/*! 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 25 A pen contains a spring with a s... [FREE SOLUTION] | 91Ó°ÊÓ

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A pen contains a spring with a spring constant of \(250 \mathrm{N} / \mathrm{m}\). When the tip of the pen is in its retracted position, the spring is compressed \(5.0 \mathrm{mm}\) from its unstrained length. In order to push the tip out and lock it into its writing position, the spring must be compressed an additional \(6.0 \mathrm{mm} .\) How much work is done by the spring force to ready the pen for writing? Be sure to include the proper algebraic sign with your answer.

Short Answer

Expert verified
The work done by the spring is \(-0.012 \, \mathrm{J}\).

Step by step solution

01

Understand the Problem

The problem involves a pen with a spring that is initially compressed and will be compressed further to move the pen's tip into the writing position. We need to calculate the work done by the spring in this compression process.
02

Define the Known Quantities

We are given the spring constant, \(k = 250\, \mathrm{N/m}\), initial compression \(x_1 = 5.0\, \mathrm{mm} = 0.005\, \mathrm{m}\), and the total compression needed for writing \(x_2 = 5.0 + 6.0 = 11.0\, \mathrm{mm} = 0.011\, \mathrm{m}\).
03

Use Work Done by a Spring Formula

The work \(W\) done by a spring force when a spring is compressed from an initial position \(x_1\) to a final position \(x_2\) is calculated using the formula: \[ W = \frac{1}{2}k(x_1^2 - x_2^2) \] where \(k\) is the spring constant, \(x_1\) and \(x_2\) are the initial and final compression distances, respectively.
04

Perform the Calculation

Substitute the known values into the formula: \[ W = \frac{1}{2} \times 250 \times (0.005^2 - 0.011^2) \] Calculate \(0.005^2 = 0.000025\) and \(0.011^2 = 0.000121\), then find \(0.000025 - 0.000121 = -0.000096\). Hence, the work done is: \[ W = \frac{1}{2} \times 250 \times (-0.000096) = -0.012\, \mathrm{J} \]
05

Interpret the Result

The negative sign indicates the work is done by the spring force, which is in the direction opposite to the compression (i.e., the spring force resists the compression).

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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 is a measure of a spring's stiffness. It tells us how much force is needed to compress or stretch the spring by a certain amount. In the given problem, the spring constant is provided as \(250 \, \text{N/m}\). This means that for every meter the spring is compressed or stretched, 250 Newtons of force are required. The spring constant, denoted by \(k\), is crucial in determining how a spring will behave under compression or tension. It is a material and design property unique to each spring. A higher spring constant signifies a stiffer spring, while a lower value indicates a more flexible spring.
Compression Distance
Compression distance refers to how much a spring is compressed from its natural length. In this problem, the spring in the pen is initially compressed by \(5.0 \, \text{mm}\) and then further compressed by an additional \(6.0 \, \text{mm}\) for a total of \(11.0 \, \text{mm}\). It is important to convert these measurements into meters when using them in calculations, making it \(0.005 \, \text{m}\) and \(0.011 \, \text{m}\), respectively. Understanding compression distance is essential because it directly affects the force exerted by the spring and the work done, as the force is proportional to the change in compression distance.
Spring Force
The spring force is the force exerted by a spring when it is compressed or stretched. According to Hooke's Law, this force is directly proportional to the displacement of the spring from its rest position and is calculated as \(F = -kx\). Here, \(k\) is the spring constant, and \(x\) is the displacement (compression or stretch distance). The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement. In our context, the spring force resists the compression as the pen is made ready for writing. This resistance is precisely what allows the pen's mechanisms to function effectively.
Work-Energy Principle
The work-energy principle states that work done by forces on an object will result in a change in energy of that object. For springs, the work done is related to the potential energy stored in the spring. The work done by a spring force can be calculated with the formula: \[ W = \frac{1}{2}k(x_1^2 - x_2^2) \] This formula considers the squares of the initial \(x_1\) and final \(x_2\) compression distances. The work is negative in this problem, meaning the spring's force is opposing the compression. This concept helps us analyze mechanical systems where energy is stored and released, such as in pen mechanisms or shock absorbers. Understanding this principle allows us to comprehend how energy conservation and transformation occur in physical systems.

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Most popular questions from this chapter

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is \(f=3.0 \mathrm{Hz} ?\)

\(\mathrm{A}\) square plate is \(1.0 \times 10^{-2} \mathrm{m}\) thick, measures \(3.0 \times 10^{-2} \mathrm{m}\) on a side, and has a mass of \(7.2 \times 10^{-2} \mathrm{kg} .\) The shear modulus of the material is \(2.0 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) One of the square faces rests on a flat horizontal surface, and the coefficient of static friction between the plate and the surface is 0.91. A force is applied to the top of the plate, as in Figure 10.29 a. Determine (a) the maximum possible amount of shear stress, (b) the maximum possible amount of shear strain, and (c) the maximum possible amount of shear deformation \(\Delta X\) (see Figure \(10.29 b\) ) that can be created by the applied force just before the plate begins to move.

An 86.0 -kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of \(1.20 \times 10^{3} \mathrm{N} / \mathrm{m} .\) He accidentally slips and falls freely for \(0.750 \mathrm{m}\) before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of \(f=2.00 \mathrm{Hz}\). On this tray is an empty cup. Obtain the coefficient of static friction between the tray and the cup, given that the cup begins slipping when the amplitude of the motion is \(5.00 \times 10^{-2} \mathrm{m}\)

Multiple-Concept Example 6 presents a model for solving this problem. As far as vertical oscillations are concerned, a certain automobile can be considered to be mounted on four identical springs, each having a spring constant of \(1.30 \times 10^{5} \mathrm{N} / \mathrm{m} .\) Four identical passengers sit down inside the car, and it is set into a vertical oscillation that has a period of 0.370 s. If the mass of the empty car is 1560 kg, determine the mass of each passenger. Assume that the mass of the car and its passengers is distributed evenly over the springs.

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