Answer :
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
[tex]d=(\dfrac{V}{N})^{\frac{1}{3}}[/tex]
[tex]d=(\dfrac{\dfrac{M}{\rho}}{N})^{\frac{1}{3}}[/tex]
Put the value into the value
[tex]d=(\dfrac{\dfrac{24}{1.74}}{6.023\times10^{23}})^{\frac{1}{3}}[/tex]
[tex]d=2.83\times10^{-8}\ cm[/tex]
We need to calculate the stress
Using formula of stress
[tex]stress=\dfrac{F}{A}[/tex]
[tex]stress =\dfrac{104\times9.8}{\pi\times(1.05\times10^{-3})^2}[/tex]
[tex]stress =294259805.89\ N/m^2[/tex]
[tex]stress =2.94\times10^{8}\ N/m^2[/tex]
We need to calculate the strain
Using formula of strain
[tex]strain=\dfrac{\Delta l}{l}[/tex]
[tex]strain=\dfrac{8.46}{1.15}[/tex]
[tex]strain=7.35[/tex]
We need to calculate the young modulus
Using formula of young modulus
[tex]Y=\dfrac{stress}{strain}[/tex]
[tex]Y=\dfrac{2.94\times10^{8}}{7.35\times10^{-3}}[/tex]
[tex]Y=4\times10^{10}\ N/m^2[/tex]
We need to calculate the approximate value of the effective spring stiffness of the inter atomic force
Using formula of effective spring stiffness
[tex]Y=\dfrac{k}{d}[/tex]
[tex]k=Y\times d[/tex]
Put the value into the formula
[tex]k=4\times10^{10}\times2.83\times10^{-8}\times10^{-2}[/tex]
[tex]k=11.32\ N/m[/tex]
Hence, The approximate value of the effective spring stiffness of the inter atomic force is 11.32 N/m.
The effective spring stiffness on the wire is 11.32 N/m.
How do you calculate the effective spring stiffness?
Given that the initial length of the wire is 1.15 m, stretched length is 8.46 m and the diameter of the wire is 2.1 mm.
A mole of magnesium has a mass of 24 grams, and its density is 1.74 g/cm3. Mass on the wire is 104 kg.
The diameter of the magnesium is calculated as given below.
[tex]d_m = (\dfrac {V}{N})^{\dfrac {1}{3}}[/tex]
Where N = number of molecules in one gram-mole = 6.022 × 10^23.
V is the volume which can be written as,
[tex]V = \dfrac {M}{\rho }[/tex]
Where M is the mass of mole and [tex]\rho[/tex] is the density.
Substituting the values, we get the diameter of the magnesium.
[tex]d_m =( \dfrac {\dfrac {M}{\rho}}{N})^{\dfrac{1}{3}[/tex]
[tex]d_m = (\dfrac {\dfrac {24}{1.74}}{6.022\times 10^{23}})^{\dfrac{1}{3}[/tex]
[tex]d_m = 2.83\times 10^{-8}\;\rm cm[/tex]
The stress on the wire is calculated as given below.
[tex]\tau = \dfrac {F}{A} = \dfrac {mg}{\pi r^2}[/tex]
Where F is the force, A is the cross-sectional area of the wire, m is the mass on the wire, g is the gravitational acceleration and r is the radius of the wire.
[tex]\tau = \dfrac {104\times 9.8}{3.14\times (1.05\times 10^{-3})^2}[/tex]
[tex]\tau = 2.94\times 10^8 \;\rm N/m^2[/tex]
The strain on the wire is calculated as given below.
[tex]\varepsilon = \dfrac {\Delta l}{l}[/tex]
Where [tex]\Delta l[/tex] is the stretched length of the wire and l is the normal length of the wire.
[tex]\varepsilon = \dfrac {8.46}{1.15}[/tex]
[tex]\varepsilon = 7.35\;\rm mm[/tex]
The elastic modulus on the wire is calculated by young's modulus formula.
[tex]E = \dfrac {\tau}{\varepsilon}[/tex]
[tex]E = \dfrac {2.94\times 10^8}{7.35\times 10^{-3}}[/tex]
[tex]E = 4\times 10^{10}\;\rm N/m^2[/tex]
The effective spring stiffness on the wire is calculated as given below.
[tex]E = \dfrac {k}{d}[/tex]
[tex]4\times 10^{10}=\dfrac {k}{2.83\times 10^{-8}\times 10^{-2}}[/tex]
[tex]k = 11.32 \;\rm N/m[/tex]
Hence we can conclude that the effective spring stiffness on the wire is 11.32 N/m.
To know more about stress and strain, follow the link given below.
https://brainly.com/question/8043545.