Answer:
Explanation:
Given that:
length l = 2.3 m
a = 0.12 cm = [tex]0.12 \times 10^{-2} \ m[/tex]
[tex]x = 1.17 \ cm = 1.17 \times 10^{-2}\ m[/tex]
m = 149 kg
[tex]\delta = 7.87 \ g/cm^3[/tex]
[tex]da = 2.28 \times 10^{-10}\ m[/tex]
[tex]F_{net} = F-mg\\ \\0 = F - mg \\ \\ F = mg \\ \\ k_sx = mg \\ \\[/tex]
∴
[tex]k_s = \dfrac{149(9.8)}{1.17 \times 10^{-2}} \\ \\ k_s = 124803.42 \ N /m[/tex]
[tex]N_{chain} = \dfrac{A_{wire}}{A_{atom}} = \dfrac{A_w}{da^2}[/tex]
[tex]N_{chain} = \dfrac{(a)^2}{(da)^2} = (\dfrac{a}{da})^2[/tex]
[tex]N_{chain} = (\dfrac{0.12 \times 10^{-2} }{2.28 \times 10^{-10}})^2[/tex]
[tex]N_{chain} = 2.77 \times 10^{13}[/tex]
[tex]N_{bond} = \dfrac{L}{da} \\ \\ = \dfrac{2.3}{2.28 \times 10^{-10}} \\ \\ N_{bond} = 1.009 \times 10^{10}[/tex]
[tex]\text{Finally; the stiffness of a single interatomic spring is:}[/tex]
[tex]k_{si} =\dfrac{N_{bond}}{N_{chain}}\times k_s[/tex]
[tex]k_{si} =\dfrac{(1.009 \times 10^{10})}{2.77*10^{13}}}\times (124803.42)[/tex]
[tex]\mathbf{k_{si} =45.46 \ N/m}[/tex]
