/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 A \(1000 \mathrm{~kg}\) car carr... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(1000 \mathrm{~kg}\) car carrying four \(82 \mathrm{~kg}\) people travels over a "washboard" dirt road with corrugations \(4.0 \mathrm{~m}\) apart. The car bounces with maximum amplitude when its speed is \(16 \mathrm{~km} / \mathrm{h}\). When the car stops, and the people get out, by how much does the car body rise on its suspension?

Short Answer

Expert verified
The car will rise by approximately 6.26 cm.

Step by step solution

01

Understanding the Problem

We need to find out how much the car body rises when the passengers exit the vehicle. This is related to how the suspension system behaves under different weights.
02

Convert Given Speed to Meters Per Second

The speed of the car is given as 16 km/h. Convert this to meters per second by using the conversion factor \[ 1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s} \]So, \[ 16 \, \text{km/h} \times \frac{1}{3.6} \approx 4.44 \, \text{m/s} \]
03

Calculate the Total Weight

Calculate the total weight of the car with passengers. The car weighs 1000 kg, and each of the 4 passengers weighs 82 kg, so:\[ \text{Total weight} = 1000 \, \text{kg} + 4 \times 82 \, \text{kg} = 1328 \, \text{kg} \]
04

Find the Natural Frequency of Bouncing

The car vibrates with the maximum amplitude when its speed is such that it resonates with the washboard bumps. The distance between the bumps is 4.0 m, so when the car travels at 4.44 m/s, the bouncing frequency resonates with the road corrugations. The frequency \( f \) can be calculated as \( f = \frac{v}{d} \) where \( v \) is the speed, and \( d \) is the distance between corrugations:\[ f = \frac{4.44}{4.0} = 1.11 \, \text{Hz} \]
05

Calculate the Change in Spring Compression

When the people exit, only the car weight remains to compress the springs. The change in spring compression results in the rise height. Knowing that the weight is proportional to the spring compression, and using the harmonic oscillator formula:\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]Solve for the spring constant \( k \) using the total initial mass:\[ k = \left(2\pi f\right)^2 \times m = \left(2\pi \times 1.11\right)^2 \times 1328 \approx 51550.86 \, \text{N/m} \]
06

Calculate the Suspension Rise

Use Hooke's Law which is \( F = k \cdot x \), where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the extension (or in this case, compression). Calculate the difference in compression when the passengers get out:\( F_{\text{initial}} = 1328 \, \text{kg} \times 9.8 \, \text{m/s}^2 \)\( F_{\text{final}} = 1000 \, \text{kg} \times 9.8 \, \text{m/s}^2 \)\( \Delta x = \frac{F_{\text{initial}} - F_{\text{final}}}{k} \)\[ \Delta x = \frac{(1328 - 1000) \, \text{kg} \times 9.8}{51550.86} \approx 0.0626 \, \text{m} \]
07

