Work Done by a Variable Force (NCERT 5.5, 5.6)
1. Why the Constant-Force Formula Is Not Enough
W = F s cosθ quietly assumes the force stays the same while the body moves. Real forces rarely cooperate: a spring pushes harder the more you compress it, a bowstring stiffens as you draw it, a rising balloon feels changing drag. Drawing a bow through 30 cm with a force that grows from 0 to 90 N — what work went in? Multiplying by any single F is wrong; the honest answer must "add up" the force's delivery moment by moment. That adding-up is NCERT 5.5, and its geometric translation — work = area under the F–x graph — is one of the most reliable marks-per-minute skills in all of Class 11 physics. NTA examines it almost exclusively through graphs, which is good news: no calculus is needed, just rectangles and triangles.
2. Complete Theory: Strips, the Integral, and the Theorem for Variable Forces
The strip argument (learn it — it is the derivation NTA paraphrases). Slice the displacement from xi to xf into many small intervals Δx. Across one interval the force is nearly constant, so the work done there is ΔW ≈ F(x) Δx — a thin rectangle under the curve. Summing: W ≈ Σ Fi Δxi. As the strips get thinner the staircase of rectangles melts into the region under the curve, and the sum becomes the integral:
W = ∫xixf F dx = area under the F–x curve
Two graph-reading rules do all the exam work. First, shapes with known areas cover most NTA graphs: rectangle (F·d), triangle (½·base·height), trapezoid (½(F₁ + F₂)·d). Second, the area is signed: a force that turns around and opposes the motion drives the curve below the axis, and that area subtracts — the body speeds up while the force helps and slows while it resists.
The theorem's variable-force form (NCERT 5.6). Apply the work-energy theorem to one strip: dK = F dx. Integrating both sides,
Kf − Ki = ∫ F dx
— the change in kinetic energy equals the signed area. This is the practical engine of graph questions: read the area, set it equal to ΔK, solve for speed. Notice what vanished: no acceleration, no time, no force algebra — endpoints only.
The canonical variable force: the spring. Hooke's law from the force catalogue, F = kx, grows linearly with compression x — so its graph is a triangle and its work is ½·(kx)·x = ½kx². That single triangle is the entire derivation of spring potential energy, delivered two pages ahead on the spring page. The strip method, not a memorised formula, is what generalises.
3. Visualising: Reading an F–x Graph Like an Accountant
Figure 5.3 — Area = work: rectangle plus triangle
Blue region: constant 12 N from 0 to 2 m → area 12 × 2 = 24 J. Green region: linear fall from 12 N to 0 over the next 2 m → area ½ × 12 × 2 = 12 J. Total shaded area = 36 J of work.
Exam read-out: the dashed strip inside the rectangle is the whole derivation — one F Δx slice. For a 2 kg block starting from rest, ΔK = 36 J gives v = √(2 × 36/2) = 6 m/s exactly, no forces summed, no acceleration found. NTA's F–x questions are this figure with rescaled numbers.
4. Solved Examples
Example 1 — The linearly stiffening force
A force starts at zero and grows linearly to 100 N while its point of application moves 0.4 m. Find the work done.
Solution (step by step): Linear growth → the F–x graph is a triangle of base 0.4 m and height 100 N. W = ½ × base × height = ½ × 0.4 × 100 = 20 J. Strip cross-check: the average force over the stroke is (0 + 100)/2 = 50 N, and 50 × 0.4 = 20 J ✓. This is precisely the spring calculation in disguise — replace "100 N at 0.4 m" with "k = 250 N/m fully compressed 0.4 m" and the same 20 J reappears as ½kx².
Example 2 — Graph to speed, endpoints only
A 2 kg block at rest on a frictionless floor is pushed by the force of Figure 5.3: constant 12 N for 2 m, then linearly decreasing to zero over the next 2 m. Find its speed after 4 m.
Solution: Total work = area = 24 + 12 = 36 J. Work-energy theorem: ΔK = 36 J with Ki = 0 → ½ × 2 × v² = 36 → v² = 36 → v = 6 m/s. Sanity check: the force never opposed the motion, so the speed grew throughout, ending maximal where the force finally died. If a third segment dragged the force below the axis, that area would be subtracted before the ΔK step — watch for it.
