The spring with a stiffness of 50 N / m lengthened by 1 cm under the action of the force.

The spring with a stiffness of 50 N / m lengthened by 1 cm under the action of the force. How much the spring with a stiffness of 100 N / m will lengthen by the action of the same force.

Data: k1 (stiffness of the first spring) = 50 N / m; Δх1 (deformation of the first spring) = 1 cm (in SI Δх1 = 0.01 m); k2 (stiffness of the second spring) = 100 N / m.

Since the force acting on the springs is constant, the deformation of the second spring can be expressed from the equality: k1 * Δх1 = F = k2 * Δх2, whence Δх2 = k1 * Δх1 / k2.

Calculation: Δх2 = 50 * 0.01 / 100 = 0.005 m (5 mm).

Answer: The second spring will lengthen by 5 mm.