What hanging mass will stretch a 2.1-m-long, 0.43 mm - diameter steel wire by 1.5 mm? the young'? s modulus of steel is 20*1010n/m2. express your answer using two significant figures?

The force (weight) due to the hanging mass is

F = mg

where

m = mass, kg

g = 9.8 m/s²

The cross sectional area of the wire is

A = (π/4) * (0.43*10⁻³ m) ² = 1.4522*10⁻⁷ m²

The stress in the wire is

σ = F/A = (9.8m N) / (1.4522*10⁻⁷ m²) = 6.7484x10⁷m Pa

The length of the wire is 2.1 m and the extension is 1.5 mm. Therefore the induced strain is

ε = (1.5x10⁻³ m) / (2.1 m) = 7.1429x10⁻⁴

By definition, Young's modulus is E = σ/ε. Therefore

(6.7484x10⁷m Pa) / (7.1429x10⁻⁴) = 20x10¹⁰ Pa

9.4477x10¹⁰m = 20x10¹⁰

m = 2.117 kg

Answer: 2.12 kg (to two sig. figures)