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 7 - Newton's Third Law - Exercises and Problems: 35

Answer

The least amount of time to travel 5.0 meters without the top block slipping is 1.75 seconds.

Work Step by Step

We can find the maximum possible force that the force of static friction can provide for the 3.0-kg block without the 4.0-kg block slipping. Let $m_u$ be the mass of the 4.0-kg block. $F_f = m_u~g~\mu_s$ $F_f = (4.0~kg)(9.80~m/s^2)(0.60)$ $F_f = 23.52~N$ We can find the acceleration of the bottom block when this force of static friction is provided by the contact between the two blocks. Let $m_b$ be the mass of the 3.0-kg block. $\sum F = m_b~a$ $F_f - (m_b+m_u)~g~\mu_k = m_b~a$ $a = \frac{F_f - (m_b+m_u)~g~\mu_k}{m_b}$ $a = \frac{(23.52~N) - (3.0~kg+4.0~kg)(9.80~m/s^2)(0.20)}{3.0~kg}$ $a = 3.27~m/s^2$ Since both blocks move together, this will be the acceleration of the system of both blocks. We can find the time it takes to travel 5.0 meters. $d = \frac{1}{2}at^2$ $t = \sqrt{\frac{2d}{a}}$ $t = \sqrt{\frac{(2)(5.0~m)}{3.27~m/s^2}}$ $t = 1.75~s$ The least amount of time to travel 5.0 meters without the top block slipping is 1.75 seconds.
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