Question

A hanging magnesium wire with diameter 1.0 mm (1.0 × 10e-3 m) is initially 1.50 m long. When a 58 kg mass is hung from it, the wire stretches an amount 27.1 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 bond.

ks= N/m

Answer 1

For calculating the effective spring stiffness of the interatomic bond, I think we only need the value of initial length of the magnesium wire and its elongation. Also, the weight of the hanging mass.

Initial length of the magnesium wire, L = 1.50 m

Elongation in the wire, x = 27.1 mm = 27.1 x 10^-3 m

Hanging weight through the wire, W = m*g = 58*9.81 N

Suppose, k = spring stiffness of the interatomic bond.

So, use the expression -

F = W = k*x

=> k = W / x = (58*9.81) / (27.1 x 10^-3) = 21000 N/m (Answer)