/*! 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 39 To compress spring 1 by \(0.20 \... [FREE SOLUTION] | 91Ó°ÊÓ

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To compress spring 1 by \(0.20 \mathrm{m}\) takes \(150 \mathrm{J}\) of work. Stretching spring 2 by \(0.30 \mathrm{m}\) requires \(210 \mathrm{J}\) of work. Which spring is stiffer?

Short Answer

Expert verified
Spring 1 is stiffer (\(k_1 = 7500\, \mathrm{N/m}\) vs. \(k_2 = 4666.67\, \mathrm{N/m}\)).

Step by step solution

01

Determine Spring Constant of Spring 1

The work done on a spring is given by the formula \( W = \frac{1}{2} k x^2 \), where \( W \) is work, \( k \) is the spring constant, and \( x \) is the displacement. For spring 1, \( W = 150 \) J and \( x = 0.20 \) m. Substituting the values, we have:\[ 150 = \frac{1}{2} k (0.20)^2 \]Solving for \( k \), we get:\[ k = \frac{2 \times 150}{0.20^2} = \frac{300}{0.04} = 7500 \, \mathrm{N/m} \]
02

Determine Spring Constant of Spring 2

Using the same formula for spring 2, where \( W = 210 \) J and \( x = 0.30 \) m, we have:\[ 210 = \frac{1}{2} k (0.30)^2 \]Solving for \( k \), we get:\[ k = \frac{2 \times 210}{0.30^2} = \frac{420}{0.09} = 4666.67 \, \mathrm{N/m} \]
03

Compare the Spring Constants

The spring constant \( k \) provides a measure of stiffness. With values found, \( k_1 = 7500 \) N/m for spring 1 and \( k_2 = 4666.67 \) N/m for spring 2. Since \( k_1 > k_2 \), spring 1 is stiffer than spring 2.

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a fundamental concept in physics. It states that work done on an object results in a change in its energy. In the context of springs, this principle is used to determine how much work is needed to compress or stretch a spring a certain distance. In essence, when work is done on a spring, such as compressing or stretching it, energy is stored in the spring as potential energy. This energy can then be calculated using the formula for work: \[ W = \frac{1}{2} k x^2 \] Here, \( W \) stands for work, \( k \) is the spring constant, and \( x \) is the displacement of the spring from its equilibrium position. This equation reveals that the work done is proportional to the square of the displacement, showing the non-linear relationship between force applied and the distance moved. Understanding this principle helps in determining the amount of work required to change the state of a spring, such as in the given example, where we calculated the work necessary to compress or stretch two different springs.
Hooke's Law
Hooke's Law is a principle that states that the force required to compress or stretch a spring is directly proportional to the displacement of the spring. Mathematically, it is expressed as: \[ F = kx \] Where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the spring's original position. It is important to note that Hooke's law only holds true for the elastic limit of the spring. This law is pivotal when calculating the stiffness of a spring. The spring constant \( k \) is a measure of how stiff the spring is. A larger \( k \) value means a stiffer spring, which requires more force to achieve the same displacement. In our exercise, we used this principle to ascertain the spring constant for each spring, helping us identify which spring is stiffer based on the spring constants calculated.
Spring Mechanics
Spring mechanics involves studying how springs respond to forces, and the properties they exhibit. Springs can be used in various applications, from mechanical watches to vehicles, due to their ability to store and release energy efficiently. Key aspects of spring mechanics include:
  • The spring constant \( k \), which characterizes the stiffness of the spring.
  • The equilibrium position, which is the natural length of the spring when no force is applied.
  • The potential energy stored in a spring, which is given by the formula \( U = \frac{1}{2} k x^2 \).
In the given problem, we explored how the mechanics of springs present a quantifiable way to compare the stiffness of two different springs based on their work-energy relationship. Understanding these mechanics allows us to design and utilize springs in practical and effective ways across numerous industries and technologies.

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