Work

NCERT Class 11 Physics Chapter 5 Section 5.3

Quick answer

Work done by a constant force F is the component of the force along the displacement times the magnitude of the displacement d: W = Fd cos θ = F·d, where θ is the angle between the force and the displacement. It is positive when θ is less than 90°, negative when θ is more than 90° (as kinetic friction does on a block sliding over a fixed floor), and zero when θ is 90°, when the force is zero, or when nothing moves. Its SI unit is the joule, 1 J = 1 N m.

With your textbook

NCERT Physics Part I (Class 11), Chapter 5 “Work, Energy and Power”, Section 5.3 “Work”, pages 73–74.

Open the official NCERT chapter (PDF, 2.0 MB, ncert.nic.in) (opens in a new tab)

Checked against the 2026-27 reprint on .

What is work?

Work in NCERT’s words

The work done by the force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement.

NCERT Physics Part I (Class 11), Section 5.3, page 74

A force does work when its point of application moves with a component along or against the force. For a constant force, multiply its component along the total displacement by the displacement’s magnitude; for a varying force, add the contributions along the actual path. Work is measured in a specified reference frame.

For a constant force F⃗\vec F and a displacement d⃗\vec d at an angle θ to each other:

Work done by a constant force

W=Fdcos⁡θ=F⃗⋅d⃗W = Fd\cos\theta = \vec F \cdot \vec d

Here F cos θ is the component of the force along the displacement. Work is the scalar product of force and displacement, so it is a scalar: it has a size and a sign, but no direction.

Positive, negative and zero work

The sign of the work comes from cos θ:

Angle θ between force and displacement Work Example
0° to less than 90° positive gravity on a falling ball; your pull on a trolley
90° zero gravity on a block sliding on a level table
more than 90° to 180° negative friction on a sliding block; brakes

Positive work is done by a force that helps the motion: it feeds energy into the body. Negative work is done by a force that opposes the motion, such as friction, which usually acts at 180° to the displacement, so cos 180° = −1.

No work is done in three cases:

  1. The point of application stays at rest. Pushing hard against a brick wall, or a weightlifter holding a heavy weight still above the head, does no work on the wall or the weight, however tiring it is. The tiredness comes from muscles working inside the body, not from work done on the load.
  2. Zero force. A block gliding on a smooth table has no horizontal force on it, yet it may travel a long way.
  3. Force perpendicular to the displacement. Gravity does no work on a block moving along a level table. If the Moon’s orbit is taken as a circle, the Earth’s pull, along the radius, is always at right angles to the Moon’s motion, along the tangent, so it does no work. The same is true of any centripetal force in uniform circular motion.

A force does no work on a point that remains at rest. Returning to the starting point is different: kinetic friction can do negative work throughout a closed path even though the total displacement is zero. A constant force, by contrast, does zero total work over any closed path.

Two cautions about signs. The normal force can do work: when a lift moves up, the floor’s push on you is along your displacement and does positive work. And friction is not always negative: static friction on a crate carried by an accelerating truck acts forward, along the crate’s motion, and does positive work on it.

Unit and dimensions of work

From W = Fd cos θ, the dimensions of work are those of force times length:

[W]=[M L T−2][L]=[M L2T−2]\begin{aligned} [W] &= [\text{M L T}^{-2}][\text{L}] \\ &= [\text{M L}^{2}\text{T}^{-2}] \end{aligned}

Work and energy have the same dimensions. Their SI unit is the joule (J), named after the British physicist James Prescott Joule:

1 J=1 N×1 m=1 kg m2 s−2\begin{aligned} 1\ \text{J} &= 1\ \text{N} \times 1\ \text{m} \\ &= 1\ \text{kg m}^{2}\ \text{s}^{-2} \end{aligned}

One joule is the work done when a force of 1 N moves its point of application through 1 m along the force. Some other units of work and energy:

Unit Value in joules
erg (CGS unit) 10−7 J
electron volt (eV) about 1.6 × 10−19 J
calorie (thermochemical) 4.184 J exactly
kilowatt hour (kWh) 3.6 × 106 J

The calorie conversion uses the thermochemical convention in NIST Special Publication 811, Appendix B.9; other calorie conventions have slightly different values.

Work done by more than one force

When several forces act on a body, each does its own work, with its own sign. The net work is the sum of these, and, for a body treated as a particle, it equals the work done by the net force. By the work-energy theorem, the net work is the change in the body’s kinetic energy.

