Problem statement

Problem 1.90. A system consists of two springs connected in series and having the stiffness coefficients k₁ and k₂. Find the minimum work to be performed in order to stretch this system by Δl.

We assume the springs obey Hooke's law, have no mass, and the joint between them has no mass. The system is pulled slowly, so nothing gains speed. This is what gives the minimum work.

We assume k₁ = N/m, k₂ = N/m and Δl = m as starting values, since the problem gives only symbols. All three are sliders.

We assume the natural length of each spring is 0.20 m in the picture. This is only for the drawing and does not change any answer.

Symbols: x₁ and x₂ are the stretches of the two springs, F the force, k the stiffness of the whole system.

The big idea in plain words

Two springs in a row act like one softer spring. The same force pulls on both, so the softer one stretches more. The work needed is the energy that ends up stored in the two springs, and pulling slowly wastes none of it as motion. The springs settle at the split of the stretch that stores the least energy, so this work is the minimum.

The solution in each frame

Only the lab frame is needed. The wall is at rest in it, and the springs act in it.

The solution step by step

Step 1, the same force. The joint has no mass, so the forces on its two sides must be equal. Both springs carry the same force F.

Step 2, the stretches. Spring 1 stretches by x₁ = F/k₁ and spring 2 by x₂ = F/k₂. The total is Δl = x₁ + x₂ = F(1/k₁ + 1/k₂).

Step 3, one equal spring. Solve for F: F = kΔl with 1/k = 1/k₁ + 1/k₂, which is k = k₁k₂/(k₁ + k₂) = N/m. At the full stretch F = , so x₁ = m and x₂ = m.

Step 4, the work. The work equals the energy stored, since nothing moves fast. The force grows from zero to kΔl along a straight line, so the work is the mean force times the stretch: A = ½kΔl² = ½ × × ² = .

Step 5, the parts. Spring 1 stores ½k₁x₁² = and spring 2 stores ½k₂x₂² = . The sum is the same A.

Step 6, why it is the minimum. Suppose the joint were forced to stay at some other place, with spring 1 stretched by x. The total energy would be ½k₁x² + ½k₂(Δl − x)². This is smallest when k₁x = k₂(Δl − x), which is the case where the forces are equal. That gives x = k₂Δl/(k₁ + k₂) = x₁ and the value A. With the joint held at the middle the energy would be , which is more.

Five methods

1. Force method

Find the force, then add force times small stretch over the whole stretch. This is steps 1 to 4.

2. Energy method

Add the two stored energies, ½k₁x₁² + ½k₂x₂². This is step 5. It works once x₁ and x₂ are known.

3. Graph method

F against Δl is a straight line with slope k. The work is the triangle under it, ½ × Δl × kΔl. The first graph on the page shades it.

4. Calculus method

A = ∫ F d(Δl) = ∫ kΔl d(Δl) from 0 to Δl = ½kΔl². Setting the slope of the energy to zero also gives the equal-force condition of step 6.

5. Check method

Units: N/m times m² is joules. Limits: if k₂ is very large, spring 2 hardly stretches and k is close to k₁. If k₁ = k₂, then k is half of one spring.

Fastest: the equal spring with the graph method. Most reliable: the force method with the numerical check.

What each slider changes

Control	Changes the answer?	What it changes
Stiffness k₁	Yes	The stretch and energy of spring 1, the combined stiffness, the force and the work
Stiffness k₂	Yes	The stretch and energy of spring 2, the combined stiffness, the force and the work
Stretch Δl	Yes	The force, both stretches and the work, as Δl²
Given-answer mode	Sets Δl	The page finds Δl for a chosen work, and warns if it is outside 0.05 m to 0.5 m
Play and stretch bar	No	Play pulls the system out slowly over 6 s, holds 1.2 s, then repeats

Parameters

Quantity	Symbol	Value used
Stiffness of spring 1	k₁	N/m
Stiffness of spring 2	k₂	N/m
Total stretch	Δl	m
Combined stiffness	k	N/m
Force at full stretch	F
Stretch of spring 1	x₁	m
Stretch of spring 2	x₂	m
Minimum work	A

Exam traps

• Adding the stiffnesses, k₁ + k₂. That is for springs side by side. In series the stiffnesses combine as 1/k = 1/k₁ + 1/k₂.
• Using only one spring, for example ½k₁Δl² = here. The stretch is shared between the springs.
• Stretching each spring by Δl. The two stretches add up to Δl, so each is smaller.
• Splitting the stretch equally between the springs. It splits in the ratio 1/k₁ to 1/k₂, so the softer spring takes more.
• Forgetting the half. Work is ½kΔl², since the force grows from zero.