Work

A weightlifter braces under a barbell loaded to the limit, arms trembling, and holds it dead still above her head for three seconds before the judges' lights flash white. Every muscle fibre is screaming, sweat is running, and by the end of those three seconds she is exhausted — yet in the strict language of physics, she has done zero work on the barbell. No displacement, no work, no matter how hard the effort feels. Meanwhile, a child kicking a football does something far less strenuous, yet the instant the boot connects and the ball rockets forward, real work has been done — energy has visibly flowed from boot to ball. This gap between the everyday meaning of "work" and its physics meaning is exactly where this chapter begins. Work, in mechanics, is not about effort or tiredness; it is a precise bookkeeping device for how a force moves energy from one place to another. It only counts when a force actually produces a displacement, and how much it counts depends sharply on the angle between the force and that displacement — full credit when they point the same way, zero credit when they are perpendicular, and negative credit when the force fights the motion. This article builds that definition from scratch: work by a constant force as a dot product, work by a force that changes as an object moves (using the area under a force-position graph), the signed nature of work, and the classic traps CBSE loves to set — the coolie who carries a load and gets tired without doing work on it, the normal reaction that never does work on a surface-constrained object, and the string tension in a pendulum that is always perpendicular to the bob's velocity. Get this single idea watertight, and kinetic energy, the work-energy theorem and power all fall into place naturally in the chapters that follow.

Key takeaways

Work is a scalar, defined for a constant force as $W = \vec F\cdot\vec d = Fd\cos\theta$ , where $\theta$ is the angle between the force and the displacement.
Three special angles decide everything: $\theta = 0°$ gives maximum positive work $Fd$ ; $\theta = 90°$ gives zero work; $\theta = 180°$ gives maximum negative work $-Fd$ .
Effort is not work. A force with no displacement (holding a weight still) or a force entirely perpendicular to the displacement (carrying a load horizontally, a string's tension on a swinging bob) does zero work, however tiring it feels.
Work can be negative. Friction, air resistance, and any force with a component opposing the motion subtract energy, not just fail to add it.
Variable forces need calculus-in-disguise. When force changes with position, work is the area under the force-versus-position graph, $W = \int F\,dx$ — which reduces exactly to $Fd\cos\theta$ when the force happens to be constant.
The SI unit is the joule ( $1\ \text{J} = 1\ \text{N}\,\cdot\,\text{m}$ ), with dimensional formula $[ML^2T^{-2}]$ — identical to energy, because work is energy in transit.
Key idea: work is the signed scalar $W = Fd\cos\theta$ (or the area under an $F$ – $x$ graph for a changing force) measuring how much of a force's push actually lands along the direction an object moves, zero when force and displacement are perpendicular and negative when they oppose.

The core insight

Work is the component of a force along an object's displacement, multiplied by that displacement — $W = Fd\cos\theta$ for a constant force, or the area under the $F$ – $x$ graph when the force varies with position.

Where it fits in the syllabus

This topic opens the CBSE Class 11 unit Work, Energy and Power, immediately after you finish the Laws of Motion. It is the natural next question once you know how to compute a force: what does that force actually do to a moving object? The prescribed scope covers work by a constant force as the dot product $W = \vec F\cdot\vec d = Fd\cos\theta$ , and work by a variable force as the area under an $F$ – $x$ graph, $W = \int F\,dx$ . Both forms are examinable, and CBSE frequently tests the variable-force idea using a simple graph rather than formal integration, so you should be comfortable reading area off a graph as much as plugging numbers into the cosine formula. This chapter's next nodes build kinetic energy and the work-energy theorem directly on top of the definition given here.

The physical picture

Picture a block being dragged across a floor by a rope pulled at some angle above the horizontal, as in the figure below. The tension $\vec F$ has two effects: a component along the floor that actually drags the block forward, and a component pointing upward that partly lifts the block off the floor but contributes nothing to the horizontal displacement. Only the first component does work. For a constant force $\vec F$ producing a displacement $\vec d$ , work is defined as the scalar (dot) product of the two vectors:

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

where: $F$ is the magnitude of the force (newtons), $d$ is the magnitude of the displacement (metres), and $\theta$ is the angle measured between the force vector and the displacement vector — never confuse this with any other angle drawn in the diagram, such as the slope of an incline. Because a dot product of two vectors yields a number with no direction attached, work is a scalar: it can be positive, negative, or zero, but it never "points" anywhere.

