Under the force of 320 N, the shock absorber spring was compressed by 9 mm

Under the force of 320 N, the shock absorber spring was compressed by 9 mm. How many millimeters will the spring be compressed under a load of 1.6 kN?

F1 = 320 N.

x1 = 9 mm = 0.009 m.

F2 = 1.6 kN = 1600 N.

x2 -?

When the spring is compressed, the load force F is compensated by the elastic force Fcont, which occurs in the spring: F = Fcont.

F1 = Fcont1.

F2 = Fcont2.

According to Hooke’s law, the elastic force Fel is directly proportional to the elongation of the spring: Fel = k * x, where k is the stiffness of the spring.

F1 = k * x1.

F2 = k * x2.

x2 = F2 / k.

k = F1 / x1.

x2 = F2 * x1 / F1.

x2 = 1600 N * 0.009 m / 320 N = 0.045 m.

Answer: the shock absorber spring will be compressed by x2 = 0.045 m under the load.