What force must be applied to the ends of a 70 kN / m wire to stretch it by 2mm?

These tasks: k (stiffness of the wire being stretched) = 70 kN / m (70 * 10 ^ 3 N / m); Δl (required spring tension) = 2 mm (in SI system Δl = 2 * 10 ^ -3 m).

The force that was applied to the ends of the wire taken is equal to the elastic force and can be determined by the formula: F = -Fel = k * Δl.

Let’s make a calculation: F = 70 * 10 ^ 3 * 2 * 10 ^ -3 = 140 N.

Answer: A force of 140 N will need to be applied to the ends of the wire.