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2 years ago

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One mole of iron (6 1023 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so thecenter-to-cent

er distance between atoms is 2.28 10-10 m. You have a long thin bar ofiron, 2.0m long, with a square cross section, 0.15 cm on a side.
You hang the rod vertically and attach a 231 kg mass to the bottom, and you observe thatthe bar becomes 1.01 cm longer. Fromthese measurements, it is possible to determine the stiffness ofone interatomic bond in iron.

What is the spring stiffness of the entire wire, considered as asingle macroscopic (large scale), very stiff spring?

1 answer:

Nezavi [6.7K]2 years ago

7 0

Answer:

(a)

Stiffness of one inter-atomic bond in iron = 5.14*10^-9 N/m

(b)

Spring stiffness of entire wire = 45.37 N/m

Explanation:

Given data:

No of atoms = = 1 mole of iron = 6*10^23

Mass = 56 g

Density = 7.87 g / cm^3

Original Length = 2 m

Hanged mass = 231 kg

Change in length = 1.01 cm

Solution:

Force applied to bar = 231 * 9.8

= 2263.8 N

Change in length = ΔL = 1.01 cm = 0.0101 m

Bar stiffness = F / ΔL = 2263.8 / 0.0101

Bar stiffness = 2.24*10^5 N/m

Cross sectional area of bar= 0.0015*0.0015

= 2.25*10^-6 m^2

No. of atoms in one layer of cross-sectional area

= Cross-Sectional area of bar / Area of one atom

No. of atoms = 2.25*10^-6 / (2.28*10^-10)^2

No. of atoms in one layer of cross-sectional area = 4.33*10^13

Length of Bar = 2 m

No. of bonds along bar length

= Bar length / Center to center distance between atoms

No. of bonds = 2 / 2.28*10^-10

No. of bonds = 8.77*10^9

Force applied to each atom

= Force / ( No. of atoms * No. of bonds )

Force applied to each bond = 2263.8 / ( 4.33*10^13 * 8.77*10^9)

Force applied to each bond = 5.96*10^-21 N

Strain on bar on bar = ΔL / L = 0.0101 / 2

= 0.0051

Strain between atom layers will be same as above.

Bond extension (elongation) = 2.28*10^-10 * 0.0051

= 1.16*10^-12 m

Bond stiffness = Force applied to each bond / Bond extension

Bond stiffness = 5.96*10^-21 / 1.16*10^-12

Bond stiffness = 5.14*10^-9 N/m

Stiffness of one inter-atomic bond in iron = 5.14*10^-9 N/m

Part (b)

Spring stiffness of entire wire = (Bar stiffness * No. of bonds) / No. of atoms

Spring stiffness = (2.24*10^5 * 8.77*10^9) / 4.33*10^13

Spring stiffness = 45.37 N/m

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4 0

2 years ago