# A car with bad shocks has a mass of 1500 kg. Before you go for a drive with three of your friends...

## Question:

A car with bad shocks has a mass of 1500 kg. Before you go for a drive with three of your friends you notice that the car sinks a distance of 6.0 cm when all four of you get in the car. You estimate that the four of you together have a mass of 271 kg. As you are driving down the highway at 65 mph you notice that the car is starting to bounce up and down with a large amplitude. You realize that there is a periodic series of small bumps and dips in the road that is driving the bouncing.

What is the distance (in meters) between adjacent bumps on the road, assuming that damping by the shocks is negligible?

## Spring Constant:

In the simple word of physics, the total force needed to extend a spring corresponding to the unit length of the spring is generally represented by spring constant. The spring constant of a spring is usually indicated by pound-force per feet.

## Answer and Explanation: 1

**Given data**

- The mass of the car with bad shock is {eq}m=1500\ \text{kg} {/eq}

- The distance sinks by car is {eq}x=6\ \text{cm}=6\times {{10}^{-2}}\ \text{m} {/eq}

- The total mass is {eq}{{m}_{t}}=271\ \text{kg} {/eq}

- The total speed in the highway is {eq}V=65\ \text{mph}=29.058\ \text{m/s} {/eq}

By using the following expression, the spring constant of the shocks is calculated as,

{eq}\begin{align*} F &=kx \\ {{m}_{t}}g &=kx \\ \left( 271\ \text{kg} \right)\left( 9.81\ \text{m/}{{\text{s}}^{2}} \right) &=k\left( 6\times {{10}^{-2}}\ \text{m} \right) \\ k &=\dfrac{\left( 271\ \text{kg} \right)\left( 9.81\ \text{m/}{{\text{s}}^{2}} \right)}{\left( 6\times {{10}^{-2}}\ \text{m} \right)} \\ k &=44308.5\ \text{N/m} \\ \end{align*} {/eq}

By using the following expression, the total mass is calculated as,

{eq}\begin{align*} {{m}_{\text{total}}} &=m+{{m}_{t}} \\ &=1500\ \text{kg}+271\ \text{kg} \\ &=1771\ \text{kg} \\ \end{align*} {/eq}

By using the following expression, the time period is calculated as,

{eq}\begin{align*} T &=2\pi \sqrt{\dfrac{{{m}_{\text{total}}}}{k}} \\ &=2\pi \sqrt{\dfrac{\left( 1771\ \text{kg} \right)}{\left( 44308.5\ \text{N/m} \right)}} \\ &=1.26\ \text{s} \\ \end{align*} {/eq}

By using the following relation, the distance between adjacent bumps is calculated as,

{eq}\begin{align*} V &=\dfrac{\text{Distance}}{T} \\ \left( 20.058\ \text{m/s} \right) &=\dfrac{\text{Distance}}{\left( 1.26\ \text{s} \right)} \\ \text{Distance} &=\left( 20.058\ \text{m/s} \right)\left( 1.26\ \text{s} \right) \\ \text{Distance} &=25.27\ \text{m} \\ \end{align*} {/eq}

**Thus, the distance between the adjacent bumps is 25.27 meter.**

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from

Chapter 17 / Lesson 11In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.