91Ó°ÊÓ

Elasticity is a fundamental property of materials that describes their ability to return to their original shape after being stretched or deformed. This feature is crucial in understanding the behavior of various materials under stress. Young's modulus, also known as the modulus of elasticity, is a quantitative measure of this property. It indicates how stiff a material is, with higher values representing more resistance to deformation.

Young's modulus is given by the formula:

• \[ Y = \frac{ ext{Stress}}{ ext{Strain}} \]

Here, stress and strain are related concepts that describe how much force is applied to a material, and how much it deforms as a result.

For our exercise, the wire's Young's modulus is quite high, meaning it's stiff and difficult to deform. This high modulus is crucial for calculating how the wire changes length when work is applied.

When a force is applied to stretch or compress a material, work is done on that material. This work is essentially the energy transferred into deforming it. The concept of work done plays a vital role in calculating the resulting deformation which is, in this case, a change in the length of a wire.

In mechanical terms, the work done can be expressed in relation to Young's modulus and the resulting strain. The formula used is:

• \[ W = \frac{1}{2} Y \left( \frac{\Delta L}{L} \right)^2 \times A \times L \]

This formula combines the stiffness of the material (Young's modulus) with the deformation ratio (strain), allowing for the calculation of energy required for deformation.

In the problem given, the work done on the wire is manipulated within this formula to reveal how the wire's length changes.

Two key concepts in elasticity are stress and strain. Stress is the internal force exerted per unit area within a material, while strain is the relative deformation experienced by the material.

Stress can be mathematically expressed as:

• \[ ext{Stress} = \frac{ ext{Force}}{ ext{Area}} \]

And strain is defined as:

• \[ ext{Strain} = \frac{\Delta L}{L} \]

In the context of the given exercise, stress helps us understand how the force applied across a tiny area affects the entire wire, and strain allows us to measure how much it lengthens or compresses.

Stress and strain are central to elasticity, providing insight into how a material will react when a force is applied.

Determining how much a wire will stretch or compress involves calculating the change in length, symbolized as \( \Delta L \), which can be deduced using the formula for work done. From the problem, by inputting all given values:

• Original length and cross-sectional area of the wire
• Work done and Young's modulus

We arrive at the calculation:

• \[ \Delta L^2 = \frac{2 \times 10^{-2}}{10 \times 10^{-16}} \]
• \[ \Delta L^2 = 2 \times 10^{14} \]

Taking the square root gives us:

• \( \Delta L = 1414 \times 10^{-4} \text{ m} \)
• Or approximately 1.414 mm

This change helps us understand how materials will respond physically to certain amounts of work, enabling practical calculations in an engineering context like that in the given exercise.