A copper wire of length 3 m is stretched by 3 mm. Its volume is 600×10⁻⁶ m³ and Young's modulus is 1.1×10¹¹ N·m⁻². If the elastic potential energy stored in the stretched wire is N joules, find N.

Question

G1 Physics · Accepted Answer

The tensile strain is strain = ΔL ÷ L = 3×10⁻³ ÷ 3 = 10⁻³. Elastic energy density in a linearly elastic material is u = (1 ÷ 2)Y(strain)². Therefore the total elastic energy is U = (1 ÷ 2)Y(strain)²×volume. Substitute: U = (1 ÷ 2)(1.1×10¹¹)(10⁻³)²(600×10⁻⁶). Thus, U = (1 ÷ 2)(1.1×10¹¹)(10⁻⁶)(600×10⁻⁶) = 33 J. Therefore N = 33.