5. Practice Questions
Q1. A crate is dragged with a constant 15 N through 6 m; the force then falls linearly to zero over the next 2 m. Find the total work.
ANSWER: Rectangle: 15 × 6 = 90 J. Triangle: ½ × 15 × 2 = 15 J. Total 105 J. Two shapes, two areas, one sum — and if a 5 N friction force had acted throughout the 8 m, subtract f·s = 40 J to get the net 65 J that would reach the kinetic-energy ledger.
Q2 (MCQ). A force–displacement graph lies entirely below the position axis over a 3 m interval with constant magnitude 8 N. The work done by this force is: (a) +24 J (b) −24 J (c) 0 (d) +8 J
ANSWER: (b) — area below the axis is negative work: W = −8 × 3 = −24 J. The body loses 24 J of kinetic energy while this force acts; the "below-axis" dip is the graph's way of writing a minus sign.
Q3. A spring-loaded toy is compressed: the resisting force grows linearly from 0 to 20 N over a 0.5 m compression. Find the work stored.
ANSWER: Triangle: W = ½ × 0.5 × 20 = 5 J — equivalently, average force (0 + 20)/2 = 10 N over 0.5 m. This 5 J is exactly the spring's potential energy ½kx² with k = F/x = 40 N/m and x = 0.5 m: ½ × 40 × 0.25 = 5 J ✓. The graph method and the spring formula are the same physics wearing two costumes.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| W = ∫ F dx | Variable force along x; equals the signed area under F–x |
| W ≈ Σ Fi Δxi | Strip approximation; thinner strips → exact |
| Rectangle / triangle / trapezoid | F·d / ½·base·height / ½(F₁ + F₂)·d — the NTA shape kit |
| Area below axis subtracts | Force opposing motion does negative work |
| Kf − Ki = ∫F dx | NCERT 5.6: work-energy theorem for variable force |
| Wspring (compressing) = ½kx² | Linear F = kx → triangular graph; stored as spring PE |
Same area logic, three graphs: F–x gives work, F–t gives impulse, v–t gives displacement. Label your axes before reading any area — the units of the area decide its meaning.
7. Frequently Asked Questions
How is work calculated when the force is not constant?
Split the displacement into many small intervals Δx. Over each one the force is nearly constant, so the small work is ΔW ≈ F Δx. Adding the strips gives W ≈ Σ F Δx, and letting the strips become infinitesimally thin turns the sum into the integral W = ∫F dx. Graphically this is the area between the F–x curve and the position axis. If the force varies in magnitude along the path, use the component along the displacement in the integral.
Why is the area under an F–x curve equal to the work done?
Because each strip of the graph is a thin rectangle of height F (the local force) and width Δx (the small displacement), so its area F Δx is exactly the work done over that strip. Summing the rectangles covers the whole curve, and in the limit of vanishing width the rectangle stack fills the region between the curve and the axis. The same area-under-a-graph logic gives displacement from a v–t graph and impulse from an F–t graph — NTA reuses all three. Areas below the axis carry a negative sign.
What does the work-energy theorem look like for a variable force?
In differential form the theorem reads dK = F dx: a tiny displacement changes kinetic energy by force times the tiny work. Integrating both sides gives K_f − K_i = ∫F dx — the change in kinetic energy equals the area under the F–x curve. This is NCERT section 5.6, and it is the practical route for spring launches and graph-based numericals: compute the area, set it equal to ΔK, and solve for the speed. No acceleration ever needs to be computed.
How do I handle an F–x graph that dips below the position axis?
Treat the graph as signed: area above the axis is positive work, area below is negative work, and the net work is their algebraic sum. A force that pushes forward over 2 m and then resists over the next 2 m does plus the first area minus the second. In kinetic-energy terms the body speeds up while the area is positive and slows while it is negative — so turning points in the motion appear exactly where the signed area stops growing, and a full stop means the net area from the start equals minus the initial kinetic energy.
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