a force of 22,000 n will cause a 1 cm x 1 cm bar of magnesium to stretch from 10 cm to 10.07 cm. calculate the modulus of elasticity, both in gpa and ksi.

Respuesta :

The elastic modulus of magnesium bar is 3.14 GPa.

We need to know about the modulus of elasticity to solve this problem. The elastic modulus of material can be measured by strain and stress. It can be determined as

B = σ / e

B = (F/A) / (ΔV/V)

where B is modulus of elasticity, σ is stress, e is strain, F is force, A is area, V is initial volume, ΔV is volume stretched.

From the question above, we know that:

F = 22000 N

A = 1cm x 1cm = 1 cm²

V = A x 10cm = 10 cm³

ΔV = (10.07-10) x A =0.07 cm³

Convert to SI unit

A = 1 x 10¯⁴ m²

V = 10¯⁵ m³

ΔV = 7 x 10¯⁴ m³

By substituting the following parameters, we get

B = (F/A) / (ΔV/V)

B = (22000 / (1 x 10¯⁴) / (7 x 10¯⁴ / 10¯⁵)

B = 3.14 x 10⁶ Pa

B = 3.14 GPa

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