Figure

A block on a floor being pulled by a rope at an angle above the horizontal, showing the force vector split into a horizontal component that does work and a vertical component that does no work since the block only moves horizontally

Three special cases are worth memorising cold, because they are the fastest CBSE and JEE trap:

$\theta = 0°$ : force fully along displacement, $W = Fd$ , the maximum possible positive work.
$\theta = 90°$ : force perpendicular to displacement, $W = 0$ , however large the force — this is why the normal reaction on a block sliding along a surface, and the tension in a string swinging a bob in a circle, never do any work.
$\theta = 180°$ : force fully opposed to displacement, $W = -Fd$ , the maximum possible negative work — this is how friction and air resistance drain energy from a moving object.

When the force is not constant — a spring's restoring force growing with stretch, or a variable applied force — you cannot simply plug numbers into $Fd\cos\theta$ , because $F$ itself keeps changing as $x$ changes. The fix is to imagine chopping the total displacement into many tiny steps $\Delta x$ , so small that $F$ is essentially constant over each one, compute the tiny work $F\,\Delta x$ for each step, and add them all up. In the limit of infinitesimally small steps this sum becomes an integral, giving the work as the area under the force-versus-position graph:

W = \int_{x_1}^{x_2} F\,dx = \text{area under the } F\text{–}x \text{ graph between } x_1 \text{ and } x_2

where: $F(x)$ is the force as a function of position along the direction of motion, and $x_1,\,x_2$ mark the start and end of the displacement. For a graph that happens to be a horizontal straight line (constant force), this area is just a rectangle of height $F$ and width $d$ , recovering $W = Fd$ exactly — so the constant-force formula is simply the special, simplest case of this more general area rule. The SI unit of work is the joule (J), where $1\ \text{J} = 1\ \text{N}\,\cdot\,\text{m} = 1\ \text{kg}\,\cdot\,\text{m}^2\text{s}^{-2}$ , giving the dimensional formula $[ML^2T^{-2}]$ — identical to the dimensional formula of energy, because work and energy are the same physical quantity viewed from two ends of a transfer: work is energy leaving one system and entering another. A subtlety worth internalising early: work is always done by a specific force, never by "the object" in some vague sense. If several forces act on a body at once, you compute the work done by each one separately using its own angle with the displacement, and the total (net) work is simply the algebraic sum of these individual works — equivalently, it equals the work done by the single resultant (net) force.

Derivation AWork by a constant force from the general variable-force rule

The goal here is to show that the familiar constant-force formula $W = Fd\cos\theta$ is not a separate law you must memorise alongside the area-under-the-graph rule for variable forces — it is the same idea, specialised to the case where the force happens to stay fixed in magnitude and direction throughout the motion. Seeing this connection means you never have to worry about which formula "applies" to a given problem; you can always fall back on summing up tiny bits of work and recover whichever form the situation calls for.

Step 1. Split the total displacement $d$ into $N$ tiny equal steps of size $\Delta x = d/N$ , each one so small that, over that single step, the force can be treated as constant. Along each step, the small work done is $\Delta W = F\cos\theta\,\Delta x$ , where $F\cos\theta$ is the component of the force along the direction of motion (this is exactly the projection idea used to define the dot product in the first place).

Step 2. Because the force is genuinely constant over the whole path in this case, $F$ and $\theta$ do not change from one tiny step to the next — every $\Delta W$ has the identical value $F\cos\theta\,\Delta x$ . Summing over all $N$ steps therefore just multiplies this one value by the number of steps: $W = \sum_{i=1}^{N} F\cos\theta\,\Delta x = N\,(F\cos\theta)\,\Delta x$ .

Step 3. Substitute $\Delta x = d/N$ back in. The $N$ in the numerator and denominator cancel exactly: $W = N\,(F\cos\theta)\,\dfrac{d}{N} = F\cos\theta\, d$ .

Step 4. Read the result. $W = Fd\cos\theta$ , precisely the constant-force formula, has emerged from the same summing process that gives $W = \int F\,dx$ for a genuinely varying force. The only difference between the two situations is whether $F\cos\theta$ stays fixed (so the "area" under the graph is a plain rectangle, height $F\cos\theta$ times width $d$ ) or changes with position (so the area has some other, more interesting shape). This is why a graph of force along the direction of motion, plotted against position, is the single unifying picture behind every work calculation in this chapter.

