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(a) Suppose you cut a massless ideal spring in half. If the full spring had a force constant k, what is the force constant of each half, in terms of \(k\)? (\(Hint\): Think of the original spring as two equal halves, each producing the same force as the entire spring. Do you see why the forces must be equal?) (b) If you cut the spring into three equal segments instead, what is the force constant of each one, in terms of \(k\)?

Short Answer

Expert verified
For half-spring, \( k_{half} = 2k \); for one-third spring, \( k_{third} = 3k \).

Step by step solution

01

Understanding Hooke's Law

Hooke's Law states that the force exerted by a spring is proportional to the displacement and the spring constant: \( F = kx \). Here, \( k \) is the spring constant that quantifies the spring's stiffness.
02

Analyzing a Spring Cut in Half

When you cut a spring in half, each half still produces the same overall force when stretched or compressed by a given amount, as the total force is always needed to restore the system to equilibrium. Thus, if the original full-length spring has a spring constant \( k \) and is viewed as two springs in series, the force constant for each half can be determined.
03

Deriving the Force Constant for Half a Spring

For two springs in series, the effective spring constant \( k_{eff} \) is given by \( \frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} \). Assuming each half has the same spring constant \( k_{half} \), we have \( \frac{1}{k} = \frac{1}{k_{half}} + \frac{1}{k_{half}} = \frac{2}{k_{half}} \). Solving this gives \( k_{half} = 2k \).
04

Spring Cut into Three Equal Segments

When the spring is cut into three equal segments, they must all work together when placed in series to equal the original spring constant \( k \).
05

Deriving the Force Constant for One Third of a Spring

Using the formula for springs in series, \( \frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} \). Assuming each of the three segments has the same spring constant \( k_{third} \), we have \( \frac{1}{k} = \frac{3}{k_{third}} \). Solving this results in \( k_{third} = 3k \).

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Key Concepts

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

Hooke's Law
Hooke's Law is a fundamental principle in physics that describes the behavior of springs. It states that the force required to compress or extend a spring is directly proportional to the displacement it experiences. This can be mathematically expressed as \( F = kx \), where \( F \) is the force applied to the spring, \( k \) is the spring constant, and \( x \) is the displacement from the spring's equilibrium position. The spring constant, \( k \), is a measure of a spring's stiffness. The larger the value of \( k \), the stiffer the spring and the more force is needed to achieve the same displacement.

Understanding Hooke's Law is crucial when working with springs, as it allows you to calculate the force needed for a specific displacement, which is essential in many engineering and physics applications. It also enables the determination of the spring constant \( k \), which is a characteristic of the spring itself and remains constant unless the spring is physically altered.
Series Springs
When springs are connected in series, their overall behavior changes compared to a single spring. In this configuration, each spring contributes to the total displacement when a force is applied. As a result, the system is less stiff as compared to individual springs alone. The effective spring constant for springs in series can be found using the formula: \[ \frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} + \ldots + \frac{1}{k_n} \] where \( k_{eff} \) is the effective spring constant of the entire system, and \( k_1, k_2, \ldots, k_n \) are the constants of individual springs.

For example, if you cut a spring into two equal halves and place them in series, the system as a whole behaves like a spring with a smaller spring constant than the original. When a spring is split into multiple segments and placed in series, each segment contributes equally to reduce the overall stiffness of the spring system. This principle was used in the exercise to determine that if a spring is cut into halves or thirds, each resulting piece has a different spring constant than any single piece would exhibit alone.
Spring Stiffness
Spring stiffness is an essential characteristic of a spring, denoted by the spring constant \( k \). This stiffness determines how much force is needed to stretch or compress a spring by a given distance. Spring stiffness directly affects how the spring behaves in various configurations, such as when springs are arranged in series or parallel.

In the exercise, when the spring is cut into different segments, the stiffness of each new spring segment changes in relation to the original spring. For instance, if the spring is cut into two halves and treated as series springs, each half has a spring constant of \( 2k \), meaning each part is independently stiffer and compensates for being in series by providing half the force. However, cutting the spring into three pieces results in each having a spring constant of \( 3k \) when put in series, showing that smaller segments proportionally increase the stiffness to maintain the spring's behavior.
  • This illustrates how manipulating spring configuration can dramatically change system attributes.
  • Understanding spring stiffness is integral to designing mechanical systems involving springs.
Overall, these factors highlight how spring stiffness and series configurations interplay to impact the mechanical behavior of springs.

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Most popular questions from this chapter

A 5.00-kg block is moving at \(\upsilon_0\) \(=\) 6.00 m/s along a frictionless, horizontal surface toward a spring with force constant \(k\) = 500 N/m that is attached to a wall (\(\textbf{Fig. P6.79}\)). The spring has negligible mass. (a) Find the maximum distance the spring will be compressed. (b) If the spring is to compress by no more than 0.150 m, what should be the maximum value of \(\upsilon_0\)?

The spring of a spring gun has force constant \(k = 400\) N/m and negligible mass. The spring is compressed 6.00 cm, and a ball with mass 0.0300 kg is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 cm long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so that the barrel is horizontal. (a) Calculate the speed with which the ball leaves the barrel if you can ignore friction. (b) Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 N acts on the ball as it moves along the barrel. (c) For the situation in part (b), at what position along the barrel does the ball have the greatest speed, and what is that speed? (In this case, the maximum speed does not occur at the end of the barrel.)

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