When the door was opened, the length of the door spring increased by 12cm

When the door was opened, the length of the door spring increased by 12cm, while the spring force was 4N. Find the stiffness of the spring.

Given:

dx = 12 centimeters = 0.12 meters – the length by which the door spring has increased;

F = 4 Newton – door spring elastic force.

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

To determine the stiffness coefficient of a door spring, you need to use Hooke’s law:

F = k * x, hence:

k = F / x = 4 / 0.12 = 33.3 Newton / meter.

Answer: The coefficient of stiffness of the spring is 33.3 Newton / meter.