it.
- A box pushed m with N against N friction: J and J, so J net.
- Gravity's sign depends on the motion: J on a kg object rising m, J on it falling.
- Brakes and friction do negative work, and the energy removed appears as heat.
- Work-energy theorem: , obtained by multiplying by .
- N over m on a kg body at rest gives J, so and m/s — matching .
- A kg car at m/s stopping in m loses J, so the braking force is N.
- A kg body going from to m/s gains J.
- Catching a kg ball at m/s means J of work by the hands, whatever the time taken.
- Force with time gives momentum; force with distance gives energy. Doubling the speed doubles the momentum and quadruples the energy — so the stopping distance quadruples from m to m.
Work out the work done by each force acting on a box you push across a rough floor, then add them and check the total against the box's change in speed — the two should agree exactly.
- Values to know: , , , .
- N over m along the motion gives J; the same at gives J.
- An N pull at over m gives J; lifting kg through m gives J.
- Three zero-work cases: , , and .
- Carrying a kg load m on level ground gives J.
- The centripetal force does no work, which is why speed is constant in uniform circular motion. The normal reaction on a horizontally pushed box does none either.
- Zero work is not zero effort — muscles spend energy, but none is transferred to the load.
- Positive work for adds energy; negative work for removes it.
- A box pushed m with N against N friction: J and J, so J net.
- Gravity's sign depends on the motion: J on a kg object rising m, J on it falling.
- Brakes and friction do negative work, and the energy removed appears as heat.
- Work-energy theorem: , obtained by multiplying by .
- N over m on a kg body at rest gives J, so and m/s — matching .
- A kg car at m/s stopping in m loses J, so the braking force is N.
- A kg body going from to m/s gains J.
- Catching a kg ball at m/s means J of work by the hands, whatever the time taken.
- Force with time gives momentum; force with distance gives energy. Doubling the speed doubles the momentum and quadruples the energy — so the stopping distance quadruples from m to m.
Work out the work done by each force acting on a box you push across a rough floor, then add them and check the total against the box's change in speed — the two should agree exactly.