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A chain of length L lies on a table with a length l hanging over the edge. How much work does it take to pull the hanging part back up slowly? The force you need changes at every moment, so W = Fd gives the wrong answer. It is a clean example of work done by a variable force and good practice in setting up an integral. This post builds the integral step by step, checks it a second way, then adds friction.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "work done by a constant force and a variable force" under Unit 4, Work, Energy and Power. The NCERT Class 11 chapter on work, energy and power covers this in Section 5.5: for a small displacement Δx the force is nearly constant, so ΔW = F(x)Δx, and adding these pieces as Δx goes to zero gives the area under the force-displacement curve. Check the syllabus for your own exam year.
Why does W = Fd fail here?
When a length y hangs over the edge, its weight is (M/L)gy, and pulling slowly means your force just balances it. As you pull, y shrinks, and so does the force. At the start you pull hardest; at the end you pull with almost no force at all. A single "F" does not exist, so you need the work integral, W = ∫F dx.
How do you set up the integral?
Use the same three questions every time.
What is the small piece? A small pull dy, during which a length dy of chain moves from hanging to lying on the table.
What is the force during that piece? When a length y hangs, F = (M/L)gy.
What are the limits? y goes from l (start) to 0 (end). Since you pull through a total distance l, it is simpler to add up from y = 0 to y = l.
W = ∫ from 0 to l of (M/L)gy dy = (M/L)g × l²/2 = Mgl²/2L.
Nguyen and Rebello (2011) found that most students they interviewed could recognise that a problem needed an integral but failed to set it up, often because they did not understand what the small term stood for. Saying step 1 in words is the fix.
Worked example 1: the numbers
A uniform chain of mass 4 kg and length 2 m lies on a smooth table with 0.5 m hanging. Take g = 10 m/s². Find the work to pull it all onto the table slowly.
W = Mgl²/2L = 4 × 10 × 0.5²/(2 × 2) = 40 × 0.25/4 = 2.5 J.
Check with the centre of mass. The hanging part has mass 4 × 0.5/2 = 1 kg, and its centre of mass is 0.25 m below the table. Raising it to table level needs 1 × 10 × 0.25 = 2.5 J. Same answer.
The centre-of-mass method is faster, but the integral is the one that works when the force is not just gravity.
Worked example 2: adding friction
Now the table is rough, with coefficient of kinetic friction 0.2. Find the total work to pull the chain on slowly.
When y hangs, the length on the table is L − y, with mass (M/L)(L − y). Friction on it is μ(M/L)g(L − y).
Your force must balance both: F = (M/L)gy + μ(M/L)g(L − y).
The gravity part gives 2.5 J, as before.
The friction part: ∫ from 0 to l of μ(M/L)g(L − y) dy = μ(M/L)g(Ll − l²/2).
Numbers: μ(M/L)g = 0.2 × 2 × 10 = 4 N/m, and Ll − l²/2 = 2 × 0.5 − 0.125 = 0.875 m². So the friction work is 4 × 0.875 = 3.5 J.
Total work = 2.5 + 3.5 = 6 J.
Here the centre-of-mass shortcut does not help with the friction part, because friction depends on how much chain lies on the table at each moment. That is why the integral method matters.
What if you let go instead?
Reverse the question: the chain starts at rest with 0.5 m hanging on a smooth table and is released. How fast is it moving when the last link leaves the table?
Take the table top as the zero of potential energy. At the start only the hanging 1 kg piece has potential energy: −1 × 10 × 0.25 = −2.5 J.
At the end the whole 4 kg chain hangs, with its centre of mass 1 m below the edge: −4 × 10 × 1 = −40 J.
Energy released: 40 − 2.5 = 37.5 J = ½ × 4 × v², so v² = 18.75 and v ≈ 4.33 m/s.
The general result is v = √(g(L² − l²)/L), which gives √(10 × 3.75/2) ≈ 4.33 m/s. Our post on the work-energy theorem has more problems of this kind.
Common mistakes and how to avoid them
Using W = Mgl or (M/L)gl × l, which forgets that the force falls as you pull.
Wrong centre of mass: l instead of l/2 below the edge.
Applying the centre-of-mass shortcut to friction. Friction needs its own integral.
Forgetting "slowly". It means no kinetic energy is added, so your work equals the gain in potential energy plus work against friction.
A practice routine
Day 1: Derive Mgl²/2L by integration and by centre of mass, from a blank page.
Day 2: Redo Worked example 2 with μ = 0.1 and with half the chain hanging.
Day 3: Two problems pulling a rope up from a well or a hanging rope onto a roof.
Day 4: Two "released chain" problems, checking each with the general formula.
If you are stuck on the calculus rather than the physics, our guide to the calculus you need for JEE physics covers setting up integrals. The Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can check each line of an integral set-up against a full solution.
Tonight, find the work to pull the chain on when half of it hangs over a smooth table, and check that it is 10 J.
