/*! 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 26 Shock absorber A heavy-duty shoc... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 26

Shock absorber A heavy-duty shock absorber is compressed \(2 \mathrm{cm}\) from its equilibrium position by a mass of \(500 \mathrm{kg} .\) How much work is required to compress the shock absorber \(4 \mathrm{cm}\) from its equilibrium position? (A mass of 500 kg exerts a force (in newtons) of \(\left.500 g, \text { where } g=9.8 \mathrm{m} / \mathrm{s}^{2} .\right)\)

Short Answer

Expert verified
Answer: The work required to compress the shock absorber by 4 cm from its equilibrium position is 196 J.

Step by step solution

01

Find the spring constant

First, we need to find the spring constant of the shock absorber (k). We know that \(F = kx\), where \(F\) is the force exerted on the shock absorber, \(k\) is the spring constant, and \(x\) is the displacement from equilibrium position. Given, Mass \(m = 500\,\text{kg}\) Gravitational acceleration \(g = 9.8\,\text{m/s}^2\) Compression \(x = 2\,\text{cm} = 0.02\,\text{m}\) Let's first find the force exerted by the mass, \(F = mg\).
02

Calculate the force exerted by the mass

We can calculate the force exerted by the mass using the formula: \(F = mg\) \(F = (500\,\text{kg})(9.8\,\text{m/s}^2) = 4900\,\text{N}\)
03

Calculate the spring constant

Now, we can use Hooke's Law to find the spring constant, k: \(F = kx\) \(k = \frac{F}{x} = \frac{4900\,\text{N}}{0.02\,\text{m}} = 245000\,\text{N/m}\)
04

Use work-energy theorem to find the work required

The work-energy theorem states that the work required to compress a spring is given by: \(W = \frac{1}{2}kx^2\), where \(W\) is the work required, \(k\) is the spring constant, and \(x\) is the displacement. We are asked to find the work required to compress the shock absorber by \(4\,\text{cm}\) (\((0.04\,\text{m})\) from its equilibrium position.
05

Calculate the work required

Let's calculate the work required using the formula: \(W = \frac{1}{2}kx^2\) \(W = \frac{1}{2}(245000\,\text{N/m})(0.04\,\text{m})^2\) \(W = \frac{1}{2}(245000\,\text{N/m})(0.0016\,\text{m}^2)\) \(W = 196\,\text{J}\) The work required to compress the shock absorber \(4\,\mathrm{cm}\) from its equilibrium position is \(196\,\text{J}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Spring Constant
The spring constant, often denoted by the symbol \( k \), represents how stiff or flexible a spring is. It is a measure of the resistance a spring has to being compressed or elongated. The unit for spring constant is \( \text{N/m} \) (newtons per meter).

In our example with the shock absorber, we used the formula from Hooke's Law, \( F = kx \), to find the spring constant. Here:
\( F \) is the force applied to the spring
\( k \) is the spring constant
\( x \) is the displacement from the equilibrium position
The force exerted by the mass of \( 500 \) kg due to gravity was \( 4900 \) N, calculated using \( F = mg \) where \( g = 9.8 \ \text{m/s}^2 \), and \( x \) was \( 0.02 \) m. This gave us a spring constant, \( k = \frac{4900}{0.02} = 245000 \ \text{N/m} \).

This high value means the shock absorber is quite stiff and requires a significant force for further compression.
Hooke's Law
Hooke's Law describes the behavior of springs and other elastic objects by relating the force exerted on them to the amount of extension or compression they undergo. Simply put, Hooke's Law states that the force required to extend or compress a spring by a distance \( x \) is proportional to that distance. Mathematically, it is expressed as:

\[ F = kx \]

It means if you double the distance \( x \) by which the spring is compressed, the force \( F \) exerted by the spring doubles as well, assuming \( k \) remains constant.

In our exercise, Hooke's Law helped to identify the spring constant of the shock absorber. After knowing the spring constant, we can predict how the spring would react to different applied forces or compression distances. This law is fundamental in understanding the mechanics of springs and is widely used in physics and engineering fields.
Work-Energy Theorem
The work-energy theorem is a fundamental principle in physics, which states that the work done on an object is equal to the change in its kinetic energy. For springs, this theorem can be used to calculate the work required to compress or stretch the spring by a particular amount.

