Answer
5.4 mm.
Work Step by Step
Find the resonant angular frequency of the spring. This is the same as the tire’s angular frequency when traveling at 90.0 km/h, which is 25 m/s.
$$\omega = \frac{v}{r}=\frac{25\;m/s}{0.5 \times 0.58\;m}=86.21\;rad/s$$
Relate the spring constant to this angular frequency, and the total mass of the tire and wheel.
$$\omega^2=\frac{k}{m}$$
$$k=\omega^2 m=(86.21\;rad/s)^2(17.0\;kg)=1.263\times10^5\;N/m$$
Use Hooke’s law, $F=kx$, to find the distance the spring is compressed when the mass is added. Consider the magnitude of the force and of the distance. Note that this spring supports only 0.25 of the total mass.
$$x=\frac{F}{k}=\frac{0.25mg}{k}=\frac{0.25(280\;kg)(9.80\;m/s^2)}{ 1.263\times 10^5\;N/m }\approx 5.4\times 10^{-3}m$$
