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Chapter 10: Problem 25

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 spring does -0.012 J of work.

Step by step solution

01

Understand the Problem

The problem involves a spring that is initially compressed and then needs to be compressed further to move the pen tip to its writing position. We are tasked with finding the work done by the spring force as it further compresses.
02

Convert Units

The original compression and additional compression of the spring are given in millimeters. Convert these to meters to use standard SI units. Thus, the initial compression is 5.0 mm = 0.005 m, and the additional compression is 6.0 mm = 0.006 m.
03

Apply Work-Energy Principle for Springs

The work done by a spring is given by the formula: \[ W = \frac{1}{2} k (x_f^2 - x_i^2) \] where \( W \) is the work done by the spring, \( k \) is the spring constant, \( x_i \) is the initial compression, and \( x_f \) is the final compression.
04

Calculate Initial and Final Compression

The initial compression \( x_i \) is 0.005 m, and the final compression \( x_f \) is 0.005 m + 0.006 m = 0.011 m.
05

Calculate the Work Done by the Spring

Substitute the known values into the work formula:\[ W = \frac{1}{2} \times 250 \times (0.011^2 - 0.005^2) \]Calculate the squares and the difference: \[ W = \frac{1}{2} \times 250 \times (0.000121 - 0.000025) \]\[ W = \frac{1}{2} \times 250 \times 0.000096 \]\[ W = 0.012 \text{ Joules} \] The work done by the spring force is -0.012 J, as the spring does work on the pen as it compresses.

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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 key concept in understanding how energy is transferred by forces acting over distances.This principle states that the work done by all forces acting on an object equals the change in its kinetic energy.
However, when it comes to springs, we focus on potential energy changes.In the realm of spring mechanics, the work done by or on a spring is related to the change in the spring's potential energy.A compressed or stretched spring stores potential energy, which can do work when the spring returns to its natural length.
The work done by a spring as it compresses or expands can be calculated using the formula:
  • \( W = \frac{1}{2} k (x_f^2 - x_i^2) \)
where:
  • \( W \) is the work done,
  • \( k \) is the spring constant,
  • \( x_i \) is the initial displacement from the unstressed position,
  • \( x_f \) is the final displacement.
When a spring is compressed further, it increases the potential energy stored. Hence, the work done by the spring is negative, indicating that energy is leaving the system to perform work on the surroundings, like moving the pen tip outwards.
Spring Constant
The spring constant, represented by the symbol \( k \), is a measure of a spring's stiffness. This constant is crucial for determining how much force is needed to compress or extend the spring by a unit length.The higher the value of \( k \), the stiffer the spring, meaning it requires more force to compress or stretch it.In our exercise, the spring constant is given as \( 250 \text{ N/m} \).This means that every meter of compression or extension requires \( 250 \text{ Newtons} \) of force.Understanding the spring constant helps in calculating how much energy is stored in the spring when it is compressed or extended.
It is a fundamental part of the work-energy equation for springs and provides insight into how much potential energy can be harnessed to do work.
Compression
Compression refers to the reduction in length or volume a spring undergoes when a force is applied.When a spring is compressed, it stores potential energy proportional to the amount of compression.The energy stored can be released to do work.In our exercise, we have two stages of compression:
  • The initial compression from the natural length, which is \( 5.0 \, \text{mm} \) or \( 0.005 \, \text{m} \).
  • An additional compression needed to lock the pen tip in place, which is \( 6.0 \, \text{mm} \) or \( 0.006 \, \text{m} \).
To find the total compression exerted on the spring, we add these values to get \( 0.011 \, \text{m} \).This total change is used in the work-energy equation to calculate the work done by the spring.
Compression in spring mechanics is essential for devices like pens, where energy must be stored and released efficiently, allowing mechanical parts to function smoothly.

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Most popular questions from this chapter

A spring lies on a horizontal table, and the left end of the spring is attached to a wall. The other end is connected to a box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. The mass of the box is \(0.80 \mathrm{kg}\), and the spring has a spring constant of \(59 \mathrm{N} / \mathrm{m}\). The coefficient of static friction between the box and the table on which it rests is \(\mu_{\mathrm{s}}=0.74 .\) How far can the spring be stretched from its unstrained position without the box moving when it is released?

A 75 -kg diver is standing at the end of a diving board while it is vibrating up and down in simple harmonic motion, as indicated in the figure. The diving board has an effective spring constant of \(k=\) \(4100 \mathrm{N} / \mathrm{m},\) and the vertical distance between the highest and lowest points in the motion is \(0.30 \mathrm{m} .\) Concepts: (i) How is the amplitude \(A\) related to the vertical distance between the highest and lowest points of the diver's motion? (ii) Starting from the top, where is the diver located one-quarter of a period later, and what can be said about his speed at this point? (iii) If the amplitude were to double, would the period also double? Explain. Calculations: (a) What is the amplitude of the motion? (b) Starting when the diver is at the highest point, what is his speed one-quarter of a period later? (c) If the vertical distance between his highest and lowest points were changed to \(0.10 \mathrm{m},\) what would be the time required for the diver to make one complete motional cycle?

A \(1.00 \times 10^{-2}\) -kg bullet is fired horizontally into a 2.50 -kg wooden block attached to one end of a massless horizontal spring \((k=845 \mathrm{N} / \mathrm{m})\). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of \(0.200 \mathrm{m} .\) What is the speed of the bullet?

To measure the static friction coefficient between a \(1.6-\mathrm{kg}\) block and a vertical wall, the setup shown in the drawing is used. A spring (spring constant \(=510 \mathrm{N} / \mathrm{m}\) ) is attached to the block. Someone pushes on the end of the spring in a direction perpendicular to the wall until the block does not slip downward. The spring is compressed by \(0.039 \mathrm{m} .\) What is the coefficient of static friction?

A rifle fires a \(2.10 \times 10^{-2}-\mathrm{kg}\) pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by \(9.10 \times 10^{-2} \mathrm{m}\) from its unstrained length. The pellet rises to a maximum height of \(6.10 \mathrm{m}\) above its position on the compressed spring. Ignoring air resistance, determine the spring constant.
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