Provide the Final Answer

The car body will rise by approximately 0.0626 m (or 6.26 cm) when the passengers exit.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Constant
In a car suspension system, the spring constant, often represented as \( k \), is a crucial factor that determines how the car responds to weight changes and road conditions. The spring constant measures the stiffness of a spring, with a higher value indicating a stiffer spring. It's expressed in units of N/m (Newtons per meter). The spring constant determines how much a spring will compress or stretch under a given force. In this exercise, the spring constant was calculated using the formula: \[ k = \left(2\pi f\right)^2 \times m \] where \( f \) is the frequency of the bouncing and \( m \) is the total mass of the car with passengers. Understanding this concept helps in knowing how a vehicle's suspension deals with the load it carries. Proper knowledge of spring constant ensures a smoother ride by maintaining optimal suspension performance.
Resonance Frequency
Resonance in a mechanical system occurs when it vibrates at its natural frequency, leading to increased amplitude. In the context of a car, the resonance frequency matches the oscillation frequency of the bumps on the washboard road at certain vehicle speeds. This resonance can cause the car to bounce with maximum amplitude—meaning the car's suspension is oscillating in sync with road irregularities. Understanding resonance frequency is crucial because it helps in designing the suspension system to avoid uncomfortable conditions like excessive bouncing. To calculate this in our problem, the speed of the car and the distance between bumps were used: \[ f = \frac{v}{d} \] where \( v \) is the velocity and \( d \) is the distance between the corrugations.
Hooke's Law
Hooke's Law is a fundamental concept in physics applicable to elastic objects like car springs. According to this law, the force required to compress or extend a spring is directly proportional to the distance it is stretched or compressed. This relation is expressed mathematically as: \[ F = k \cdot x \] where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement of the spring from its equilibrium position.
For the car suspension, Hooke's Law helps to determine how much the car's body rises when the passengers exit. The change in force exerted by the car's weight difference before and after passengers leave the vehicle helps calculate the change in spring compression, showing by how much the car body rises.
Harmonic Oscillator
The car suspension system functions as a harmonic oscillator, which is a system that, once displaced, will experience periodic oscillations. The suspension can be modeled as a mass-spring-damper system, where the mass equates to the car and its passengers, the spring represents the suspension's springs, and damping accounts for energy losses like friction. A key feature of harmonic oscillators is their natural frequency, which is the frequency at which the system tends to oscillate in the absence of any driving or damping force. With the conditions given in the exercise, the car acts as a simple harmonic oscillator resonating due to its movement over corrugated terrain, showing maximum amplitude at its natural frequency.
Natural Frequency
Natural frequency is the rate at which a system oscillates in the absence of any external force or damping. For a car's suspension system, the natural frequency is vital for determining how the car will respond to different driving conditions and road surfaces. In this scenario, the car resonates when its speed synchronized with the corrugations on the road, leading us to calculate the natural frequency using the derived formula involving spring constant and mass: \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \] Here, \( k \) is the spring constant and \( m \) is the mass of the loaded car. Knowing this frequency can help engineers design suspensions that mitigate uncomfortable bouncing and improve ride comfort by aligning poorly with the natural frequency, thus minimizing resonance with typical road features.

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Most popular questions from this chapter

A \(50.0 \mathrm{~g}\) stone is attached to the bottom of a vertical spring and set vibrating. If the maximum speed of the stone is \(15.0 \mathrm{~cm} / \mathrm{s}\) and the period is \(0.500 \mathrm{~s}\), find the (a) spring constant of the spring, (b) amplitude of the motion, and (c) frequency of oscillation.

A \(10 \mathrm{~g}\) particle undergoes SHM with an amplitude of \(2.0 \mathrm{~mm}\), a maximum acceleration of magnitude \(8.0 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}\), and an unknown phase constant \(\phi\). What are (a) the period of the motion, (b) the maximum speed of the particle, and (c) the total mechanical energy of the oscillator? What is the magnitude of the force on the particle when the particle is at (d) its maximum displacement and (e) half its maximum displacement?

When the displacement in SHM is one-half the amplitude \(x_{m}\), what fraction of the total energy is (a) kinetic energy and (b) potential energy? (c) At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy?

A simple harmonic oscillator consists of an \(0.80 \mathrm{~kg}\) block attached to a spring \((k=200 \mathrm{~N} / \mathrm{m}) .\) The block slides on a horizontal frictionless surface about the equilibrium point \(x=0\) with a total mechanical energy of \(4.0 \mathrm{~J} .\) (a) What is the amplitude of the oscillation? (b) How many oscillations does the block complete in \(10 \mathrm{~s}\) ? (c) What is the maximum kinetic energy attained by the block? (d) What is the speed of the block at \(x=0.15 \mathrm{~m} ?\)

A \(55.0 \mathrm{~g}\) block oscillates in SHM on the end of a spring with \(k=1500 \mathrm{~N} / \mathrm{m}\) according to \(x=x_{m} \cos (\omega t+\phi) .\) How long does the block take to move from position \(+0.800 x_{m}\) to (a) position \(+0.600 x_{m}\) and (b) position \(-0.800 x_{m} ?\)

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