What is the force constant for the suspension system of a car that settles 1.20 cm when an 80.0-kg person gets in?

Strategy

Consider the car to be in its equilibrium position x = 0 before the person gets in. The car then settles down 1.20 cm, which means it is displaced to a position x = -1.20 x 10^-2 m. At that point, the springs supply a restoring force F equal to the person's weight w = mg =(80.0 kg) (9.80 m/s²)= 784 N. We take this force to be F in Hooke's law. Knowing F and x, we can then solve the force constant k.