/*! 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 75 A light spring with spring const... [FREE SOLUTION] | 91Ó°ÊÓ

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A light spring with spring constant \(1.20 \times 10^{3} \mathrm{~N} / \mathrm{m}\) hangs from an elevated support. From its lower end hangs a second light spring, which has spring constant \(1.80 \times 10^{3} \mathrm{~N} / \mathrm{m}\). A \(1.50-\mathrm{kg}\) object hangs at rest from the lower end of the second spring. (a) Find the total extension distance of the pair of springs. (b) Find the effective spring constant of the pair of springs as a system. We describe these springs as being in series. Hint: Consider the forces on each spring scparately.

Short Answer

Expert verified
The total extension of the two springs is 0.02042 m, and the effective spring constant of the pair is 720 N/m.

Step by step solution

01

Find the Force Exerted by the Hanging Object

The object with mass 1.50 kg exerts a force downwards due to gravity. This force can be calculated using the equation for gravitational force, \( F = mg \), where m is the mass of the object and g is acceleration due to gravity, which is approximately 9.8 m/s^2. The force equals \( 1.50 kg * 9.8 m/s^2 = 14.7 N \). This is the total force stretching both springs.
02

Determine the Extension of Each Spring

Since the total force stretching both springs is equal to the weight of the hanging object (14.7 N), each spring will extend according to its spring constant (k) and the exerted force. Using Hooke’s law (\( F=kx \)), where x is the extension of the spring, we can determine the extension of each spring by rearranging the equation to \(x=F/k\). The extension of the first spring is \(14.7 N / 1200 N/m = 0.01225 m\) and the second spring is \(14.7 N / 1800 N/m = 0.00817 m\).
03

Calculate the Total Extension

The total extension of the pair of springs is the sum of the extensions of each individual spring, which equals \(0.01225 m + 0.00817 m = 0.02042 m \).
04

Calculate the Effective Spring Constant

When two springs are in series, their effective spring constant \( k_{eff} \) is given by \(1 / k_{eff} = 1/k1 + 1/k2\). So, \(1 / k_{eff} = 1/1200 N/m + 1/1800 N/m\). Thus, the effective spring constant is \(k_{eff} = 1/(1/1200 N/m + 1/1800 N/m) = 720 N/m\).

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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, denoted as \( k \), is a value that describes the stiffness of a spring. It's an essential component in understanding spring mechanics. The stiffer the spring, the higher the spring constant. This is because a stiff spring requires more force to stretch or compress it by a certain distance.

- Measured in newtons per meter (\( N/m \)), the spring constant tells us how much force is needed to change the length of the spring.- For example, in our exercise, the first spring has a spring constant of \( 1.20 \times 10^{3} \mathrm{~N/m} \). This means it takes 1200 N of force to stretch that spring by one meter.

Understanding the spring constant helps us determine how a spring will react under different forces.
Hooke's Law
Hooke's Law is a fundamental concept when studying spring mechanics and describes how the force applied to a spring relates to the distance the spring stretches or compresses. This relationship is given by the equation \( F = kx \), where:
  • \( F \) is the force applied to the spring, measured in newtons (N).
  • \( k \) is the spring constant, a measure of the spring's stiffness.
  • \( x \) is the distance the spring is stretched or compressed, in meters (m).
This law tells us that the force needed to stretch or compress a spring is directly proportional to its extension or compression. Hence, the further you pull, the more force is required. In our exercise, Hooke's Law helps us calculate the extension of each spring in response to the gravitational force applied by the mass hanging from the springs.
Effective Spring Constant
When dealing with systems that involve multiple springs, determining the effective spring constant is crucial for understanding the system's overall behavior. In our exercise, two springs are arranged in series. The effective spring constant \( k_{eff} \) in such a system can be found using the formula:\[\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}\]
  • Here, \( k_1 \) and \( k_2 \) are the individual spring constants of the springs in series.
  • The effective spring constant provides a single value that represents how the system behaves as a whole.
Using this principle, we see that even though each spring has its own stiffness, the system behaves as if it were a single spring with a lower spring constant, reflecting the combined effect of both springs.
Gravitational Force
Gravitational force is the force exerted by gravity on an object, directly related to its mass. This force can be calculated using the formula \( F = mg \), where:
  • \( F \) is the gravitational force in newtons (N).
  • \( m \) is the mass of the object, measured in kilograms (kg).
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \).
In the context of our exercise, the hanging object exerts a gravitational force downwards, which is responsible for stretching the springs. This force is crucial for determining how much the springs will extend under the object's weight, as it translates to the total force applied to the spring system.

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