Chapter 3.5, Problem 38AE

Runaway Car Using Hooke's Law, we can show that the work done in compressing a spring a distance of $x$ feet from its at-rest position is $W\text{ = }\frac{1}{2}k{x}^2$, where $k$ is a stiffness constant depending on the spring. It can also be shown that the work done by a body in motion before it comes to rest is given by $\tilde{W}\text{ = }\frac{w}{2g}{v}^2$, Where $w\text{ =}$ weight of the object (in Ibs), $g\text{ =}$ acceleration due to gravity( $32.2ft/{\sec }^2$), and $v\text{ =}$ object’s velocity (in ft/sec). A parking garage has a spring shock absorber at the end of a ramp to stop runaway cars. The spring has a stiffness constant $k=9450lb/ft$ and must be able to stop a 4000-lb car traveling at 25 mph. What is the least compression required of the spring? Express your answer using feet to the nearest tenth.

[Hint: Solve $W>\tilde{W},x\ge \text{ 0}$].

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