What tensile force should be applied to a steel wire (E = 2 * 10 ^ 11Pa) 7.5m long and 1.5mm ^ 2 in order to lengthen it by 0.5mm.

Data: l0 (steel wire length) = 7.5 m; S (wire cross section) = 1.5 mm2 (1.5 * 10 ^ -6 m); Δl (required steel wire elongation) = 0.5 mm (0.5 * 10 ^ -3 m).
Constants: according to the condition E (Young’s modulus for steel) = 2 * 10 ^ 11 Pa.
The tensile force that needs to be applied to the steel wire is determined from the equality: Δl / l0 = F / (E * S), whence F = Δl * E * S / l0.
Calculation: F = 0.5 * 10 ^ -3 * 2 * 10 ^ 11 * 1.5 * 10 ^ -6 / 7.5 = 20 N.
Answer: You need to apply a tensile force of 20 N.