The formula derived from this theorem for a spring is:

\[ W = \frac{1}{2}kx^2 \]

\( W \) is the work done, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position. This equation tells us that the work required increases with the square of the displacement, which means small increases in compression lead to larger increases in work required due to the squaring effect.

In the task at hand, we were asked to determine the work needed to compress the shock absorber by 4 cm. Using the work-energy theorem and the spring constant from earlier calculations, we found that \( 196 \) joules of work is needed to compress the shock absorber to this extent. These calculations help design efficient systems in vehicles, ensuring safety and comfort by controlling forces during motion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Compressing and stretching a spring Suppose a force of \(15 \mathrm{N}\) is required to stretch and hold a spring \(0.25 \mathrm{m}\) from its equilibrium position. a. Assuming the spring obeys Hooke's law, find the spring constant \(k\) b. How much work is required to compress the spring \(0.2 \mathrm{m}\) from its equilibrium position? c. How much additional work is required to stretch the spring \(0.3 \mathrm{m}\) if it has already been stretched \(0.25 \mathrm{m}\) from its equilibrium position?

Calculating work for different springs Calculate the work required to stretch the following springs \(1.25 \mathrm{m}\) from their equilibrium positions. Assume Hooke's law is obeyed. a. A spring that requires \(100 \mathrm{J}\) of work to be stretched \(0.5 \mathrm{m}\) from its equilibrium position b. A spring that requires a force of \(250 \mathrm{N}\) to be stretched \(0.5 \mathrm{m}\) from its equilibrium position

Without evaluating integrals, prove that $$\int_{0}^{2} \frac{d}{d x}\left(12 \sin \pi x^{2}\right) d x=\int_{0}^{2} \frac{d}{d x}\left(x^{10}(2-x)^{3}\right) d x$$.

Surface-area-to-volume ratio (SAV) In the design of solid objects (both artificial and natural), the ratio of the surface area to the volume of the object is important. Animals typically generate heat at a rate proportional to their volume and lose heat at a rate proportional to their surface area. Therefore, animals with a low SAV ratio tend to retain heat, whereas animals with a high SAV ratio (such as children and hummingbirds) lose heat relatively quickly. a. What is the SAV ratio of a cube with side lengths \(R ?\) b. What is the SAV ratio of a ball with radius \(R ?\) c. Use the result of Exercise 38 to find the SAV ratio of an ellipsoid whose long axis has length \(2 R \sqrt[3]{4},\) for \(R \geq 1\) and whose other two axes have half the length of the long axis. (This scaling is used so that, for a given value of \(R,\) the volumes of the ellipsoid and the ball of radius \(R\) are equal.) The volume of a general ellipsoid is \(V=\frac{4 \pi}{3} A B C,\) where the axes have lengths \(2 A, 2 B,\) and \(2 C .\) d. Graph the SAV ratio of the ball of radius \(R \geq 1\) as a function of \(R\) (part (b)) and graph the SAV ratio of the ellipsoid described in part (c) on the same set of axes. Which object has the smaller SAV ratio? e. Among all ellipsoids of a fixed volume, which one would you choose for the shape of an animal if the goal were to minimize heat loss?

Work by two different integrals A rigid body with a mass of 2 kg moves along a line due to a force that produces a position function \(x(t)=4 t^{2},\) where \(x\) is measured in meters and \(t\) is measured in seconds. Find the work done during the first 5 s in two ways. a. Note that \(x^{\prime \prime}(t)=8 ;\) then use Newton's second law \(\left(F=m a=m x^{m}(t)\right)\) to evaluate the work integral \(W=\int_{x_{0}}^{x_{1}} F(x) d x,\) where \(x_{0}\) and \(x_{f}\) are the initial and final positions, respectively. b. Change variables in the work integral and integrate with respect to \(t .\) Be sure your answer agrees with part (a).
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.