Work Done by a Force

In short

Work done by a constant force is W = Fd\cos\theta, where F is the magnitude of the force, d is the magnitude of the displacement, and \theta is the angle between them. Work is positive when the force has a component along the motion, negative when it opposes the motion, and zero when the force is perpendicular to the motion. The SI unit of work is the joule (1 J = 1 N·m).

A coolie at a railway station lifts a 40 kg suitcase onto his head and walks 200 metres along the platform. He exerts a large upward force — about 400 N to support the suitcase against gravity. He covers a large distance. Yet physics says the work he does on the suitcase while walking is zero. Not small — zero.

That sounds wrong. The coolie is clearly exerting effort. He is clearly tired at the end. But "work" in physics is not the same as "effort" in everyday language. Physics cares about something very specific: the direction of the force relative to the direction of the motion. The coolie's force is straight up. His displacement is horizontal. Those two directions are perpendicular — and when force and displacement are perpendicular, the work is exactly zero.

Understanding why this is true — and what "work" actually measures in physics — is the starting point for all of energy physics.

The direction matters

Start with the simplest case. You push a heavy wooden almirah across your room. Your hands press horizontally, the almirah slides horizontally, and both the force and the displacement point in the same direction. The harder you push and the farther the almirah slides, the more work you do. This much is intuitive.

Now change the angle. Instead of pushing horizontally, you push at a downward angle — your hands are on the top of the almirah, and you're leaning into it. Part of your force drives the almirah forward, and part of it presses the almirah into the floor. Only the forward part — the component of your force along the displacement — actually contributes to moving the almirah. The downward part just increases the friction.

A force $\vec{F}$ applied at angle $\theta$ to the displacement $\vec{d}$. Only the component along the displacement — $F\cos\theta$ — does work. The perpendicular component $F\sin\theta$ pushes sideways and contributes nothing to the displacement.

This is the central idea: only the component of force along the displacement counts as work. The perpendicular component, no matter how large, does nothing to move the object in the direction it is actually moving.

Now the coolie makes sense. His force is entirely perpendicular to the displacement. The component of his upward force along the horizontal direction is F\cos 90° = 0. So the work is zero — not because his force is small, but because all of it acts at right angles to the motion.

And friction makes sense too. When a cricket ball slides along the pitch after a throw from the boundary, the friction force points backward — directly opposing the motion. The angle between friction and displacement is 180°, and \cos 180° = -1. The work done by friction is negative. Negative work means the force is taking energy away from the object, slowing it down.

The formula

These three cases — force along the motion, force perpendicular to it, force opposing it — are all captured by one formula.

Work done by a constant force

The work done by a constant force \vec{F} on an object that undergoes a displacement \vec{d} is:

W = Fd\cos\theta

where F is the magnitude of the force, d is the magnitude of the displacement, and \theta is the angle between the force and the displacement vectors.

Read the formula piece by piece. F is how hard you push. d is how far the object moves. \cos\theta is the fraction of your force that points in the direction of motion. Multiply them together and you get the work — a single number (a scalar, not a vector) measured in joules.

The three cases fall out immediately:

Angle \theta	\cos\theta	Sign of work	Physical meaning
0°	+1	Positive	Force fully along motion — accelerates the object
Between 0° and 90°	Positive	Positive	Force has a forward component
90°	0	Zero	Force entirely perpendicular — does not speed up or slow down
Between 90° and 180°	Negative	Negative	Force has a backward component — decelerates
180°	-1	Negative	Force directly opposes motion — maximum deceleration

The unit: the joule

Work is force times distance, so its unit is the newton-metre. This combination has its own name: the joule (symbol J).

1 \text{ J} = 1 \text{ N·m} = 1 \text{ kg·m}^2/\text{s}^2

Why: you will also see the older CGS unit, the erg (1 erg = 10^{-7} J), in some Indian textbooks. The erg is tiny — a mosquito flapping its wings once does about 1 erg of work. Stick to joules.

A sense of scale: a 1 kg coconut falling from a height of 1 metre has gravity doing about 9.8 J of work on it. A person climbing one floor of stairs (roughly 3 metres, body mass 60 kg): gravity does about -1764 J of work on them (negative because gravity opposes the upward motion), and their muscles do +1764 J to overcome it.

Explore the angle

The interactive figure below lets you drag the angle \theta from 0° to 180° and watch the work change. The force is fixed at 50 N and the displacement at 3 m, so W = 150\cos\theta J.

