Question

A hanging iron wire with diameter 0.5 mm (0.5 × 10-3 m) is initially 1.10 m long. When a 108 kg mass is hung from it, the wire stretches an amount 29.6 mm. A mole of iron has a mass of 56 grams, and its density is 7.87 g/cm3. Find the approximate value of the effective spring stiffness of the interatomic force.

NOTE: The effective spring stiffness of the force between two atoms, not the whole system, thanks!

Answer 1

Solution)

We know,

Y = σ/ε

σ =applied stress

ε=resulting strain

So,

σ = F/A = m*g/(π*d²/4) = 108*9.8/(π*0.5*10^-3²/4) N/m²=5.39*10^9 N/m^2

ε = ∆Lw/Lw = 29.6*10^-3/1.10=0.0269

Now, Y= (5.39*10^9)/0.0269 = 2.03*10^11 N/m^2

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