Question

Challenge Problem Runaway Car Using Hooke's Law, we can show that the work \(W\) 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 \(\tilde{W}=\frac{w}{2 g} v^{2},\) where \(w=\) weight of the object (in Ib), \(g=\) acceleration due to gravity \(\left(32.2 \mathrm{ft} / \mathrm{s}^{2}\right),\) and \(v=\) object's velocity \((\mathrm{in} \mathrm{ft} / \mathrm{s})\). A parking garage has a spring shock absorber at the end of a ramp to stop runaway cars. The spring has a stiffness constant \(k=9450 \mathrm{lb} / \mathrm{ft}\) and must be able to stop a \(4000-\mathrm{lb}\) car traveling at \(25 \mathrm{mph}\). What is the least compression required of the spring? Express your answer using feet to the nearest tenth. Source: www.sciforums com

Short answer

The least compression required of the spring is 4.9 feet.

Step-by-step solution

Convert car's velocity to feet per second

First, convert the car's velocity from miles per hour (mph) to feet per second (ft/s). Given 1 mile = 5280 feet and 1 hour = 3600 seconds:\[ v = 25 \text{ mph} \times \frac{5280 \text{ ft}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ s}} = 36.67 \text{ ft/s} \]

Calculate the work done by the car in motion

Using the formula \( \tilde{W} = \frac{w}{2g} v^2 \), substitute the known values: weight \( w = 4000 \, \text{lb} \), gravitational acceleration \( g = 32.2 \text{ ft/s}^2 \), and velocity \( v = 36.67 \text{ ft/s} \):\[ \tilde{W} = \frac{4000}{2 \times 32.2} \times (36.67)^2 \] Calculating this, we get:\[ \tilde{W} = \frac{4000}{64.4} \times 1344.15 = 83.2 \times 1344.15 = 111818.88 \text{ ft-lb} \]

Apply Hooke's Law to find spring compression

Using the work done in compressing a spring formula \( W = \frac{1}{2} k x^2 \), set it equal to the work done by the car in motion: \( \tilde{W} = W \):\[ \frac{1}{2} k x^2 = 111818.88 \text{ ft-lb} \] Substituting the stiffness constant \( k = 9450 \text{ lb/ft} \), we get:\[ \frac{1}{2} \times 9450 \times x^2 = 111818.88 \] Solving for \( x^2 \):\[ 4725 x^2 = 111818.88 \quad \rightarrow \quad x^2 = 23.67 \] Taking the square root of both sides:\[ x = \sqrt{23.67} = 4.9 \text{ feet} \]

Key concepts

Kinematics

Kinematics is the branch of physics that focuses on the concepts of motion without needing to worry about the forces causing it. In this problem, we first used kinematics to convert the car’s speed from miles per hour (mph) to feet per second (ft/s).

The conversion is crucial because our other formulas use feet and seconds as units. We converted 25 mph to ft/s by noting that 1 mile equals 5280 feet and 1 hour equals 3600 seconds. Therefore:

• Multiply 25 mph by 5280 ft/mile to get total feet/hour.
• Then, divide by 3600 seconds/hour to find feet/second.

This brings the speed to approximately 36.67 ft/s, setting the stage for applying energy-related formulas.

Energy Conservation

The principle of energy conservation states that energy cannot be created or destroyed, only transformed from one form to another. In our exercise, two primary forms of energy are considered:

• Kinetic Energy: This is the energy a body possesses due to its motion. It's given by the formula \[\tilde{W} = \frac{w}{2g} v^2\]
• Elastic Potential Energy: This is stored energy in a spring when compressed, described by Hooke’s Law as \[W = \frac{1}{2} k x^2\]

We used the kinetic energy of the car right before it stops to find out how much energy the spring needs to absorb in the form of elastic potential energy. By equating these two energies, \[ \frac{w}{2 g} v^2 = \frac{1}{2} k x^2\], we set up the equation to solve for the spring compression distance, \(x\).

Spring Compression

Spring compression deals with how much a spring must be compressed to absorb a certain amount of energy. According to Hooke’s Law,\( W = \frac{1}{2} k x^2 \), where:

• \(W\) is the work done in compressing the spring.
• \(k\) is the spring constant, which depends on the stiffness of the spring.
• \(x\) is the distance the spring is compressed.

To find the compression needed to stop the car, we set the work done by the moving car (kinetic energy) equal to the work done in compressing the spring (elastic potential energy). Substituting the given values and solving for \(x\), we obtained:

\[ x = \frac{1}{\text{stiffness constant}} \times \text{energy absorbed} \ = \frac{1}{9450} \times 111818.88 = 4.9 \text{ feet} \]

This tells us that the spring must be compressed 4.9 feet to safely stop the car, demonstrating a practical application of Hooke's Law in energy transformation scenarios.