Identical springs of steel and copper are equally stretched. On which, more work will have to be done?

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Q. Identical springs of steel and copper are equally stretched. On which, more work will have to be done?

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Concept understanding — Youngs Modulus

Young’s Modulus: The Stretchiness of a Solid

When you pull on a rubber band, it stretches easily. When you pull on a steel rod of the same size, it barely moves. Both are elastic — they return to their original shape when you let go — but they resist stretching very differently. Young’s modulus is the number that tells you exactly how much a material resists being stretched or compressed lengthwise.

The Intuition: Stiffness per Unit Size

Think of a spring. A stiff spring requires a large force to stretch it a little. A soft spring stretches a lot with a small force. Young’s modulus is like the “stiffness” of a material, but it’s cleverly designed to be independent of the object’s shape and size.

If you take a thick steel rod and a thin steel wire of the same length, the rod is harder to stretch. That’s because you’re pulling on more material. Young’s modulus removes this size effect — it tells you the stiffness of the material itself, not the particular piece you’re holding.

The Precise Definition

Young’s modulus ( $E$ or $Y$ ) is defined as the ratio of tensile stress to tensile strain, as long as the material obeys Hooke’s law (the deformation is reversible and proportional to the force).

Y = \frac{\text{Tensile Stress}}{\text{Tensile Strain}}

Let’s break down the two parts.

Tensile Stress ( $\sigma$ ) is the force per unit area. If you pull with a force $F$ on a rod of cross-sectional area $A$ , the stress is:

\sigma = \frac{F}{A}

Stress has units of pressure — pascals (Pa) or $\text{N/m}^2$ . It tells you how “intense” the pulling is, regardless of the rod’s thickness.

Tensile Strain ( $\varepsilon$ ) is the fractional change in length. If the original length is $L_0$ and it stretches by $\Delta L$ , the strain is:

\varepsilon = \frac{\Delta L}{L_0}

Strain is a pure number — it has no units. A strain of 0.01 means the rod stretched by 1% of its original length.

Putting it together:

Y = \frac{F/A}{\Delta L / L_0} = \frac{F L_0}{A \Delta L}

What the Number Tells You

A high Young’s modulus means the material is very stiff — it takes a huge stress to produce even a tiny strain. Steel has $Y \approx 200 \times 10^9$ Pa. A low Young’s modulus means the material is easily stretched. Rubber has $Y \approx 0.01 \times 10^9$ Pa — about 20,000 times smaller than steel.

Note

Young’s modulus is only valid in the elastic region — where the material returns to its original shape after the force is removed. If you stretch too far (past the elastic limit), the material deforms permanently or breaks, and Young’s modulus no longer applies.

A Worked Example

A steel wire of length 2.0 m and cross-sectional area $1.0 \times 10^{-6} \text{ m}^2$ is pulled by a force of 100 N. How much does it stretch? (Young’s modulus of steel = $2.0 \times 10^{11}$ Pa)

From $Y = \frac{F L_0}{A \Delta L}$ , rearrange:

$\Delta L = \frac{F L_0}{A Y} = \frac{100 \times 2.0}{(1.0 \times 10^{-6}) \times (2.0 \times 10^{11})} = \frac{200}{2.0 \times 10^5} = 1.0 \times 10^{-3} \text{ m} = 1.0 \text{ mm}$ …

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For the same stretch, the stiffer spring (higher Young’s modulus) stores more elastic potential energy. Steel has a higher Young’s modulus than copper, so more work is done on the steel spring.

Why Young’s modulus decides the work

When you stretch a spring, you do work against the restoring force. That work is stored as elastic potential energy. For a spring obeying Hooke’s law, the work done is

W = \frac{1}{2} k x^2

where $k$ is the spring constant and $x$ is the extension. The question says the springs are identical in geometry (same length, same cross‑sectional area) and are equally stretched (same $x$ ). So the only difference is the material — and that enters through $k$ .

For a wire or spring of length $L$ and cross‑section $A$ , the spring constant is related to Young’s modulus $Y$ by

k = \frac{Y A}{L}

k = \frac{Y A}{L}

Since $A$ and $L$ are the same for both springs, $k$ is directly proportional to $Y$ . Steel has a Young’s modulus of about $2.0 \times 10^{11} \ \text{N/m}^2$ , while copper’s is about $1.1 \times 10^{11} \ \text{N/m}^2$ — roughly half. So the steel spring is stiffer.

Step‑by‑step reasoning

Work done on a spring For a stretch $x$ from natural length, the work done equals the elastic potential energy stored:

W = \frac{1}{2} k x^2

This is the area under the force‑extension graph.

Spring constant from material properties For a wire of length $L$ and area $A$ , the force needed for an extension $x$ is $F = \frac{Y A}{L} x$ …

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Q.Identical springs of steel and copper are equally stretched. On which, more work will have to be done?

Why Young’s modulus decides the work

Step‑by‑step reasoning

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