7.2 Hooke’s law: stress and strain revisited  (Page 2/7)

How stiff are car springs?

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

Solution

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

   Substitute known values and solve $k$ :

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

Discussion

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.

Energy in hooke’s law of deformation

In order to produce a deformation, work must be done. That is, a force must be exerted through a distance, whether you pluck a guitar string or compress a car spring. If the only result is deformation, and no work goes into thermal, sound, or kinetic energy, then all the work is initially stored in the deformed object as some form of potential energy. The potential energy stored in a spring is ${\text{PE}}_{\text{el}}=\frac{1}{2}{\mathrm{kx}}^2$ . Here, we generalize the idea to elastic potential energy for a deformation of any system that can be described by Hooke’s law. Hence,

where ${\text{PE}}_{\text{el}}$ is the elastic potential energy    stored in any deformed system that obeys Hooke’s law and has a displacement $x$ from equilibrium and a force constant $k$ .

It is possible to find the work done in deforming a system in order to find the energy stored. This work is performed by an applied force $F_{\text{app}}$ . The applied force is exactly opposite to the restoring force (action-reaction), and so $F_{\text{app}}=\text{kx}$ . [link] shows a graph of the applied force versus deformation $x$ for a system that can be described by Hooke’s law. Work done on the system is force multiplied by distance, which equals the area under the curve or $(1/2){\mathrm{kx}}^2$ (Method A in the figure). Another way to determine the work is to note that the force increases linearly from 0 to $\mathrm{kx}$ , so that the average force is $(1/2)\mathrm{kx}$ , the distance moved is $x$ , and thus $W=F_{\text{app}}d=[(1/2)\text{kx}](x)=(1/2){\mathrm{kx}}^2$ (Method B in the figure).