Work Done In Stretching A Wire

Stretching a wire involves the application of force to elongate it, a concept that intertwines physics with everyday experiences. When you pull on a rubber band or stretch a spring, you are engaging in a practical demonstration of work done in stretching a wire. This phenomenon is governed by principles of material science and mechanics, where energy is transferred to overcome internal resistances and change the shape of the wire. In real-life applications, understanding this concept is crucial, whether it's designing resilient structures, creating durable materials, or even in the mechanics of everyday objects like car springs and bicycle cables. The work done in stretching a wire not only highlights fundamental physical principles but also underscores its importance in engineering and technology. In this article, we will cover the concept of Work Done In Stretching A Wire. This concept falls under the broader category of Properties of Solids and Liquids.

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This Story also Contains

1. Work Done in Stretching a Wire
2. Solved Examples Based on Work Done in Stretching a Wire
3. Summary

Work Done in Stretching a Wire

When a body is in its natural shape, its potential energy corresponding to the molecular forces is minimal. When deformed, internal forces appear and work has to be done against these forces. Thus, the potential energy of the body is increased. This is called the elastic potential energy.

Suppose a wire having natural length L and cross-sectional area A is fixed at one end and is stretched by an external force applied at the other end. When the extension is x,the wire is under a longitudinal stress F/A. The strain is x/L.

If Young's modulus is Y, then

$
\frac{F / A}{x / L}=Y \Rightarrow F=\frac{A Y}{L} x
$

So, the work done for an additional small increase dx in length will be:-
$
d W=F d x \Rightarrow d W=\frac{A Y}{L} x d x
$

The total work done by the external force in increasing the length from 0 to $\Delta L$ will be:-
$
W=\int_0^{\Delta L} \frac{A Y}{L} x d x=\frac{1}{2} \frac{Y A}{L}(\Delta L)^2
$

This work is stored in the wire as its elastic potential energy. So, the elastic potential energy of the stretched wire is:
$
U=\frac{1}{2} \frac{Y A}{L}(\Delta L)^2
$

We can also write,

$
\begin{aligned}
& W=\frac{1}{2}\left[\frac{Y A \Delta L}{L}\right] \Delta L=\frac{1}{2}(\text { maximum stretching force }) \times \text { extension } \\
& W=\frac{1}{2} Y(A L)\left[\frac{\Delta L}{L}\right]^2=\frac{1}{2} \times Y \times \text { Volume } \times(\text { strain })^2 \\
& W=\frac{1}{2}\left[\frac{\Delta L}{L}\right]\left[Y \frac{\Delta L}{L}\right](A L)=\frac{1}{2} \times \text { strain } \times(Y \text { strain }) \times \text { Volume } \\
& W=\frac{1}{2}\left[\frac{\Delta L}{L}\right]\left[Y \frac{\Delta L}{L}\right](A L)=\frac{1}{2} \times \text { strain } \times \text { stress } \times \text { Volume }
\end{aligned}
$

Also, Potential energy per unit volume $=\frac{1}{2} \times \operatorname{strain} \times$ stress

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Solved Examples Based on Work Done in Stretching a Wire

Example 1: Two wires of the same materials (Young modulus) and the same length L but radii R and 2R respectively are joined end to end and a weight W is suspended from the combination as shown in the figure.

The elastic PE stored in the system is

1) $\frac{3 w^2 L}{4 \pi^2 R Y}$
2) $\frac{5 w^2 L}{8 \pi R^2 y}$
3) $\frac{3 w^2 L}{8 \pi_R^2 y}$
4) $\frac{\omega^2 L}{8 \pi R^2 y}$

Solution:

Using,

$
U=\frac{1}{2} k x^2=\frac{k}{2}\left(\frac{F}{k}\right)^2=\frac{F^2}{2 k}
$

We have
$
\begin{aligned}
U & =U_1+U_2 \\
& =\frac{F^2}{2 k_1}+\frac{F^2}{2 k_2} \quad(F=W) \\
& =\frac{W^2}{2}\left[\frac{1}{k_1}+\frac{1}{k_2}\right]
\end{aligned}
$
and
$
\begin{aligned}
& k_1=\frac{Y A}{L}=\frac{Y \pi(2 R)^2}{L}, \\
& k_2=\frac{Y \pi R^2}{L}
\end{aligned}
$

Put the values $k_1 \& k_2$ in equation (i)
$
\begin{aligned}
U & =\frac{W^2}{2}\left[\frac{L}{4 \pi R^2 Y}+\frac{L}{\pi R^2 Y}\right] \\
U & =\frac{5 \omega^2 L}{8 \pi R^2 Y}
\end{aligned}
$

Hence, the answer is the option (2).

