A steel wire of length 20 m stores 80 J elastic energy when stretched by 2 cm. If Y = 2.0 × 10¹¹ N m⁻², find its cross-sectional area in mm².

Question

G1 Physics · Accepted Answer

Elastic energy: U = (1/2)kx² Given U=80 J and x=0.02 m: 80 = (1/2)k(0.02)² k = 4×10⁵ N m⁻¹ For a wire: k = YA/L Thus, A = kL/Y = (4×10⁵)(20)/(2×10¹¹) = 4×10⁻⁵ m² = 40 mm² Final value: 40