Q. 38 - Precalculus Enhanced with Graphing Utilities

Question

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 $w=\frac{1}{2}k{x}^2$ ,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 $\tilde{w}=\frac{w}{2g}{v}^2$ where w = weight of the object (lb), g = acceleration due to gravity (32.2 ft/sec2), and v = 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 = 9450 lb/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.

Step-by-Step Solution

Answer

The least compression required for spring is $4.20ft$

Step 1. Given information

A parking garage has a spring shock absorber at end of ramp to stop runway cars. The spring has stiffness constant $k=9450lb/ft$and must be able to stop a 4000-lb car travelling at $25mph$ .

Step 2. Find the least compression required for spring

Using Hook's Law, we can show that the work done in compressing a spring a distance of x feet from it's at rest position is $w=\frac{1}{2}k{x}^2$

where, k is stiffness constant depending on the spring. 

The work done by a body in motion before it comes to rest is given by

$\tilde{w}=\frac{w}{2g}{v}^2$

where, w is weight of object (Ib), g is acceleration due to gravity , $(32.2ft/se{c}^2)$and v is object's velocity

$$\begin{gathered} v=25mph \\ =25\times 1.47ft/sec \\ =36.75ft/sec \end{gathered}$$

The work done in compressing a spring a distance of x feet from it's at rest position is

$$\begin{gathered} W=\frac{1}{k{x}^2} \\ =\frac{1}{2}\times 9450\times x^2 \\ =4725 x^2 \end{gathered}$$

The velocity is 

$$\begin{gathered} v=25mph \\ =25\times 1.47 ft/sec \\ =36.75ft/sec \end{gathered}$$

Step 3. Finding the work done by a body in motion before it comes to rest

$$\begin{gathered} \tilde{w}=\frac{w}{2g}{v}^2 \\ =\frac{4000}{2\times 32.2}\times 36.7\times 36.7 \\ =\frac{2693780}{32.2} \\ =83657.76 \end{gathered}$$

For the least compression the work done in compressing a spring shouls be greater that the work done by the body in motion.So,

$$\begin{gathered} w>\tilde{w} \\ 4725x^2>83657.76 \\ {x}^2>\frac{83657.76}{4725} \\ {x}^2>17.70 \\ x>\pm \sqrt{17.70} \\ x>\pm 4.20 \end{gathered}$$

Hence,the least compression required for spring is $4.20ft$