Example 2: Two wires of the same material and the same length L but radii R and 2R respectively, are joined end to end and a weight w is suspended from the combination as shown in the figure. The elastic potential energy in the system is

1) $\frac{3 w^2 L}{4 \pi R^2 Y}$
2) $\frac{3 w^2 L}{8 \pi R^2 Y}$
3) $\frac{5 w^2 L}{8 \pi R^2 Y}$
4) $\frac{w^2 L}{\pi R^2 Y}$

Solution:

$\begin{aligned} & \Delta l_1=\frac{w L}{\left(4 \pi R^2\right) Y}, \Delta l_2=\frac{w L}{\pi R^2 Y} \\ & \therefore \quad U=\frac{1}{2} k_1\left(\Delta l_1\right)^2+\frac{1}{2} k_2\left(\Delta l_2\right)^2 \\ & U=\frac{1}{2} \times Y\left(\frac{4 \pi R^2}{L}\right) \times\left[\frac{w L}{4 \pi R^2 Y}\right]^2+\frac{1}{2} \times \frac{Y\left(\pi R^2\right)}{L} \times\left[\frac{w L}{\pi R^2 Y}\right]^2 \\ & U=\frac{5 w^2 L}{8 \pi R^2 Y}\end{aligned}$

Hence, the answer is the option (3).

Example 3: An elastic material with Young's modulus y is subjected to a tensile stress S elastic energy stored per unit volume of the material is

1) $\frac{y S}{2}$
2) $y^{-1} S^2$
3) $\frac{S^2}{2 y}$
4) $\frac{S}{2 y}$

Solution:

$\begin{aligned} & \text { strain }=\frac{\text { stress }}{y}=\frac{S}{y} \text { Now, energy stored per unit volume } \\ & =\frac{1}{2} \times \text { stress } \times \text { strain } \\ & =\frac{1}{2} \times S \times \frac{S}{y}=\frac{S^2}{2 y}\end{aligned}$

Hence, the answer is the option (3).

Example 4: A uniform rod is kept at a smooth horizontal surface, and a constant force is applied on the rod in the horizontal direction at end A. Find the ratio of energy stored per unit volume at end A to the energy stored per unit volume in the middle of the rod.

1) 2

2) 4

3) 8

4) 10

Solution:

$\begin{aligned} & U=\frac{1}{2} \frac{\text { stress }}{Y} \\ & \text { stress }=\frac{\text { tension }}{\text { Area }} \\ & \frac{U_A}{U_{\text {middle }}}=\frac{\left(T_A\right)^2}{\left(T_{\text {middle }}\right)^2}=\frac{F^2}{\left(\frac{F}{2}\right)^2}=4\end{aligned}$

Hence, the answer is the option (2).

Example 5:

In an experiment, brass and steel wires of length 1 m each with areas of cross-section 1 mm² are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and the other end is subjected to elongation. The stress (in N/m²) required to produce a net elongation of 0.2mm is, [Given, the Young's Modulus for steel and brass are respectively, $120 \times 10^9 \mathrm{~N} / \mathrm{m}^2$ and $60 \times 10^9 \mathrm{~N} / \mathrm{m}^2$].

1) 1200000

2) 8000000

3) 1800000

4) 200000

Solution:

$\begin{aligned} & k_1=\frac{y_1 A_1}{l_1}=\frac{120 \times 10^9 \times A}{1} \\ & k_2=\frac{y_2 A_2}{l_2}=\frac{60 \times 10^9 \times A}{1} \\ & K_{e q}=\frac{K_1 K_2}{K_1+K_2}=\frac{120 \times 60}{180} \times 10^9 \times A \\ & \Rightarrow K_{e q}=40 \times 10^9 \times A \\ & F=k_{e q}(x) \\ & F=\left(40 \times 10^9\right) A_0\left(0.2 \times 10^{-3}\right) \\ & \Rightarrow \frac{F}{A}=8 \times 10^6 \mathrm{~N} / \mathrm{m}^2\end{aligned}$

Hence, the answer is the option (2).

Summary

The concept of work done in stretching a wire is essential for understanding elastic potential energy and its real-world applications. By analyzing the equations and examples, we see that the work done to stretch a wire can be calculated using Young's modulus and other parameters. The elastic potential energy stored in the wire is determined by its extension and material properties. Examples illustrate practical scenarios involving different materials and configurations, demonstrating how these principles are applied in engineering and everyday objects.