/*! 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 69 A spring of negligible mass is a... [FREE SOLUTION] | 91影视

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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 \(\frac{9g}{4}\).

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