A force of 30 N stretches the spring 5 cm. What is the force that pulls the spring 8 cm?

F1 = 30 N.

x1 = 5 cm = 0.05 m.

x2 = 8 cm = 0.08 m.

F2 -?

According to Hooke’s law F, the elastic force is determined by the formula: F = – k * x, where k is the stiffness of the spring, x is the absolute elongation of the spring, the sign “-” means that the elastic force is directed in the opposite direction to the elongation.

F1 = k * x1, F2 = k * x2.

k = F1 / x1.

F2 = F1 * x2 / x1.

F2 = 30 N * 0.08 m / 0.05 m = 48 N.

Answer: the force stretching the spring is F2 = 48 N.