A lesson from the third law is worth noting. When a cyclist skids to a stop, the road pushes back on the cycle and does negative work on it. The cycle pushes on the road with an equal and opposite force, but the road does not move, so that force does no work. NCERT Example 5.3 (page 74) works this out: a 200 N stopping force over 10 m does −2000 J of work on the cycle, while the cycle does no work on the road. Equal and opposite forces do not always do equal and opposite work.

Work in daily life

  • Lifting a load. Lifting a 10 kg bag through 1 m at a steady speed needs an upward force equal to its weight, about 98 N, so you do about 98 J of work on it, while gravity does about −98 J.
  • Pulling with a rope at an angle. Only the part of your pull along the ground does work on a sledge. The upward part does no work, because the sledge does not move upward; it only makes the sledge press less on the ground, which reduces friction.
  • Carrying a bag along a level corridor. At a steady speed, the upward force of your hand on the bag is at right angles to its motion, so it does no work on the bag.
  • Brakes and friction. They do negative work, taking kinetic energy away and turning it into heat.

Solved examples

Example 1: Pulling a box with a rope

A box is pulled 10 m along a level floor by a rope that makes 60° with the floor. The tension in the rope is 50 N. Find the work done by the rope on the box.

W=Fdcos⁡θ=50×10×cos⁡60∘=50×10×0.5=250 J\begin{aligned} W &= Fd\cos\theta \\ &= 50 \times 10 \times \cos 60^\circ \\ &= 50 \times 10 \times 0.5 \\ &= 250\ \text{J} \end{aligned}

Answer: 250 J. A pull of 50 N straight along the floor would have done 500 J; at 60° only half of the pull is along the motion.

Example 2: Every force on a sliding block

A 2 kg block slides 5 m down a rough slope inclined at 30°. The coefficient of kinetic friction is 0.2; only weight, the normal force and friction act. Find the work done by gravity, by friction and by the normal force, and the net work. Take g = 9.8 m s−2.

Gravity: the component of the weight along the slope is mg sin 30°:

Wg=mgsin⁡30∘×d=2×9.8×0.5×5=49 J\begin{aligned} W_g &= mg\sin 30^\circ \times d \\ &= 2 \times 9.8 \times 0.5 \times 5 \\ &= 49\ \text{J} \end{aligned}

Friction: the normal force is mg cos 30°, about 16.97 N, so friction is 0.2 × 16.97, about 3.39 N, acting up the slope:

Wf=−fd=−0.2×16.97×5≈−17.0 J\begin{aligned} W_f &= -f d \\ &= -0.2 \times 16.97 \times 5 \\ &\approx -17.0\ \text{J} \end{aligned}

Normal force: it is perpendicular to the motion, so WN = 0.

Net work:

Wnet=49−17.0+0≈32.0 J\begin{aligned} W_{\text{net}} &= 49 - 17.0 + 0 \\ &\approx 32.0\ \text{J} \end{aligned}

Answer: gravity +49 J, friction about −17.0 J, normal force 0, net about +32.0 J. If the block is able to start sliding from rest, the work-energy theorem gives its speed at the bottom: 12(2)v2\tfrac{1}{2}(2)v^{2} equals 32.0 J, so v2 is 32.0 m s, and v is 32.0\sqrt{32.0}, about 5.66 m s.

Example 3: Lifting at a steady speed

A person lifts a 10 kg bag through 1.5 m at a steady speed. Find the work done on the bag by the person, by gravity, and in total. Take g = 9.8 m s−2.

At a steady speed the person’s upward force equals the weight, mg = 98 N.

Wperson=98×1.5=147 JW_{\text{person}} = 98 \times 1.5 = 147\ \text{J} Wg=−98×1.5=−147 JW_g = -98 \times 1.5 = -147\ \text{J}

Answer: the person does +147 J, gravity −147 J, and the net work is zero, so the bag’s kinetic energy does not change, as the work-energy theorem requires.

Common mistakes to avoid

Common mistake

Using the whole force instead of its component along the motion

Only F cos θ does work. In Example 1, using 50 N × 10 m gives 500 J, twice the right answer.

Common mistake

Taking θ as the angle with the vertical or the slope

θ is the angle between the force and the displacement. On an incline, gravity makes 90° − α with the motion down a slope of angle α, which is why its work is mgd sin α.

Common mistake

Getting the sign of friction’s work wrong

Air resistance in still air, retarding braking forces, and kinetic friction on a body sliding over a fixed surface do negative work. Leaving out the minus sign makes the block in Example 2 gain energy from friction. But friction is not always negative: static friction on a crate carried by an accelerating truck does positive work on the crate, and ideal static friction on a wheel rolling without slipping over a fixed rigid surface does no work, because the contact point is instantaneously at rest. A moving surface can transfer energy through friction.

