Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 9 - Work and Kinetic Energy - Exercises and Problems: 58

Answer

The mass $m_2$ hangs below the ceiling a distance of $L_1+\frac{(m_1+m_2)~g}{k_1}+L_2+\frac{(m_2)~g}{k_2}$

Work Step by Step

The force of the second spring is equal to the weight of the hanging mass $m_2$. Since the second spring pulls down on the hanging mass $m_1$ with the weight of the hanging mass $m_2$, the force of the first spring is equal to the sum of the weights of both hanging masses. We can find the distance $x_1$ that the first spring stretches. $k_1~x_1 = (m_1+m_2)~g$ $x_1 = \frac{(m_1+m_2)~g}{k_1}$ We can find the distance $x_2$ that the second spring stretches. $k_2~x_2 = (m_2)~g$ $x_2 = \frac{(m_2)~g}{k_2}$ We can find the total distance $d$ that the mass $m_2$ hangs below the ceiling. $d = L_1+x_1+L_2+x_2$ $d = L_1+\frac{(m_1+m_2)~g}{k_1}+L_2+\frac{(m_2)~g}{k_2}$ The mass $m_2$ hangs below the ceiling a distance of $L_1+\frac{(m_1+m_2)~g}{k_1}+L_2+\frac{(m_2)~g}{k_2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.