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A force-displacement graph question gives you a shape, not a formula, and asks for work or a final speed. Students who know W = Fd for a constant force often stall here, or get the area right but miss a sign. This post shows how to read the area properly, with two worked examples and the mistakes examiners look for.
Why is work the area under the graph?
NCERT Class 11 Chapter 5, Work, Energy and Power, section 5.5, splits the displacement into small steps Δx. Over each step the force is nearly constant, so the work is F(x)Δx, a thin rectangle. Adding the rectangles and letting Δx shrink gives the area under the curve, which "is exactly equal to the work done by F(x)". The book's worked example adds that "the area on the negative side of the force axis has a negative sign".
The JEE (Main) 2026 syllabus lists "Work done by a constant force and a variable force" and the CBSE 2026-27 curriculum lists the same in Chapter 5. Check your year's syllabus.
Why do students find area so hard?
Nguyen and Rebello (2011) interviewed students in a calculus-based physics course. Only a few recognised that area under a curve applied to the physics problems, and although all could say that the integral equals the area under the curve, they struggled to pick the graph whose area matched a given integral. A study by Susac and colleagues (2018) found students solved slope questions better than area questions, and concluded that area under a graph is a difficult concept that seems unlikely to develop on its own. In short: practise it directly. Our guide to reading physics graphs covers slopes; this post covers areas.
Worked example 1: a block pushed by a changing force
A 2.0 kg block starts at rest on a smooth floor. The force on it along x is:
from x = 0 to 2 m: rises steadily from 0 to 10 N;
from 2 to 5 m: stays at 10 N;
from 5 to 6 m: falls steadily from 10 N to 0;
from 6 to 8 m: stays at −5 N.
Split the graph into shapes:
Triangle, 0 to 2 m: ½ × 2 × 10 = 10 J.
Rectangle, 2 to 5 m: 3 × 10 = 30 J.
Triangle, 5 to 6 m: ½ × 1 × 10 = 5 J.
Rectangle below the axis, 6 to 8 m: 2 × (−5) = −10 J.
Total work from 0 to 8 m: 10 + 30 + 5 − 10 = 35 J.
Now the speeds. By the work-energy theorem, the work equals the change in kinetic energy.
At x = 6 m: ½ × 2.0 × v² = 45 J, so v = √45 ≈ 6.7 m/s.
At x = 8 m: ½ × 2.0 × v² = 35 J, so v = √35 ≈ 5.9 m/s.
The speed is greatest at x = 6 m, where the force changes sign. Before that the force speeds the block up; after it, the force slows it down.
Worked example 2: stretching a spring further
A spring with k = 400 N/m is already stretched by 2.0 cm. How much work must you do to stretch it to 5.0 cm?
The force you apply is kx, a straight line through the origin. The work is the area of the trapezium between x = 0.020 m and x = 0.050 m:
Forces at the ends: 400 × 0.020 = 8.0 N and 400 × 0.050 = 20 N.
Area = average force × width = ½(8.0 + 20) × 0.030 = 14 × 0.030 = 0.42 J.
Check with the formula: ½k(x2² − x1²) = 200 × (0.0025 − 0.0004) = 200 × 0.0021 = 0.42 J. The spring force does −0.42 J over the same stretch, matching NCERT's result that the spring's own work is negative when it is being stretched.
A common slip is ½k(x2 − x1)² = 200 × (0.030)² = 0.18 J. That is the area of a triangle starting at the 2 cm mark, which is not what the graph shows. For energy problems built on this, see our guide to spring-block systems.
Which mistakes cost marks?
Adding sizes instead of signed areas. Example 1 would give 55 J instead of 35 J.
Mixing units on the axes. If x is in cm, convert to m before computing areas in joules.
Using ½k(x2 − x1)² for a spring that is already stretched. Use the trapezium or ½k(x2² − x1²).
Counting grid squares without the scale. Find the work represented by one square first.
Reading a force-time graph as if it were force-displacement. Area under F-t is impulse, not work.
A practice routine
Draw three F-x graphs of your own with one section below the axis, and find the total work.
For each, find the speed at two points using the work-energy theorem.
Do two spring problems that start from a stretched position.
Finish with one past-paper question and write the shapes you split it into before calculating.
When the shapes are clear but the answer still does not match, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Next step: redo example 1 with the last section at −10 N and find where the block stops.
