Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 4 - Dynamics: Newton's Laws of Motion - General Problems - Page 106: 84

Answer

The elevator's acceleration must have been between $0.771 ~m/s^2$ and $1.03 ~m/s^2$

Work Step by Step

The line holding the 2.10-kg ball broke. Let's suppose that the elevator's acceleration caused the tension force in the fishing line to be equal to the limit of 22.2 N: $ma = F_T - mg$ $a = \frac{F_T - mg}{m} = \frac{22.2 ~N - (2.10 ~kg)(9.80 ~m/s^2)}{2.10 ~kg}$ $a = 0.771 ~m/s^2$ Since the fishing line broke, the elevator's acceleration must have been at least $a = 0.771 ~m/s^2$. The line holding the 2.05-kg ball didn't break. Let's suppose that the elevator's acceleration caused the tension force in the fishing line to be equal to the limit of 22.2 N. $ma = F_T - mg$ $a = \frac{F_T - mg}{m} = \frac{22.2 ~N - (2.05 ~kg)(9.80 ~m/s^2)}{2.05 ~kg}$ $a = 1.03 ~m/s^2$ Since the fishing line didn't break, the elevator's acceleration must have been at most $a = 1.03 ~m/s^2$ Therefore, the elevator's acceleration must have been between $0.771 ~m/s^2$ and $1.03 ~m/s^2$.
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