/*! 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 6 A spring \((k=830 \mathrm{N} / \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 6

A spring \((k=830 \mathrm{N} / \mathrm{m})\) is hanging from the ceiling of an elevator, and a 5.0 -kg object is attached to the lower end. By how much does the spring stretch (relative to its unstrained length) when the elevator is accelerating upward at \(a=0.60 \mathrm{m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
The spring stretches approximately 6.27 cm.

Step by step solution

01

Understand the forces involved

To find how much the spring stretches, we must consider the forces acting on the object. When the elevator accelerates upward, there are two forces: the gravitational force (weight) and the force due to the elevator's acceleration.
02

Calculate the gravitational force

The gravitational force on the object is calculated using the equation: \[ F_g = m \cdot g \]where \( m = 5.0 \text{ kg} \) is the mass of the object and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. Substitute the values: \[ F_g = 5.0 \times 9.8 = 49 \text{ N} \]
03

Calculate the additional force due to elevator acceleration

The force due to the elevator's acceleration is given by: \[ F_a = m \cdot a \]where \( a = 0.60 \text{ m/s}^2 \) is the acceleration of the elevator. Substitute the known values: \[ F_a = 5.0 \times 0.60 = 3.0 \text{ N} \]
04

Calculate the total force acting on the spring

The total force acting on the spring is the sum of the gravitational force and the additional force due to the elevator's acceleration: \[ F_{total} = F_g + F_a = 49 \text{ N} + 3.0 \text{ N} = 52 \text{ N} \]
05

Calculate the spring stretch using Hooke's Law

Hooke's Law relates the force on a spring to its stretch (or compression): \[ F = k \cdot x \]where \( k = 830 \text{ N/m} \) is the spring constant and \( x \) is the stretch of the spring.Solve for \( x \):\[ x = \frac{F_{total}}{k} = \frac{52}{830} \approx 0.06265 \text{ m} \]
06

Convert the stretch to centimeters

To express the stretch in centimeters, convert the measurement from meters: \[ x = 0.06265 \times 100 = 6.27 \text{ cm} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 that describes how springs work. It tells us how much a spring will stretch or compress when a force is applied to it.
It's expressed by the formula: \[ F = k \cdot x \]
  • \( F \) represents the force applied to the spring.
  • \( k \) is the spring constant, a measure of the spring's stiffness.
  • \( x \) is the amount the spring stretches or compresses.

Essentially, the law implies that the force needed to stretch or compress a spring is directly proportional to the distance it is stretched or compressed, given within the elastic limit of the spring.
For example, if you apply a larger force to the spring, it will stretch more, assuming all other factors remain constant.
spring constant
The spring constant \( k \) is a crucial part of understanding how springs react to forces.
It defines how stiff or flexible a spring is. A higher spring constant means a stiffer spring that requires more force to stretch or compress by a given distance.
  • Measured in newtons per meter (N/m), it shows the force that results in a one-meter change in the spring's length.
  • In the given exercise, \( k = 830 \text{ N/m} \), indicating the required force to stretch the spring by one meter is 830 N.

This rigidity factor is important in engineering and physics when it comes to designing systems that involve spring-like behavior.
By knowing the spring constant, you can predict how much a spring will stretch under a specific load.
gravitational force
Gravitational force is the pull exerted by the Earth on an object, drawing it towards the center of the Earth.
It can be calculated using the formula:\[ F_g = m \cdot g \]
  • \( m \) is the mass of the object.
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \) on Earth.

In the context of the exercise, a 5.0 kg object hanging from a spring inside an elevator experiences a downward force of 49 N due to gravity.
Gravitational force is crucial when calculating how much a spring stretches because it directly impacts the overall force acting on the system.
object acceleration
When an object accelerates, it experiences additional forces in the direction of the acceleration.
According to Newton's second law, these forces can be calculated by:\[ F_a = m \cdot a \]
  • \( m \) is the object's mass.
  • \( a \) is the acceleration (change in velocity) of the object.

In the elevator scenario, with an upward acceleration of \( 0.60 \text{ m/s}^2 \), this results in an extra 3.0 N force added to the gravitational pull.
Understanding this concept helps explain how motion and forces combine to impact the total stretch of a spring in dynamic situations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A simple pendulum is made from a \(0.65-\mathrm{m}\) -long string and a small ball attached to its free end. The ball is pulled to one side through a small angle and then released from rest. After the ball is released, how much time elapses before it attains its greatest speed?

In a room that is 2.44 m high, a spring (unstrained length \(=0.30 \mathrm{m} )\) hangs from the ceiling. A board whose length is 1.98 \(\mathrm{m}\) is attached to the free end of the spring. The board hangs straight down, so that its \(1.98-\mathrm{m}\) length is perpendicular to the floor. The weight of the board (104 \(\mathrm{N} )\) stretches the spring so that the lower end of the board just extends to, but does not touch, the floor. What is the spring constant of the spring?

The fan blades on a jet engine make one thousand revolutions in a time of 50.0 \(\mathrm{ms}\) . Determine (a) the period (in seconds) and (b) the frequency (in \(\mathrm{Hz} )\) of the rotational motion. (c) What is the angular frequency of the blades?

A hand exerciser utilizes a coiled spring. A force of 89.0 \(\mathrm{N}\) is required to compress the spring by 0.0191 \(\mathrm{m} .\) Determine the force needed to compress the spring by 0.0508 \(\mathrm{m} .\)

When an object of mass \(m_{1}\) is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequency of 12.0 \(\mathrm{Hz}\) . When another object of mass \(m_{2}\) is hung on the spring along with the first object, the frequency of the motion is 4.00 \(\mathrm{Hz}\) . Find the ratio \(m_{2} / m_{1}\) of the mases.
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Nuclear Physics

Read Explanation

Waves Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Magnetism

Read Explanation

Geometrical and Physical Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.