A steel wire has Young's modulus 2.0×10¹¹ N m⁻². Its Poisson ratio is 0.2 and the magnitude of transverse strain is 10⁻³. The elastic potential-energy density i s ______ ×10⁵ in SI units. Find the numerical value in the blank.

Question

G1 Physics · Accepted Answer

Poisson ratio is ν = |transverse strain|/|longitudinal strain|. Given ν = 0.2, transverse strain = 10⁻³. Therefore longitudinal strain is ε = 10⁻³/0.2 = 5×10⁻³. Elastic strain-energy density for a linearly elastic wire is U = (1/2)Yε². Substitute Y=2.0×10¹¹ N m⁻²: U = (1/2)(2.0×10¹¹)(5×10⁻³)² = 1.0×10¹¹×25×10⁻⁶ = 25×10⁵ J/m³. Therefore the required numerical value is 25.