Common mistake

Thinking that effort means work

Holding a heavy bag still, or carrying it along a level path at steady speed, does no work on the bag in the physics sense, even though it is tiring.

Quick revision summary

  • Work done by a constant force is the component of the force along the displacement times the size of the displacement.
  • Work is a scalar. It is positive when the angle is less than 90°, zero at 90° and negative when the angle is more than 90°.
  • A force does no work if its point of application stays at rest, if the force is zero, or if it is perpendicular to the motion at every instant. Zero total displacement alone is not enough for a varying force.
  • The SI unit of work and energy is the joule, 1 J = 1 N m; the dimensions are [M L T].
  • The net work is the sum of the works of all the forces; equal and opposite forces need not do equal and opposite work.

Formula sheet

  • W = Fd cos θ = F·d
  • 1 J = 1 N m = 1 kg m2 s−2
  • [W] = [M L2 T−2]
  • Wnet = W1 + W2 + … (sum of the works of all the forces)
  • 1 kWh = 3.6 × 106 J, 1 eV ≈ 1.6 × 10−19 J

Questions students ask

Is work a scalar or a vector quantity?

Work is a scalar quantity. For a constant force it is the scalar (dot) product of two vectors, W = F · d = Fd cos θ, so it has a size and a sign (positive, negative or zero) but no direction.

What is the difference between work and energy?

Energy is the capacity of a body to do work, something the body has; work is energy transferred to or from a body by a force acting through a displacement, something that happens. They have the same SI unit, the joule, and the same dimensions, [M L T]. By the work-energy theorem, the net work done on a body equals the change in its kinetic energy.

What is the work done by gravity when a body is lifted or lowered?

When a body of mass m is raised through a height h near the Earth’s surface, gravity does −mgh of work; when it comes down through h, gravity does +mgh. The work depends only on the change in height, not on the path, because gravity is a conservative force (see path independence). If you lift the body at a steady speed, you do +mgh of work against gravity.

What is the work done by friction?

The work done by kinetic friction on a body sliding a distance d over a fixed surface is W = −μkNd, because the friction force μkN acts against the motion; on a level floor with only horizontal pushes, N = mg and W = −μkmgd. This negative work takes kinetic energy away and turns it into heat. Friction’s work is not always negative, though: a box dropped onto a moving conveyor belt is dragged forward by friction, which does positive work on it until the box moves with the belt.

Why is the work done by a centripetal force zero?

The work done by a centripetal force is zero because the force points along the radius, at right angles to the velocity, which is along the tangent. With θ = 90°, Fd cos 90° = 0 for every small step, so a body in uniform circular motion keeps its kinetic energy. For the same reason, the magnetic force on a moving charge, always perpendicular to its velocity, does no work.

Practice questions

Try each question first, then open the answer to check your working.

  1. Q1. A force of 20 N acts on a body that moves 4 m in a direction making 60° with the force. Find the work done.

    Show answerHide answer W=20×4×cos⁡60∘=40 J\begin{aligned} W &= 20 \times 4 \times \cos 60^\circ \\ &= 40\ \text{J} \end{aligned}
  2. Q2. A stone tied to a string moves in a circle at constant speed on a smooth horizontal table. How much work does the tension do in one full revolution? Explain.

    Show answerHide answer

    Zero. The tension acts along the string, towards the centre, while the stone moves along the tangent. The force is always at right angles to the displacement, so it does no work at any moment.

  3. Q3. A 5 kg block is dragged 8 m at constant speed along a rough level floor by a horizontal force. The coefficient of kinetic friction is 0.25. Find the work done by the applied force and by friction. Take g = 9.8 m s−2.

    Show answerHide answer

    At constant speed the applied force equals friction:

    f=0.25×5×9.8=12.25 N\begin{aligned} f &= 0.25 \times 5 \times 9.8 \\ &= 12.25\ \text{N} \end{aligned}

    The applied force does 12.25 × 8, that is 98 J, and friction does −98 J. The net work is zero, so the speed does not change.

More practice questions

Sources and review

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NCERT source
NCERT, Physics Part I (Class 11), reprint 2026-27. Chapter 5 “Work, Energy and Power”, Section 5.3 “Work”, pages 73–74. Checked on . Official chapter PDF on ncert.nic.in (opens in a new tab) (2.0 MB).
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