/*! 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 force is \(-0.012\) J.

Step by step solution

01

Understand Work done by a Spring

The work done by a spring force is calculated using the formula: \[ W = - \frac{1}{2} k (x_f^2 - x_i^2) \]where \( k \) is the spring constant, \( x_f \) is the final compression from the unstrained position, and \( x_i \) is the initial compression.
02

Convert Units

Since the compression distances are given in millimeters, first convert them to meters:- Initial compression \( x_i = 5.0 \) mm = \( 0.005 \) m.- Additional compression \( = 6.0 \) mm, so final compression \( x_f = 11.0 \) mm = \( 0.011 \) m.
03

Substitute Values into the Work Formula

Substitute the given values into the work done formula:\[ W = -\frac{1}{2} \times 250 \times ((0.011)^2 - (0.005)^2) \]
04

Calculate the Work Done

First, calculate \( (0.011)^2 \) and \( (0.005)^2 \):- \((0.011)^2 = 0.000121\)- \((0.005)^2 = 0.000025\)Substitute these into the formula:\[ W = -\frac{1}{2} \times 250 \times (0.000121 - 0.000025) \]\[ W = -\frac{1}{2} \times 250 \times 0.000096 \]
05

Final Calculation

Calculate \( W \) as follows:\[ W = -125 \times 0.000096 = -0.012 \text{ J} \]The negative sign indicates that the work is done by the spring force.

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Key Concepts

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

Spring Constant
In the realm of mechanics, springs are fascinating components. They have a special trait known as "spring constant," denoted as \( k \). The spring constant reflects the force required to compress or extend a spring by a unit distance. It's essentially a measure of the spring's stiffness. A higher \( k \) value indicates a stiffer spring. For instance, in the provided exercise, the spring constant is 250 \( \, \text{N/m} \). This means that 250 Newtons of force is needed to compress the spring by one meter.

To understand this better:
  • The spring constant remains constant for a specific spring, regardless of the amount of force applied within its elastic limit.
  • This constant helps in calculating forces acting on the spring and the work done by or on it.
  • Knowing \( k \) allows us to predict how the spring will behave under different forces and conditions.
Compression Distance
Compression distance is a critical component when dealing with springs. It refers to how much a spring is compressed or stretched from its original, unstrained position. In many applications including pen mechanisms, understanding compression is key to determining functionality.

For example, in the exercise, the spring is initially compressed by 5.0 mm, which is equal to 0.005 m. An additional compression of 6.0 mm (0.006 m) is needed to push the pen's tip out. Hence, the total compression becomes 11.0 mm (0.011 m).

Here's how you approach conversion and understanding:
  • Always convert units to meters when dealing with force-related calculations, as these follow the SI unit system.
  • Additional compression is often added to the initial compression to get the total compression distance, which is key for calculating the work done.
  • Observing changes in compression can help predict mechanical wear and anticipate the force needed for certain operations.
Work Formula
Calculating work done by a spring involves using a specific formula that takes into account the spring constant and the compression distance. The formula is: \[W = - \frac{1}{2} k (x_f^2 - x_i^2)\]where:
  • \( W \) is the work done by the spring.
  • \( k \) is the spring constant (250 \( \, \text{N/m} \) in this case).
  • \( x_i \) is the initial compression distance (0.005 m).
  • \( x_f \) is the final compression distance (0.011 m).
This formula essentially calculates the difference in potential energy of the spring as it moves from one compression state to another.

The negative sign is important as it indicates that the spring is doing work on another object, typically releasing stored energy. The step-by-step calculation involves:
  • Plugging in all the values correctly to maintain unit consistency.
  • Ensuring square values of distances are computed correctly to get accurate results.
  • Recognizing the outcome will tell how much energy is transferred from the spring to aid in motion or change state.
Energy Conservation
Energy conservation is a fundamental principle in physics that states energy cannot be created or destroyed, only transformed from one form to another. In the context of springs, energy is stored as potential energy when the spring is compressed or stretched, and it is released as work is done by the spring. The work done by the spring is the transformation of stored elastic potential energy into kinetic energy as the spring returns to its equilibrium position.

Consider the pen example from the exercise:
  • The spring initially stores energy when it's compressed.
  • As it returns to its uncompressed state, this stored energy is transformed into work that aids in moving the pen tip outwards.
  • This work represents the energy transferred out of the spring.
Understanding this transformation highlights the elegance of energy conservation laws, as they help in efficiently planning and solving mechanical systems. Whether in everyday tools like pens or more complex systems, recognizing how springs conserve and transform energy is key to understanding mechanics.

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Most popular questions from this chapter

A vertical ideal spring is mounted on the floor and has a spring constant of 170 \(\mathrm{N} / \mathrm{m}\) . A 0.64 \(\mathrm{kg}\) block is placed on the spring in two different ways. \((\mathrm{a})\) In one case, the block is placed on the spring and not released until it rests stationary on the spring in its equilibrium position. Determine the amount (magnitude only) by which the spring is compressed. (b) In a second situation, the block is released from rest immediately after being placed on the spring and falls downward until it comes to a momentary halt. Determine the amount (magnitude only by which the spring is now compressed.

A spring is hung from the ceiling. A \(0.450-\mathrm{kg}\) block is then attached to the free end of the spring. When released from rest, the block drops 0.150 \(\mathrm{m}\) before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.

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 m. What is the coefficient of static friction?

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?

Two metal beams are joined together by four rivets, as the drawing indicates. Each rivet has a radius of \(5.0 \times 10^{-3} \mathrm{m}\) and is to be exposed to a shearing stress of no more than \(5.0 \times 10^{8} \mathrm{Pa}\) . What is the maximum tension \(\overrightarrow{\mathrm{T}}\) that can be applied to each beam, assuming that each rivet carries one-fourth of the total load?

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