Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 162: 32


(a) Zach's apparent weight is 784 N (b) Zach's apparent weight is 1050 N

Work Step by Step

Let $F_N$ be the normal force of the elevator pushing up on Zach. Note that this force $F_N$ will be equal to Zach's apparent weight. (a) When the elevator is descending at a constant speed, Zach's acceleration is zero. Then $F_N$ will be equal in magnitude to mg. $F_N = mg$ $F_N = (80~kg)(9.80~m/s^2)$ $F_N = 784~N$ Zach's apparent weight is 784 N. (b) We can find the acceleration directed upward when the elevator is braking. $a = \frac{v_f-v_0}{t}$ $a = \frac{0-10~m/s}{3.0~s}$ $a = -3.3~m/s$ The magnitude of the acceleration is $3.3~m/s^2$ and it is directed upward. We can use the acceleration to find $F_N$. $\sum F = ma$ $F_N - mg = ma$ $F_N = m(g+a)$ $F_N = (80~kg)(9.80~m/s^2+3.3~m/s^2)$ $F_N = 1050~N$ Zach's apparent weight is 1050 N.
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