Work done by a constant force

Push a stalled car and it does not budge — you strain, you sweat, but physics says you have done zero work on the car, because it has not moved. Push it again and it rolls forward five metres, and now you have done real, measurable work, whether you pushed dead level or at a slant. The physics definition of "work" is stricter and stranger than the everyday word: it depends not just on how hard you push but on the angle between your push and the resulting displacement, and it can even come out negative. This idea opens Unit 4, Work, Energy and Power, of ISC Class 11 Physics. You are expected to define work done by a constant force as the dot product of force and displacement, compute it numerically — including when the force acts at an angle — and judge by inspection whether a given force does positive, negative, or zero work. This article builds that definition from the vector dot product and works through the standard cases (force along, across, and against the motion), extending the Pass-1 examples into a fuller treatment of signs, components, and the traps examiners like to set.

The core insight

Work done by a constant force is the scalar quantity W=F⃗⋅S⃗=FScos⁡θW = \vec F \cdot \vec S = FS\cos\theta, positive when the force helps the displacement, zero when it is perpendicular to it, and negative when it opposes it.

The core idea

Definition. When a constant force F⃗\vec F acts on a body while it undergoes a displacement S⃗\vec S, the work done by that force is defined as the scalar (dot) product of the two vectors:

W=F⃗⋅S⃗=FScos⁡θW = \vec F \cdot \vec S = FS\cos\theta

where: FF is the magnitude of the force, SS is the magnitude of the displacement, and θ\theta is the angle between the direction of F⃗\vec F and the direction of S⃗\vec S. Because a dot product of two vectors is always a scalar, work itself is a scalar — it has magnitude but no direction, unlike force or displacement. Its SI unit is the joule, 1 J=1 N⋅m=1 kg⋅m2⋅s−21\ \text{J} = 1\ \text{N}\cdot\text{m} = 1\ \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}.

Reading the angle as a component. An equivalent way to see the formula is that only the component of force along the displacement contributes to work:

W=(Fcos⁡θ) SW = (F\cos\theta)\,S

where: Fcos⁡θF\cos\theta is the component of F⃗\vec F resolved along the direction of S⃗\vec S. Any part of the force perpendicular to the motion — for example, the sideways pull of a string tilted above the horizontal — does not move the body in its own direction and so contributes nothing to the work.

The three sign regimes. These follow directly from cos⁡θ\cos\theta:

  • θ=0°\theta = 0°: force and displacement point the same way, cos⁡θ=1\cos\theta = 1, so W=FSW = FS — the maximum possible positive work for that force and displacement.
  • θ=90°\theta = 90°: force is perpendicular to displacement, cos⁡θ=0\cos\theta = 0, so W=0W = 0 regardless of how large the force is. A person carrying a bag at constant height across a level floor does zero work against gravity on that stretch, because gravity acts vertically while the walk is horizontal; the normal reaction of a floor on a sliding block is another everyday zero-work force.
  • 90°<θ≤180°90° < \theta \le 180°: cos⁡θ\cos\theta is negative, so W<0W < 0 — the force has a component that opposes the motion. Friction resisting a sliding block, or any braking force that decelerates a moving body, does negative work on it.

Several forces at once. When two or more constant forces act on a body simultaneously, the total work done on it is the sum of the works done by each force separately, and this sum equals the work done by the single resultant (net) force over the same displacement.

Figure
A block pulled by a taut string at angle theta above the horizontal, with the force vector resolved into a horizontal component F cos theta that does work and a vertical component F sin theta that does no work along the horizontal displacement S
Only the component of force along the displacement does work block F theta F cos theta (does work) F sin theta (no work) displacement S
Derivation AWork done by several forces equals work done by their resultant

The goal is to show, using only vector algebra, that when constant forces F⃗1,F⃗2,…\vec F_1, \vec F_2, \dots act together on a body through the same displacement S⃗\vec S, the sum of the individual works equals the single work done by their vector sum, the resultant force — this is the fact used silently whenever a problem adds up "work by applied force" and "work by friction" and expects the total to match the work of the net force.

Step 1. Write the total work as the sum of the individual dot products. By definition, Wtotal=W1+W2+⋯=F⃗1⋅S⃗+F⃗2⋅S⃗+⋯W_{\text{total}} = W_1 + W_2 + \cdots = \vec F_1 \cdot \vec S + \vec F_2 \cdot \vec S + \cdots, one term for each force acting through the same displacement S⃗\vec S.

Step 2. Apply the distributive property of the dot product: a⃗⋅c⃗+b⃗⋅c⃗=(a⃗+b⃗)⋅c⃗\vec a \cdot \vec c + \vec b \cdot \vec c = (\vec a + \vec b)\cdot \vec c, exactly as ordinary multiplication distributes over addition of numbers. Applying this repeatedly gives Wtotal=(F⃗1+F⃗2+⋯ )⋅S⃗W_{\text{total}} = (\vec F_1 + \vec F_2 + \cdots)\cdot \vec S.

Step 3. Identify the resultant. The sum F⃗1+F⃗2+⋯\vec F_1 + \vec F_2 + \cdots is, by definition, the resultant (net) force F⃗net\vec F_{\text{net}} on the body, so Wtotal=F⃗net⋅S⃗W_{\text{total}} = \vec F_{\text{net}} \cdot \vec S.