Drag the red point to change the angle $\theta$ between force and displacement. The curve shows $\cos\theta$ — the fraction of the force that contributes to work. At $\theta = 0°$, the full force contributes. At $90°$, nothing. Past $90°$, the work turns negative. The readout uses $F = 50$ N and $d = 3$ m.

Notice the symmetry: \cos 60° = 0.5 and \cos 120° = -0.5. At 60°, half the force drives the motion forward. At 120°, half the force pulls it backward. The sign tells you whether energy is being transferred to the object or away from it.

Work done by common forces

Every force in a problem contributes its own work, independently. Here are the forces you encounter most often and what their work looks like.

Gravity

Gravity acts straight down with magnitude mg. If an object falls through a vertical height h, the angle between the gravitational force (downward) and the displacement (downward) is 0°:

W_{\text{gravity}} = mgh \quad \text{(object falling — positive work)}

Why: gravity and displacement point in the same direction, so \cos 0° = 1. Gravity is transferring gravitational potential energy into kinetic energy.

If the object rises through a height h, the displacement is upward but gravity still points down. The angle is 180°:

W_{\text{gravity}} = -mgh \quad \text{(object rising — negative work)}

Why: now gravity opposes the motion. It takes energy away from the object, slowing it down (or requiring another force to keep it going up).

A crucial fact: the work done by gravity depends only on the vertical height change, not on the path taken. Whether you walk up a straight staircase or a winding ramp, gravity does the same work as long as the total height gained is the same. This will become important when you study conservative forces and potential energy.

Normal force

The normal force is always perpendicular to the surface, and the displacement is always along the surface. The angle between them is 90°:

W_{\text{normal}} = N \cdot d \cdot \cos 90° = 0 \quad \text{(always)}

The normal force never does work on an object sliding along a surface. It supports the object against falling through the surface, but it does not speed it up or slow it down.

Friction

Kinetic friction acts along the surface, directly opposing the direction of motion. The angle between friction and displacement is 180°:

W_{\text{friction}} = f_k \cdot d \cdot \cos 180° = -f_k \cdot d \quad \text{(always negative)}

Why: \cos 180° = -1. Friction always opposes the motion and always removes kinetic energy from the object, converting it to heat. Rub your palms together fast — the warmth you feel is friction's negative work being converted into thermal energy.

Applied force at an angle

When you push or pull at an angle \theta to the horizontal (like pulling a suitcase by its strap), only the horizontal component F\cos\theta drives the object along the floor:

W_{\text{applied}} = Fd\cos\theta

This is the general formula. The special cases above (gravity, normal, friction) are all instances of W = Fd\cos\theta with specific values of \theta.

Worked examples

Example 1: Pulling a suitcase across a railway platform

A traveller pulls a 20 kg suitcase along a flat platform using a strap that makes a 30° angle with the horizontal. The applied force along the strap is 60 N, and the suitcase slides 25 m. The coefficient of kinetic friction between the suitcase wheels and the platform is \mu_k = 0.15. Find the work done by (a) the applied force, (b) gravity, (c) the normal force, and (d) friction.

Free body diagram of the suitcase. The applied force $F$ pulls at $30°$ above horizontal, gravity pulls down, the normal force pushes up, and friction opposes the horizontal motion.

Step 1. Work done by the applied force.

The angle between the applied force (along the strap at 30° above horizontal) and the displacement (horizontal) is \theta = 30°.

W_{\text{applied}} = Fd\cos\theta = 60 \times 25 \times \cos 30° = 1500 \times \frac{\sqrt{3}}{2} \approx 1299 \text{ J}

Why: only the horizontal component of the pull (60\cos 30° \approx 52 N) does work. The vertical component (60\sin 30° = 30 N) lifts the suitcase slightly, reducing the normal force, but does not contribute to horizontal work.

Step 2. Work done by gravity.

Gravity acts vertically downward. The displacement is horizontal. The angle is 90°.

W_{\text{gravity}} = mgd\cos 90° = 0

Why: the suitcase does not rise or fall — it moves along a flat platform. Gravity is perpendicular to the motion the entire time.

Step 3. Work done by the normal force.

The normal force acts vertically upward. The displacement is horizontal. The angle is 90°.

W_{\text{normal}} = Nd\cos 90° = 0

Why: same reasoning as gravity. The normal force supports the suitcase against the floor; it never pushes the suitcase forward or backward.

Step 4. Work done by friction.

First, find the friction force. The normal force is not simply mg here — the vertical component of the applied force partially lifts the suitcase:

N = mg - F\sin 30° = 20 \times 9.8 - 60 \times 0.5 = 196 - 30 = 166 \text{ N}

f_k = \mu_k N = 0.15 \times 166 = 24.9 \text{ N}

Friction opposes the motion, so \theta = 180°.