Worked examples

Board level - dragging a crate with a rope at an angle

A crate is pulled $12\ \text{m}$ across a horizontal floor by a rope that makes an angle of $30°$ with the floor. The tension in the rope is $150\ \text{N}$ . Find the work done by the tension.

Step 1. Identify the quantities from the statement: $F = 150\ \text{N}$ , $d = 12\ \text{m}$ , and $\theta = 30°$ , the angle between the rope (the force) and the floor (the direction of the displacement).

Step 2. Apply the constant-force formula directly: $W = Fd\cos\theta = 150 \times 12 \times \cos 30°$ . Since $\cos 30° \approx 0.866$ , this becomes $W = 1800 \times 0.866$ .

Step 3. Compute and round sensibly. $W = 1558.8\ \text{J} \approx 1.56\times10^{3}\ \text{J}$ .

The tension does about 1559 J of positive work on the crate. The horizontal component of the tension, $150\cos30° \approx 129.9\ \text{N}$ , acts in the same direction as the crate's motion, which is exactly why the work comes out positive; the vertical component of the tension (which would partly lift the crate) contributes nothing to this number because it has no component along the floor.

Competitive level - work by gravity, the normal force, and a position-dependent applied force

A $2\ \text{kg}$ block slides down a frictionless incline, dropping a vertical height of $3\ \text{m}$ while moving $5\ \text{m}$ along the slope. Take $g = 9.8\ \text{m/s}^2$ . Find (a) the work done by gravity, (b) the work done by the normal reaction, and (c) suppose separately a horizontal force pulls a different block along a straight track where the force varies with position as $F(x) = (10 - 2x)\ \text{N}$ for $x$ from $0$ to $5\ \text{m}$ — find the work done by this variable force over that stretch.

Step 1. Work done by gravity on the incline. Gravity acts vertically downward with magnitude $mg = 2 \times 9.8 = 19.6\ \text{N}$ . Since gravity is a constant force, only the net vertical displacement (the height dropped) matters — not the length of the slanted path — so $W_g = mgh = 19.6 \times 3 = 58.8\ \text{J}$ , positive because the block moves in the same sense that gravity pulls.

Step 2. Work done by the normal reaction on the incline. The normal force $N$ is always perpendicular to the incline's surface, while the block's displacement is along that surface, so $\theta = 90°$ between $N$ and the displacement. Hence $W_N = Nd\cos90° = 0$ : the normal force, however large it needs to be to keep the block on the surface, never does any work on an object constrained to move along that surface.

Step 3. Work by the variable force, part (c). Since $F(x) = (10-2x)\ \text{N}$ changes with position, use the area-under-the-graph rule: $W = \int_0^5 (10-2x)\,dx$ . This is the area of a trapezium under a straight line dropping from $F(0)=10\ \text{N}$ to $F(5)=0\ \text{N}$ over a width of $5\ \text{m}$ .

Step 4. Evaluate the area (or the integral) directly. Area of the trapezium $= \tfrac12(F_{\text{start}}+F_{\text{end}})\times \text{width} = \tfrac12(10+0)\times5 = 25\ \text{J}$ . Equivalently, integrating gives $W = \left[10x - x^2\right]_0^5 = (50-25)-(0-0) = 25\ \text{J}$ .

So: work by gravity is 58.8 J, work by the normal reaction is exactly 0 J, and the variable applied force does 25 J of work over the 5 m stretch. Timing note: In a 3-mark competitive question, do not waste time setting up the full integral with limits if the $F$ – $x$ graph is a straight line — recognise it instantly as a triangle or trapezium and compute the area geometrically; save the integral notation only for curved or piecewise graphs where geometry alone will not give the area.

