Steel spring under the action of a force of 120 N lengthened by 15 mm. Find its stiffness.

Given:

F = 120 Newton – the force under the action of which the steel spring was lengthened;

dx = 15 mm (millimeters) – the amount of spring elongation.

It is required to determine k (Newton / meter) – the coefficient of stiffness of the steel spring.

We translate the units of measurement of length into the SI system:

dx = 15 mm = 15 * 10 ^ -3 = 15/1000 = 0.015 meters.

Then, to determine the stiffness coefficient, you must use the following formula:

F = dx * k, from here we find that:

k = F / dx = 120 / 0.015 = 8000 Newton / meter.

Answer: The coefficient of stiffness of a steel spring is 8000 Newton / meter.