University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 4 - Newton's Laws of Motion - Problems - Exercises - Page 127: 4.44

Answer

(a) $\sum F = F_T - mg$ The maximum acceleration is $a = 2.93~m/s^2$. (b) On the moon, the maximum acceleration is $a = 11.1~m/s^2$.

Work Step by Step

(a) $\sum F = F_T - mg$ Let's assume that $F_T = 28,000~N$. We can find the maximum acceleration of the elevator. $F_T - mg = ma$ $a = \frac{F_T-mg}{m} = \frac{(28,000~N)-(2200~kg)(9.80~m/s^2)}{2200~kg}$ $a = 2.93~m/s^2$ The maximum acceleration is $a = 2.93~m/s^2$. (b) $F_T - mg = ma$ $a = \frac{F_T-mg}{m} = \frac{(28,000~N)-(2200~kg)(1.62~m/s^2)}{2200~kg}$ $a = 11.1~m/s^2$ On the moon, the maximum acceleration is $a = 11.1~m/s^2$.
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