How Stiff Are Car Springs?

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 $x=-1\text{.}\text{20}\times {\text{10}}^{-2}\text{m}$. At that point, the springs supply a restoring force $F$ equal to the person’s weight $w=\text{mg}=\left(\text{80}\text{.}0\mathrm{<mspace><mspace\; width="0.25em"><mtext>\; kg</mtext></mspace></mspace>}\right)\left(9\text{.}\text{80}\mathrm{<mspace\; width="0.25em"><msup><mtext>m/s</mtext><mrow><mn>2</mn></mrow></msup></mspace>}\right)=\text{784}\mathrm{<mspace\; width="0.25em"><mtext>N</mtext></mspace>}$. We take this force to be $F$ in Hooke’s law. Knowing $F$ and $x$, we can then solve the force constant $k$.

1. Solve Hooke’s law, $F=-\text{kx}$, for $k$:
   $$k=-\frac{F}{x}.$$

   16.2

   Substitute known values and solve $k$:

   $$\begin{array}{lll}k& =& -\frac{\text{784}\text{N}}{-1\text{.}\text{20}\times {\text{10}}^{-2}\text{m}}\\ & =& 6\text{.}\text{53}\times {\text{10}}^4\text{N/m}.\end{array}$$

   16.3

Note that $F$ and $x$ have opposite signs because they are in opposite directions—the restoring force is up, and the displacement is down. Also, note that the car would oscillate up and down when the person got in if it were not for damping (due to frictional forces) provided by shock absorbers. Bouncing cars are a sure sign of bad shock absorbers.