How much will the spring with a stiffness of 100 N / m lengthen under the action of a force of 20 N, 8N, 2N?

Given:

k = 100 N / m (Newton per meter) – the coefficient of stiffness of a certain spring;

F1 = 20 Newton – the magnitude of the first force;

F2 = 8 Newton – the value of the second force;

F3 = 2 Newton – the value of the third force.

It is required to determine dx1, dx2, dx3 (meter) – how much the springs will lengthen under the action of forces F1, F2 and F3.

Under the action of the first force, the spring will lengthen by:

dx1 = F1 / k = 20/100 = 0.2 meters.

Under the action of the second force, the spring will lengthen by:

dx2 = F2 / k = 8/100 = 0.08 meters.

Under the action of the third force, the spring will lengthen by:

dx3 = F3 / k = 2/100 = 0.02 meters.

Answer: under the action of a force of 20 Newton, the spring will lengthen by 0.2 meters, 8 Newton – by 0.08 meters, 2 Newton – by 0.02 meters.