Question

A hanging magnesium wire with diameter 2.1 mm (2.1 × 10-3 m) is initially 1.15 m long. When a 104 kg mass is hung from it, the wire stretches an amount 8.46 mm. A mole of magnesium has a mass of 24 grams, and its density is 1.74 g/cm3. Find the approximate value of the effective spring stiffness of the interatomic force.

Answer 1

Answer:

The approximate value of the effective spring stiffness of the inter atomic force is 11.32 N/m.

Explanation:

Given that,

Diameter of wire = 2.1 mm

Initial length = 1.15 m

Mass = 104 kg

Stretching length = 8.46 mm

Mass of mole of magnesium =24 g

Density = 1.74 g/cm³

We need to calculate the diameter of magnesium

Using formula of diameter

$d=(\dfrac{V}{N})^{\frac{1}{3}}$

$d=(\dfrac{\dfrac{M}{\rho}}{N})^{\frac{1}{3}}$

Put the value into the value

$d=(\dfrac{\dfrac{24}{1.74}}{6.023\times10^{23}})^{\frac{1}{3}}$

$d=2.83\times10^{-8}\ cm$

We need to calculate the stress

Using formula of stress

$stress=\dfrac{F}{A}$

$stress =\dfrac{104\times9.8}{\pi\times(1.05\times10^{-3})^2}$

$stress =294259805.89\ N/m^2$

$stress =2.94\times10^{8}\ N/m^2$

We need to calculate the strain

Using formula of strain

$strain=\dfrac{\Delta l}{l}$

$strain=\dfrac{8.46}{1.15}$

$strain=7.35$

We need to calculate the young modulus

Using formula of young modulus

$Y=\dfrac{stress}{strain}$

$Y=\dfrac{2.94\times10^{8}}{7.35\times10^{-3}}$

$Y=4\times10^{10}\ N/m^2$

We need to calculate the approximate value of the effective spring stiffness of the inter atomic force

Using formula of effective spring stiffness

$Y=\dfrac{k}{d}$

$k=Y\times d$

Put the value into the formula

$k=4\times10^{10}\times2.83\times10^{-8}\times10^{-2}$

$k=11.32\ N/m$

Hence,  The approximate value of the effective spring stiffness of the inter atomic force is 11.32 N/m.