Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 69

A spring of negligible mass is attached to the ceiling of an elevator. When the elevator is stopped at the first floor, a mass \(M\) is attached to the spring, stretching the spring a distance \(D\) until the mass is in equilibrium. As the elevator starts upward toward the second floor, the spring stretches an additional distance \(D / 4\). What is the magnitude of the acceleration of the elevator? Assume the force provided by the spring is linearly proportional to the distance stretched by the spring.

Short Answer

Expert verified
Answer: The acceleration of the elevator as it moves upwards is \(a = \frac{9g}{4}\), where \(g\) is the acceleration due to gravity.

Step by step solution

01

Calculate the spring constant

To calculate the spring constant, we will first have to find the tension in the spring when the elevator is at rest. At equilibrium, the weight of the mass, \(Mg\), is balanced by the spring force, \(F_s\). Using Hooke's Law and Newton's Second Law of Motion, we have: \(F_s = -kD = -Mg\) Now, we solve for the spring constant, \(k\): \(-kD = -Mg\) \(k = \frac{Mg}{D}\)
02

Analyze the situation when the elevator is moving upwards

When the elevator moves upwards, the mass attached to the spring also moves upwards, causing the spring to stretch an additional \(D/4\). In this state, we have two forces acting on the mass: the force due to gravity (\(Mg\)) and the spring force (\(F_s\)). Using Newton's Second Law of Motion, the net force acting on the mass is \(F_{net} = Ma\), where \(a\) is the acceleration of the elevator upwards.
03

Calculate the net force acting on the mass

As the mass stretches the spring by an additional \(D/4\), the new displacement (from the spring's equilibrium position) is \(x = D + \frac{D}{4} = \frac{5D}{4}\). We can now find the net force acting on the mass using Hooke's Law: \(F_s = -kx = -\frac{Mg}{D}\left(\frac{5D}{4}\right)\) Now, we can calculate the net force: \(F_{net} = Mg - F_s\) \(F_{net} = Mg - (-\frac{5Mg}{4}) = Mg + \frac{5Mg}{4} = \frac{9Mg}{4}\)
04

Apply Newton's Second Law of Motion to find the acceleration

Now that we have the net force acting on the mass, we can find the acceleration of the elevator: \(F_{net} = Ma\) \(\frac{9Mg}{4} = Ma\) We can solve for the acceleration, \(a\): \(a = \frac{9g}{4}\) So, the magnitude of the acceleration of the elevator as it moves upwards is \(\frac{9g}{4}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

On the bunny hill at a ski resort, a towrope pulls the skiers up the hill with constant speed of \(1.74 \mathrm{~m} / \mathrm{s}\). The slope of the hill is \(12.4^{\circ}\) with respect to the horizontal. A child is being pulled up the hill. The coefficients of static and kinetic friction between the child's skis and the snow are 0.152 and 0.104 , respectively, and the child's mass is \(62.4 \mathrm{~kg}\), including the clothing and equipment. What is the force with which the towrope has to pull on the child?

Two blocks are stacked on a frictionless table, and a horizontal force \(F\) is applied to the top block (block 1). Their masses are \(m_{1}=2.50 \mathrm{~kg}\) and \(m_{2}=3.75 \mathrm{~kg}\). The coefficients of static and kinetic friction between the blocks are 0.456 and 0.380 , respectively a) What is the maximum applied force \(F\) for which \(m_{1}\) will not slide off \(m_{2} ?\) b) What are the accelerations of \(m_{1}\) and \(m_{2}\) when \(F=24.5 \mathrm{~N}\) is applied to \(m_{1}\) ?

4.57 Your refrigerator has a mass of \(112.2 \mathrm{~kg}\), including the food in it. It is standing in the middle of your kitchen, and you need to move it. The coefficients of static and kinetic friction between the fridge and the tile floor are 0.460 and 0.370 , respectively. What is the magnitude of the force of friction acting on the fridge, if you push against it horizontally with a force of each magnitude? a) \(300 \mathrm{~N}\) b) \(500 \mathrm{~N}\) c) \(700 \mathrm{~N}\)

The gravitational acceleration on the Moon is a sixth of that on Earth. The weight of an apple is \(1.00 \mathrm{~N}\) on Earth. a) What is the weight of the apple on the Moon? b) What is the mass of the apple?

Three objects with masses \(m_{1}=36.5 \mathrm{~kg}, m_{2} 19.2 \mathrm{~kg},\) and \(m_{3}=12.5 \mathrm{~kg}\) are hanging from ropes that run over pulleys. What is the acceleration of \(m_{1} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Fields in Physics

Read Explanation

Famous Physicists

Read Explanation

Dynamics

Read Explanation

Translational Dynamics

Read Explanation

Engineering Physics

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.

Sign-up for free