W_{\text{friction}} = f_k \cdot d \cdot \cos 180° = 24.9 \times 25 \times (-1) = -622.5 \text{ J}

Why: friction removes about 623 J of kinetic energy from the suitcase, converting it to heat between the wheels and the platform surface.

Result: W_{\text{applied}} \approx 1299 J, W_{\text{gravity}} = 0, W_{\text{normal}} = 0, W_{\text{friction}} \approx -623 J. The net work on the suitcase is 1299 - 623 = 676 J — this is the energy that goes into speeding up the suitcase.

What this shows: Of the four forces acting on the suitcase, two do zero work (gravity and normal — both perpendicular to the motion), one does positive work (the applied pull), and one does negative work (friction). The net work determines whether the suitcase speeds up or slows down.

Example 2: Block sliding down a smooth incline

A 5 kg wooden block slides 4 m down a frictionless incline that makes 37° with the horizontal (\sin 37° = 0.6, \cos 37° = 0.8). Find the work done by gravity and by the normal force.

The block's weight $mg$ acts straight down. Its component along the incline ($mg\sin 37°$) pulls the block down the slope. The normal force $N$ acts perpendicular to the surface. The angle between $mg$ and the displacement $d$ (along the incline) is $90° - 37° = 53°$.

Step 1. Work done by gravity.

Gravity acts straight down with magnitude mg = 5 \times 9.8 = 49 N. The displacement is 4 m along the incline, directed downhill. The angle between gravity (downward) and the displacement (down the slope at 37° below horizontal) is 90° - 37° = 53°.

W_{\text{gravity}} = mgd\cos 53° = 49 \times 4 \times 0.6 = 117.6 \text{ J}

Why: you can verify this using the component method. The component of gravity along the incline is mg\sin 37° = 49 \times 0.6 = 29.4 N. This component acts over 4 m: 29.4 \times 4 = 117.6 J. Both methods give the same answer — they must, because \cos 53° = \sin 37° = 0.6.

Step 2. Work done by the normal force.

The normal force is perpendicular to the incline surface. The displacement is along the incline. The angle between them is 90°.

W_{\text{normal}} = Nd\cos 90° = 0

Why: the normal force pushes the block away from the surface but never along it. Perpendicular forces cannot do work on objects moving along the surface.

Step 3. Check using the vertical height.

The vertical height the block descends is h = d\sin 37° = 4 \times 0.6 = 2.4 m. So:

W_{\text{gravity}} = mgh = 49 \times 2.4 = 117.6 \text{ J}

Why: gravity's work depends only on the vertical height change, not on the path. Whether the block slides 4 m down a 37° ramp or falls 2.4 m straight down, gravity does the same 117.6 J of work.

Result: W_{\text{gravity}} = 117.6 J (positive — gravity accelerates the block down the slope). W_{\text{normal}} = 0.

What this shows: On an incline, gravity does positive work even though the displacement is not vertical — because gravity has a component along the slope. The steeper the incline, the larger the component, and the more work gravity does over the same distance. At 90° (vertical drop), you get \sin 90° = 1 and gravity does its maximum work mgd. At 0° (flat ground), \sin 0° = 0 and gravity does no work at all.

Work as energy transfer

Work is not just a formula — it is the mechanism by which energy moves from one form to another. When gravity does positive work on a falling coconut, gravitational potential energy converts into kinetic energy. When friction does negative work on a sliding cricket ball, kinetic energy converts into thermal energy. When you pull a suitcase, chemical energy in your muscles converts into kinetic energy of the suitcase (via the positive work of your pull) and thermal energy (via the negative work of friction).

The total work done by all forces on an object equals the change in its kinetic energy. This is the work-energy theorem, which you will derive in a later article. For now, the key insight is:

Positive work = energy flows into the object's motion (it speeds up)
Negative work = energy flows out of the object's motion (it slows down)
Zero work = the force does not change the object's speed

This is why the coolie at the railway station does zero work on the suitcase while walking horizontally. His upward force supports the suitcase against gravity, but it does not speed up or slow down the suitcase's horizontal motion. His muscles burn energy — they get tired — but that energy goes into maintaining posture and overcoming internal physiological processes, not into changing the suitcase's kinetic energy.

