Work Done by a Constant Force (NCERT 5.3)
1. What "Work" Means in Physics
In everyday speech, holding a heavy suitcase is exhausting work; in physics it is zero work, because nothing moves. The chapter's first discipline is this redefinition: work is done only when a force succeeds in displacing its point of application along some component of itself. The precise measure is the dot product of force and displacement:
W = F · s = F s cos θ
where θ is the angle between the force vector and the displacement vector. The definition is lean but loaded: it prices a force's effort by how much of it points along the motion. A perfect pull delivers F s; a perpendicular push delivers nothing; a opposing drag delivers negative work — it takes energy out. Master the angle and the sign, and half of this chapter's MCQs answer themselves.
2. Complete Theory: Dot Product, Units, and the Sign Discipline
Why a dot product. Work must reduce two vectors to one number, and the scalar (dot) product is the operation built for exactly that: it keeps the component of F along s and discards the rest. This is also why work is a scalar — a signed number, not a direction — which later lets the works of several forces simply add with no geometry at all. Dimensions: [W] = M L² T⁻², unit joule (1 J = 1 N m), CGS unit erg with 1 J = 10⁷ erg.
The sign table — the real content of NCERT 5.3. For 0° ≤ θ < 90°, cos θ > 0 and work is positive (the force helps the motion: gravity on a falling body, the rope on a sledge). At θ = 90°, work is zero — the zero-work trio: no force (F = 0), no displacement (s = 0, the held suitcase), or force ⊥ displacement (normal force on a level floor, gravity on level walking, the centripetal force in uniform circular motion at every instant). For 90° < θ ≤ 180°, work is negative (the force fights the motion: kinetic friction on a sliding block, gravity on a rising ball).
Work by the resultant = sum of the works. Since each force's work is a scalar, the net work is plain arithmetic: Wnet = W₁ + W₂ + …, each with its own sign. This line connects directly to the work-energy theorem, and it turns every multi-force problem into a bookkeeping exercise: applied force delivers, gravity plus-or-drops mgh per level change, normal typically idles, friction bills −f·s.
Frame dependence (the fine print). Work depends on the frame because displacement does. In the frame of a moving train, the seat's static friction does positive work on a suitcase riding along — in the platform frame the suitcase barely moves and the work is nearly zero. The fix is procedural, not conceptual: pick one inertial frame, do the whole audit in it, and never mix.
3. Visualising the Angle That Decides the Work
Figure 5.2 — Only the component along s bills any work
Blue dashed arrow: F cos 37° = 40 N, parallel to s — this part works. Orange dashed arrow: F sin 37° = 30 N, perpendicular to s — this part idles. Green arrow: displacement s = 10 m.
Exam read-out: the vertical 30 N does zero work — so does gravity and so does the normal — and the entire 400 J is billed by the horizontal component. NTA's incline and rope questions are this figure with different cosines; always decompose before you multiply.
4. Solved Examples
Example 1 — The rope at an angle
A block is dragged 10 m along a level floor by a rope pulled with 50 N at 37° above the horizontal. Find the work done by the rope.
Solution (step by step): θ between rope (force) and floor-displacement is 37°, so W = F s cos θ = 50 × 10 × cos 37° = 50 × 10 × 0.8 = 400 J. Cross-check by components: only F cos 37° = 40 N acts along s, giving 40 × 10 = 400 J ✓. Note what did not bill anything: the vertical 30 N component (⊥ s), gravity (⊥ s), and the normal (⊥ s) — three idlers that NTA loves to list as "workers" in wrong options.
Example 2 — The full multi-force audit
A 2 kg block is pulled 5 m across a rough floor by a 10 N force at 37° above the horizontal. Kinetic friction is 2 N. Find the work done by each force and the net work.
Solution: Pull: W = 10 × 5 × 0.8 = 40 J. Friction: W = −f s = −2 × 5 = −10 J. Gravity (20 N, ⊥ s): W = 0. Normal (⊥ s): W = 0. Net: W = 40 − 10 + 0 + 0 = 30 J. By the work-energy theorem, the block's kinetic energy grew exactly 30 J — from rest, ½ × 2 × v² = 30 → v ≈ 5.5 m/s. Six lines, one audit table in your head; no free-body algebra required.
