/*! 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] | 91Ó°ÊÓ

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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 \(196\,\text{J}\).

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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.

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