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: 14

Answer

(a) The person's apparent weight is 690 N (b) The person's apparent weight is 740 N (c) The person's apparent weight is 690 N

Work Step by Step

Let $F_N$ be the normal force of the elevator floor pushing up on the person. Note that $F_N$ is equal in magnitude to the person's apparent weight. (a) If the elevator is at rest, then the acceleration is zero. Then $F_N$ is equal in magnitude to the person's weight. $F_N = mg$ $F_N = (60~kg)(9.80~m/s^2)$ $F_N = 690~N$ The person's apparent weight is 690 N (b) We can find the acceleration of the elevator. $a = \frac{v_f}{t}$ $a = \frac{10~m/s}{4.0~s}$ $a = 2.5~m/s^2$ We can use the acceleration to find the person's apparent weight. $\sum F = ma$ $F_N- mg = ma$ $F_N = m(g+a)$ $F_N = (60~kg)(9.80~m/s^2+ 2.5~m/s^2)$ $F_N = 740~N$ The person's apparent weight is 740 N (c) If the elevator is moving at a constant speed, then the acceleration is zero. Then $F_N$ is equal in magnitude to the person's weight. $F_N = mg$ $F_N = (60~kg)(9.80~m/s^2)$ $F_N = 690~N$ The person's apparent weight is 690 N
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