5. Practice Questions
Q1. A porter stands for 10 s with a 20 kg suitcase, then walks 5 m along a level platform holding it. Find the work done by the porter's upward force on the suitcase in each stage.
ANSWER: Zero in both stages. Standing: s = 0. Walking: the porter's force is vertically up, the displacement is horizontal, θ = 90° → W = 0. The force fights gravity the whole time (keeping the suitcase up against mg), but "holding a force up" is not "doing work on" — no displacement along the force, no work.
Q2 (MCQ). A bead moves in a circle on a smooth wire, with the normal force from the wire acting radially. The work done by the normal force over half a revolution is: (a) positive (b) negative (c) zero (d) depends on the radius
ANSWER: (c) — the normal is radial and the velocity (hence the tiny displacement) is tangential at every instant: θ = 90° throughout, W = 0 over any arc length, half-circle or full. Same logic as uniform circular motion's zero-work result.
Q3. A 2 kg body falls freely through 5 m, is caught, and thrown back up through the same 5 m. Find the work done by gravity in the descent, the ascent, and the round trip (g = 10 m s⁻²).
ANSWER: Descent: gravity is along the displacement → W = +mgh = 2 × 10 × 5 = +100 J. Ascent: gravity opposes → W = −100 J. Round trip: 100 − 100 = 0 — gravity is a conservative force, and this round-trip cancellation is the seed of the potential-energy idea two pages ahead.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| W = F s cos θ | Constant force, straight-line displacement; θ between F and s |
| 0° ≤ θ < 90° → W > 0 | Force helps motion: gravity on a fall, rope on a sledge |
| θ = 90° → W = 0 | Normal on level floor, gravity on level walking, centripetal force, held load |
| 90° < θ ≤ 180° → W < 0 | Force fights motion: kinetic friction, gravity on a rise |
| Wnet = ΣWi | Scalar addition — the work-energy theorem's fuel |
| Wgravity = ± mgh | Sign by level change; zero over any round trip on level ground |
| 1 J = 10⁷ erg | [W] = M L² T⁻²; SI joule, CGS erg |
Every work computation is three lookups: which force, which displacement, which angle between them. Get θ from the geometry (not from habit), attach the sign, and let the scalars add.
7. Frequently Asked Questions
Define work in physics and give its SI unit.
Work is the scalar product of the force applied and the displacement produced: W = F·s = F s cosθ, where θ is the angle between the force and the displacement. Its SI unit is the joule (J), with 1 J = 1 N m = 1 kg m² s⁻²; the CGS unit is the erg, and 1 J = 10⁷ erg. Work is a signed scalar: positive when the force has a component along the displacement, negative when it opposes it.
When is the work done by a force zero?
Three cases, the zero-work trio: (1) the force itself is zero; (2) the displacement is zero — holding a load overhead involves effort but no work in the physics sense; (3) the force is perpendicular to the displacement (θ = 90°), as with the normal reaction on a level floor, gravity on level walking, or the centripetal force in uniform circular motion. NTA's favourite is the third case, dressed up as a moving bead on a wire or a satellite in circular orbit.
Can the work done by friction be positive?
Yes — the sign of friction's work depends on the displacement of the body it acts on, not on a universal rule. When a box rests on the accelerating bed of a truck and is carried forward, static friction acts forward along the box's displacement and does positive work. On a sliding block kinetic friction usually opposes the slide and does negative work, and on a block pressed against a wall it may do zero work if there is no vertical displacement. Friction converts mechanical energy to heat at the rate of |f| × relative sliding distance, but its work sign is situational.
Why is work a scalar although force and displacement are vectors?
Because the dot product collapses the vector information into a single number: W = F·s keeps only the component of force along the displacement. Directions matter only through cosθ — that is why works from different forces simply add algebraically with no geometry, why the work-energy theorem is a scalar equation, and why reversing the displacement (W = −F s for an opposing force) shows up as a sign rather than a new direction. Energy bookkeeping is scalar bookkeeping; that is its power and also why it cannot by itself tell you which way things moved.
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