A wire of length L and cross-sectional area A is made of a material of Young's modulus Y. If the wir

Q: Why is the work done not simply Force times displacement, i.e., W = F * x?

The stretching force is not constant; it starts at zero and increases linearly with elongation ($F \propto y$). Therefore, we must integrate the variable force over the displacement, which introduces the factor of 1/2.

Q: Does this formula hold true for any amount of stretching?

No, this formula is derived assuming Hooke's Law holds, which is only valid within the elastic limit of the material. Beyond this limit, the deformation becomes plastic, and the relationship is no longer linear.

Q: Where does the work done go? Is it lost as heat?

Within the elastic limit, the work done is stored completely as elastic potential energy in the intermolecular bonds of the material. When the force is removed, the wire returns to its original shape, releasing this stored energy.

Q: How does the cross-sectional area affect the work required to stretch the wire?

Since the equivalent spring constant is $k = YA/L$, a larger cross-sectional area $A$ increases the stiffness of the wire. Consequently, more work is required to stretch a thicker wire by the same distance $x$ compared to a thinner one.

Q: What is the physical significance of the 1/2 factor in the energy density formula?

The factor of 1/2 represents the average force acting during the stretching process. Since the force increases linearly from $0$ to $F$, the average force is $F/2$, making the work done $W = (F/2)x$.

Question

A wire of length $L$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$. If the wire is stretched by an amount $x$, the work done is ......

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine suspending a uniform metal wire of natural length $L$ and cross-sectional area $A$ vertically from a rigid ceiling.

In its natural, unstretched state, no external forces act on it, and its elongation is exactly zero.

This baseline state serves as our reference point for measuring any future deformation.

### The Physics of Elasticity

When we apply an external tensile force $F$ to stretch this wire, the material's internal structure resists this deformation.

This resistance is quantified by **Young's Modulus** ($Y$), which is defined as the ratio of tensile stress to tensile strain within the elastic limit:

$$Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta L/L}$$

Here, stress is the force per unit area ($F/A$), and strain is the fractional change in length ($\Delta L/L$).

### The Spring Analogy

Let us find the restoring force developed in the wire when it is stretched by an instantaneous displacement $y$.

By rearranging the definition of Young's Modulus, we can express the force $F$ as:

$$F = \left(\frac{YA}{L}ight) y$$

This linear relationship between force and displacement is mathematically identical to Hooke's Law for an ideal spring ($F = ky$).

Thus, we can define an **equivalent spring constant** ($k$) for the wire:

$$k = \frac{YA}{L}$$

This elegant analogy simplifies our understanding of elastic deformation, showing that a thicker or shorter wire is stiffer and harder to stretch.

### Calculating the Work Done

To find the total work done in stretching the wire by a final amount $x$, we must integrate the work done during tiny, incremental displacements.

For an infinitesimal elongation $dy$, the small work done $dW$ by the external force is:

$$dW = F \cdot dy = \left(\frac{YA}{L} yight) dy$$

To find the total work $W$, we integrate this expression from $y = 0$ to $y = x$:

$$W = \int_{0}^{x} \frac{YA}{L} y \, dy$$

Since $Y$, $A$, and $L$ are constant physical parameters of the wire, we can pull them out of the integral:

$$W = \frac{YA}{L} \int_{0}^{x} y \, dy$$

Evaluating this simple integral yields:

$$W = \frac{YA}{L} \left[ \frac{y^2}{2} ight]_{0}^{x} = \frac{1}{2} \left(\frac{YA}{L}ight) x^2$$

This is the total work done in stretching the wire, which is stored entirely as **elastic potential energy** within the material.

### Alternative Form and Physical Insights

We can rewrite this final expression to gain deeper physical insights:

$$W = \frac{1}{2} \left(\frac{F}{y}ight) x^2 = \frac{1}{2} F x$$

This shows that the work done is equal to half the product of the maximum stretching force and the total elongation.

Furthermore, dividing the total energy by the volume of the wire ($V = A 	imes L$) gives the **energy density** ($u$):

$$u = \frac{W}{AL} = \frac{1}{2} \left(\frac{F}{A}ight) \left(\frac{x}{L}ight) = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain}$$

This universal relation highlights that the energy stored per unit volume depends solely on the stress and strain developed within the material.

Animated Solution for Physics - Properties of Solids and Liquids: A wire of length $L$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$ . If the wire is stretched by an amount $x$ , the work done is ......

Visualized Solution

Visualizing the Unstretched Wire

Understanding Young's Modulus

Relating Force and Elongation

Analogy with a Spring

Setting Up the Work Integral

Integrating to Find Total Work

Evaluating the Integral

Alternative Form: Energy Density

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

Analyzing the Setup

The Physics of Elasticity

The Spring Analogy

Calculating the Work Done

Alternative Form and Physical Insights

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