11 min readConcept sheetSection A MCQ (1 mark); Section B very short answer (2 marks), often with the force-extension graph; also a NEET favourite

Work done by a varying force; work done in stretching a spring

For a varying force, work = area under the force-displacement graph. Stretching a spring by x needs work ½kx², stored as elastic potential energy.

Work by a varying forceW = ΣF(x) Δx, the area under the F-x graph between the starting and final positions

Hooke's law for a springFₛ = −kx: the restoring force is proportional to the extension or compression x, and opposite to it

Spring constant kRestoring force per unit extension; SI unit N m⁻¹. For a hanging load, k = mg/s

Work in stretching a spring by xW = ½kx², done by the external force; the spring force does −½kx²

Work from a graph

When the force changes with position, W = Fd cannot be used directly. Split the displacement into many tiny steps Δx. Make each step so small that the force is almost constant over it. The work in one step is ΔW = F(x) Δx, which is the area of a thin strip under the F-x curve. Add all the strips: W = ΣF(x) Δx. As Δx is made smaller and smaller, the sum becomes exactly the area under the F-x graph between the two positions x₁ and x₂. So, work done = area under the force-displacement graph. For a constant force, this area is just a rectangle, F × d.

The spring force and Hooke's law

Fix one end of a light spring to a wall. Tie a block to the other end on a smooth table. Pull the block out by x: the spring pulls it back. Push it in by x: the spring pushes it out. This restoring force is Fₛ = −kx, for small x (Hooke's law). Here k is the spring constant, which tells how stiff the spring is; unit N m⁻¹. The minus sign shows that the spring force is always opposite to the displacement.

Derivation: work done in stretching a spring by x

1
The spring force rises evenly from 0 (at x = 0) to kx (at extension x).
2
Because it rises in a straight line, the average force is (0 + kx)/2 = ½kx.
3
Work = average force × displacement = ½kx × x.
4★
So W = ½kx².
5
Graph check: the F-x graph is a straight line through the origin. The area of the triangle under it = ½ × base × height = ½ × x × kx = ½kx².
6
So, the external force does +½kx² of work, and the spring force does −½kx². The result is the same for a compression x.

★ key step

How to draw the force-extension graph of a spring

Draw the extension x on the horizontal axis and the spring force F on the vertical axis.
Draw a straight line from the origin rising to the right. Label it F = kx and write slope = k beside it.
Mark xₘ on the x-axis and the matching height kxₘ on the F-axis.
Shade the triangle under the line from 0 to xₘ. Label it area = work done = ½kxₘ².
For the restoring force Fₛ = −kx, draw instead a straight line through the origin sloping downward: positive x gives negative Fₛ, and negative x gives positive Fₛ.

Activity: measuring the spring constant

1
Hang a spring from a stand. Note the position of its lower end against a metre scale.
2
Hang a known mass m, say 200 g, from it. The spring stretches and comes to rest.
3
Measure the extension s.
4
At rest, the spring force balances the weight: ks = mg, so k = mg/s.
5
Repeat with two or three different masses. Record m, s and k for each, and take the average k.

Worked example: stretching a spring

Problem

A spring stretches by 2 cm when a 0.5 kg mass hangs from it. Find its spring constant, and the work needed to stretch it by 6 cm from its natural length. Take g = 9.8 m s⁻².

Working

k = mg/s = (0.5 kg × 9.8 m s⁻²) ÷ 0.02 m = 4.9 N ÷ 0.02 m = 245 N m⁻¹.
Extension x = 6 cm = 0.06 m.
W = ½kx² = ½ × 245 N m⁻¹ × (0.06 m)².
W = 0.5 × 245 × 0.0036 = 0.441 J.

Answer

k = 245 N m⁻¹; work ≈ 0.44 J

In real life

The springs in a scooter's suspension and in a mattress store energy as they are pressed and give it back. A shopkeeper's spring balance uses Hooke's law: the stretch is proportional to the weight hung on it.

Exam angle

2 marks: “A spring of constant k is stretched by x. Sketch how its restoring force changes with x, and use the sketch to find the energy stored.” Fₛ = −kx gives a straight line through the origin sloping downward. The triangle between the line and the x-axis has area ½ × x × kx, so the energy stored is ½kx².

2 marks: “Define spring constant and give its SI unit.” The restoring force per unit extension of a spring, k = F/x. SI unit: N m⁻¹.

3 marks: “A spring of constant 200 N m⁻¹ is stretched by 5 cm from its natural length. Find the work done. How much more work is needed to stretch it further, to 10 cm?” W₁ = ½ × 200 × (0.05)² = 0.25 J. Work from 0 to 10 cm = ½ × 200 × (0.10)² = 1.0 J. Extra work = 1.0 − 0.25 = 0.75 J. It is more than 0.25 J because the force grows with x.

Remember it

Spring: force kx (a straight line), work ½kx² (the triangle under it).