Common traps

Believing that applying a force automatically means work is done. Without any displacement at all, or with the force entirely perpendicular to whatever displacement exists ( $\theta = 90°$ ), the work is exactly zero — no partial credit.
Forgetting that work is a signed quantity: any force with a component opposing the motion, such as friction, air resistance, or a retarding push, does negative work, not merely "no" work.
Confusing the angle $\theta$ in $W = Fd\cos\theta$ with some other angle drawn in the figure, such as the angle of an incline itself — always identify the angle strictly between the force vector and the displacement vector.
Treating a curved or slanted path as if it changes the work done by gravity — for gravity (a constant force), only the net vertical height dropped matters, regardless of the actual shape of the path taken.
Trying to force $W = Fd\cos\theta$ onto a problem where the force clearly changes with position (a spring, a graph that is not a horizontal line) — that formula only applies when $F$ is truly constant; otherwise you must find the area under the $F$ – $x$ graph.

Common trapThe normal reaction and any force perpendicular to the motion never do work, no matter how large they are

Students very often assume that the normal reaction from a floor or incline, or the tension in a string swinging a bob in a circle, must be "doing something" because it is large and clearly necessary to keep the object moving in that path. It does no work whatsoever, because it is always perpendicular to the object's instantaneous displacement, so $\cos 90° = 0$ kills the entire term. The same logic explains why a coolie carrying a load horizontally on his head does zero work against gravity even though he is exhausted at the end of the walk — the force (weight, supported vertically) and the displacement (horizontal) are at $90°$ to each other. His fatigue comes from internal muscular work maintaining posture, not from work done on the load in the physics sense.

You should be able to…

State the definition $W = \vec F\cdot\vec d = Fd\cos\theta$ for a constant force, and explain precisely why work is classified as a scalar despite being built from two vectors.
Identify by inspection whether a given force does zero, positive, or negative work on an object, using only the angle between the force and the displacement.
Explain, with at least one concrete example, why exerting a force is not the same thing as doing work in the physics sense.
Compute the work done by gravity on an object moving along an incline or any curved path, using only the net height dropped.
Compute the work done by a position-dependent force using the area under its $F$ – $x$ graph, for graphs made of straight-line segments.
State the SI unit and dimensional formula of work, and explain in one sentence why they match those of energy.

Ask the tutor: Still unsure why the normal force and tension in circular motion always do zero work even though they are clearly "doing something" to keep the object moving — ask the AI tutor to walk through the perpendicular-force argument with a live diagram.

If work is Fd cos theta, why does carrying a heavy bag horizontally tire me out even though the physics says zero work is done on the bag?

Because the "work" your muscles do to hold your arm rigid and keep the bag from sagging is internal biological work, not mechanical work done on the bag. Your muscle fibres are constantly contracting and relaxing at a microscopic level even while your arm appears still, burning chemical energy as heat — that is where the fatigue comes from. But from the bag's point of view, the force holding it up is vertical while its displacement (as you walk) is horizontal, so $\theta = 90°$ and $W = Fd\cos90° = 0$ on the bag itself. Physics-work and feeling-tired are simply measuring two different things.

Can work be negative, and what does negative work actually mean physically?

Yes. Negative work happens whenever a force has a component pointing opposite to the object's displacement, such as friction on a sliding block or air resistance on a falling ball. Physically, it means that force is removing mechanical energy from the object rather than adding it — the object slows down because of it. The work-energy theorem, coming next, shows the total net work on an object exactly equals its change in kinetic energy, so net negative work always means the object loses speed.

Why does gravity's work only depend on height and not on the actual path taken, like a slide versus a straight vertical drop?

Because gravity is a constant force pointing straight down, and work only cares about the component of displacement along the direction of the force. Whether an object falls straight down, slides down a curved ramp, or zig-zags its way down, its net vertical displacement (the height dropped) is the same, and that is the only piece of the motion gravity's work formula sees — any horizontal or sideways wandering contributes nothing because it is perpendicular to gravity. This special property, where the work done depends only on the start and end points and not on the path between them, is what makes gravity a "conservative" force, a term you will meet formally in the next few chapters.

How is the area-under-the-graph method for variable forces actually used in exams?

Most CBSE and competitive problems keep the $F$ – $x$ graph to shapes with elementary areas — straight lines forming triangles, trapeziums, or rectangles — precisely so you compute work with basic geometry instead of formal integration. Read the force values at the start and end positions, identify the shape enclosed between the graph, the $x$ -axis, and the two limits, and apply the standard area formula for that shape. True integration, using an explicit algebraic $F(x)$ , is reserved for problems that hand you a formula rather than a picture.