Question

A hanging copper wire with diameter 1.4 mm $\left(1.4\times {10}^{-3}\right)\mathit{m}$ is initially 0.95m long. When a 36kg mass is hung from it, the wire stretches an amount 1.83mm, and when a 72kg mass is hung from it, the wire stretches an amount 3.66mm. A mole of copper has a mass of 63g, and its density is$\mathbf{9}\mathbf{}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$ Find the approximate value of the effective spring stiffness of the interatomic force.

Short answer

The approximate value of stiffness of interatomic force is$27N/m$ .

Step-by-step solution

Identification of given data

The given data can be listed below,

• The diameter of copper wire is$D=1.4\times {10}^{-3}m$ ,
• The length of the copper wire is,$L=0.95m$.
• The mass of first hanging object is,$m_1=36kg$
• The mass of second hanging object is, $m_2=72kg$.
• The length of wire stretched due to first mass is, $l_1=1.83mm$.
• The length of wire stretched due to second mass is,$l_2=3.66mm$ .
• The mass of one mole copper is, $M=63g$
• The density of copper is,$\rho =9g/c{m}^3$

Concept/Significance of stiffness of spring

The capacity of a spring to withstand a force is called spring stiffness. The force required to compress or stretch the spring increases as the spring stiffens.

Determination of the approximate value of the effective spring stiffness of the interatomic force.

The cross-sectional area of the wire is given by,

$A=\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^2$

Here, d is the diameter of the wire.

Substitute all the values in the above,

$$\begin{gathered} A=3.14\times \left(\frac{1.4\times {10}^{-3}}{2}\right) \\ =1.538\times {10}^{-3}{m}^2 \\ =1.54\times {10}^{-6}{m}^2 \end{gathered}$$

The volume of a copper atom is given by,

$v=\frac{M}{\rho \times N_A}$

Here, M is the mass of one mole of copper, is the density of copper and is the Avogadro number whose value is $6.023\times {10}^{23}$.

Substitute all the values in the above expression.

$$\begin{gathered} V=\frac{63g}{9g/c{m}^3\times 6.023\times {10}^{23}atom} \\ =1.17\times {10}^{-23}c{m}^3/atom \end{gathered}$$

The interatomic distance between the atoms is given by,

$d=V^{1/3}$

Substitute value in the above,

$$\begin{gathered} d_a=\sqrt{1.17\times {10}^{-23}c{m}^3{\left(\frac{1m}{100cm}\right)}^3} \\ =2.26\times {10}^{-10}m \end{gathered}$$

The stiffness of the wire due to first mass is given by,

$k=\frac{m_{1}g}{l_1}$

Here,$m_1$is the mass of first hanging mass and $l_1$is the length stretched due to mass.

Substitute all the values in the above,

$$\begin{gathered} k_1=\frac{\left(36kg\right)\left(9.8m/s^2\right)}{1.83\times {10}^{-3}m} \\ =192.78\times {10}^{3}N/m \end{gathered}$$

The number of side-by-side atoms in chain is given by,

$$\begin{gathered} N_{chain}=\frac{A_{wire}}{A_{atom}} \\ =\frac{\mathrm{\pi }{\left(\mathrm{d}/2\right)}^2}{\left(d_a\right)} \end{gathered}$$

Here, d is the diameter of wire and is the interatomic distance of atoms.

Substitute all the values in the above,

$$\begin{gathered} N_{chain}=\frac{\mathrm{\pi }{\left(1.4\times {10}^{-3}\mathrm{m}/2\right)}^2}{{\left(2.26\times {10}^{-10}m\right)}^2} \\ =30.14\times {10}^{12}chains \end{gathered}$$

The number of bonds in an atomic chain of wire is given by,

$N_{bond}=\frac{L}{d_a}$

Here, Lis the length of copper wire.

Substitute values in the above equation.

$$\begin{gathered} N_{bond}=\frac{3m}{2.26\times {10}^{-10}m} \\ =4.20\times {10}^{9}bonds \end{gathered}$$

The stiffness of a single interatomic bond is very smaller than the stiffness of whole copper wire it can be given by,

$k_{aj}=\frac{N_{bond}{k}_1}{N_{chain}}$

Here, $N_{bond}$ is the number of bonds in atomic chain, $k_1$is the stiffness of first mass and $N_{chain}$ is the number of chains.

Substitute all the values in the above expression,

$$\begin{gathered} K_{aj}=\frac{\left(4.20\times {10}^9\right)\left(192.78\times {10}^{3}N/m\right)}{30.14\times {10}^{12}} \\ =27N/m \end{gathered}$$

Thus, the approximate value of stiffness of interatomic force is$27N/m$ .