Find the stiffness of a spring that has been lengthened by 10cm under a force of 5H.

Given:

dx = 10 centimeters = 0.1 meter – the amount by which the spring lengthened under the action of the force F;

F = 5 Newton – the force acting on the spring.

It is required to determine the stiffness of the spring k (N / m).

According to Hooke’s Law:

k = F / dx = 5 / 0.1 = 50 Newton / meter.

Answer: the spring rate is 50 N / m.