Step 4. Read the result. It does not matter whether you add up the works of individual forces or first find the resultant and take one dot product — both routes give the same total. The worked example below checks this numerically: 240 J240\ \text{J} (applied) +(−240 J)+ (-240\ \text{J}) (friction) +0 J+0 J+ 0\ \text{J} + 0\ \text{J} (gravity, normal) =0 J= 0\ \text{J}, matching zero net force at constant velocity.

Worked example

A block of mass 10 kg10\ \text{kg} is pushed 8 m8\ \text{m} along a rough horizontal floor at constant velocity by a horizontal force of 30 N30\ \text{N}. Find the work done by the applied force, the work done by friction, the work done by gravity and the normal reaction, and confirm the total is consistent with the block's unchanging speed.

Step 1. Find the work done by the applied force. The push is horizontal and the displacement is horizontal, so the angle between them is θ=0°\theta = 0°: Wapplied=FScos⁡0°=30×8×1=240 JW_{\text{applied}} = FS\cos 0° = 30 \times 8 \times 1 = 240\ \text{J}.

Step 2. Find the friction force, then its work. Constant velocity means the net force on the block is zero, so friction must exactly balance the 30 N30\ \text{N} applied force: f=30 Nf = 30\ \text{N}. Friction acts opposite to the motion, so θ=180°\theta = 180°: Wfriction=fScos⁡180°=30×8×(−1)=−240 JW_{\text{friction}} = fS\cos 180° = 30 \times 8 \times (-1) = -240\ \text{J}.

Step 3. Find the work done by gravity and by the normal reaction. Gravity (mgmg, downward) and the normal reaction (NN, upward) both act perpendicular to the horizontal displacement, so θ=90°\theta = 90° for each: Wgravity=Wnormal=mgScos⁡90°=0 JW_{\text{gravity}} = W_{\text{normal}} = mgS\cos 90° = 0\ \text{J}.

Step 4. Add up and check consistency. Wtotal=240+(−240)+0+0=0 JW_{\text{total}} = 240 + (-240) + 0 + 0 = 0\ \text{J}. The applied force does +240 J+240\ \text{J}, friction does −240 J-240\ \text{J}, gravity and the normal reaction each do 0 J0\ \text{J}, and the net work on the block is 0 J0\ \text{J} — matching the fact that its kinetic energy does not change while it moves at constant velocity, a preview of the work–energy theorem.

Competitive stretch

Now tilt the pull: a boy pulls a 20 N20\ \text{N} string at 60°60° above the horizontal, moving the cart 5 m5\ \text{m} horizontally. Directly, W=FScos⁡θ=20×5×cos⁡60°=20×5×0.5=50 JW = FS\cos\theta = 20 \times 5 \times \cos 60° = 20 \times 5 \times 0.5 = 50\ \text{J}. Using the component view instead: the horizontal component of the pull is Fcos⁡60°=10 NF\cos 60° = 10\ \text{N}, so W=10×5=50 JW = 10 \times 5 = 50\ \text{J} — the same answer, confirming that resolving the force and using the angle directly are the same calculation. Had the string been horizontal (θ=0°\theta = 0°), the work would have been the larger value 100 J100\ \text{J}; tilting the pull away from the direction of motion always reduces work, never increases it.

Common traps

  • Measuring θ\theta as the angle a force makes with some reference line (like the vertical or a nearby surface) instead of the actual angle between the force vector and the displacement vector.
  • Assuming work is always positive. Work is negative whenever the force has a component opposing the displacement — the commonest source of sign errors with friction and other retarding forces.
  • Concluding a large force must do a large amount of work. A force perpendicular to the displacement — the normal reaction on a level floor, the tension in circular motion — does exactly zero work no matter how large it is.
  • Mixing up "work done by a force" with "work done against a force." The two have equal magnitude and opposite sign; a question asking for one is not answered by quoting the other.

Ask the tutor: Ask the AI tutor to walk through a few more angled-force examples, including an inclined plane, so you can practise picking the correct theta every time.

If I push on a wall as hard as I can and nothing moves, have I really done zero work, even though I am exhausted afterwards?

Yes, in the physics sense, the work done on the wall is exactly zero, because the displacement SS in W=FScos⁡θW = FS\cos\theta is zero — no motion, no work, no matter how large the force. Your tiredness comes from continuous, costly contractions inside your muscles, a biological process that consumes chemical energy even with no external displacement, which is a completely different kind of "work" from the mechanical definition used here. The two uses of the word measure different things, and physics problems always mean the mechanical one.

Why does a force perpendicular to the displacement do zero work instead of some small positive amount?

Because cos⁡90°=0\cos 90° = 0 exactly, not approximately, so the formula gives zero regardless of how strong the force is. Physically, a perpendicular force changes only the direction the body would tend to move in, never how far it actually travels along its existing path — think of the normal reaction of a floor holding up a sliding block, or the string tension in circular motion pulling toward the centre while the object moves along the circle's edge. Neither force speeds the body up or slows it down along its path, which is exactly what "zero work" captures.

When a problem gives me a force and a displacement in different directions, how do I decide the correct theta to plug into W equals FS cos theta?

Draw both vectors starting from the same point — sketch the force and the displacement as two arrows from the object's position — and read off the angle between those two arrows; that is the theta you need, regardless of any other angle mentioned in the problem (like a slope's incline or a string's angle to a wall). A common shortcut is to resolve the force into a component along the displacement and a component perpendicular to it; only the along-displacement component, Fcos⁡θF\cos\theta, ever contributes to the work, which is the same number you would get from the full formula.

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Where this leads