Common confusions

"Work means effort." In everyday language, holding a heavy bag while standing still is "hard work." In physics, the work done on the bag is zero — there is no displacement. Your muscles do internal work (they contract and relax repeatedly to maintain tension), which is why you get tired. But the work done on the bag by your hands is zero.
"A force always does work." A force does work only if the object moves and the force has a component along the displacement. The normal force on a book resting on a table exerts a large force but does zero work because there is no displacement. A satellite in circular orbit has gravity acting on it constantly, but the force is always perpendicular to the velocity, and the work done by gravity per orbit is zero — the satellite neither speeds up nor slows down.
"Negative work means something went wrong." Negative work is not an error or a failure. It means the force opposes the motion. Brakes on an autorickshaw do negative work — that is exactly their purpose. Negative work by friction is what stops a cricket ball on the pitch. Negative work removes kinetic energy, and that is often precisely what you want.
"You need to memorise which forces do positive or negative work." Apply W = Fd\cos\theta every time. If \theta < 90°, the work is positive. If \theta > 90°, the work is negative. If \theta = 90°, the work is zero. The formula handles all cases.
"Work done by gravity is always mgh." The magnitude is mgh where h is the vertical height change, but the sign depends on direction. Gravity does positive work (+mgh) when the object moves downward (falls, slides down a ramp) and negative work (-mgh) when the object moves upward (thrown up, climbs stairs). Always check whether gravity helps or opposes the motion.

If you came here to understand what work is, compute it for constant forces, and solve problems involving gravity, friction, and applied forces, you have everything you need. What follows is for readers who want the vector formulation and the concept of net work.

Work as a dot product

The formula W = Fd\cos\theta is the definition of the dot product (also called the scalar product) of the force and displacement vectors:

W = \vec{F} \cdot \vec{d} = Fd\cos\theta

If you know the force and displacement in component form — \vec{F} = F_x\hat{i} + F_y\hat{j} and \vec{d} = d_x\hat{i} + d_y\hat{j} — then the work is:

W = F_x d_x + F_y d_y

Why: the dot product in component form multiplies matching components and adds. This is equivalent to Fd\cos\theta but is often easier to compute, because you do not need to find the angle explicitly — you just need the components.

For example, if \vec{F} = (30\hat{i} + 40\hat{j}) N and \vec{d} = (5\hat{i} + 0\hat{j}) m, then:

W = 30 \times 5 + 40 \times 0 = 150 \text{ J}

The y-component of force (40 N upward) contributes nothing because the displacement has no y-component. This is the component method in action — the perpendicular part of the force does zero work.

Net work and the superposition principle

When multiple forces act on an object, each force does its own work independently. The net work is the sum:

W_{\text{net}} = W_1 + W_2 + W_3 + \cdots = \sum_i \vec{F}_i \cdot \vec{d}

Equivalently, you can compute the net force first and then find its work:

W_{\text{net}} = \vec{F}_{\text{net}} \cdot \vec{d} = \left(\sum_i \vec{F}_i\right) \cdot \vec{d}

Both methods give the same answer. The first (adding individual works) is useful when you want to track how much energy each force contributes or removes. The second (net force first) is useful when you only care about the total effect on the object's motion.

In Example 1 above, the net work was 1299 - 623 = 676 J. This net work equals the change in the suitcase's kinetic energy — a fact you will prove rigorously in the article on the work-energy theorem.

Why work is a scalar

Force is a vector. Displacement is a vector. But work is a scalar — a single number with a sign. The dot product extracts this scalar from two vectors, and the scalar has a clear physical meaning: it measures how much the force contributes to moving the object in the direction it is moving.

This is why work can be negative. Negative work is not "less than nothing" — it is the dot product of two vectors that point in roughly opposite directions (\theta > 90°). The sign encodes the direction of energy flow: positive means energy goes into the object's motion, negative means energy comes out.

When the formula breaks down

The formula W = Fd\cos\theta requires three conditions: (1) the force is constant in magnitude and direction over the entire displacement, (2) the displacement is a straight line, and (3) the force acts on a rigid body (no deformation). When any of these fail — a spring whose force increases with stretch, a ball moving along a curve, a gas being compressed — you need the more general formula:

W = \int_{\text{path}} \vec{F} \cdot d\vec{s}

This is the subject of the next article, Work Done by Variable Forces.

Where this leads next

Work Done by Variable Forces — what happens when the force changes as the object moves, and why you need integration to compute the work.
Kinetic Energy — the energy an object has because of its motion, and why it equals \frac{1}{2}mv^2.
Work-Energy Theorem — the proof that net work equals change in kinetic energy, connecting force, motion, and energy.
Potential Energy — how gravity and springs store energy, and the connection between conservative forces and potential energy.
Scalar Product of Vectors — the mathematical tool behind the work formula, with geometric and algebraic perspectives.