## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 155: 19

#### Answer

(a) The person's apparent weight is 784 N (b) The person's apparent weight is 1050 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) When the elevator is moving at a constant speed, $F_N$ is equal in magnitude to the gravitational force on the person. $F_N = mg$ $F_N = (80~kg)(9.80~m/s^2)$ $F_N = 784~N$ The person's apparent weight is 784 N (b) We can find the acceleration while the elevator is braking. $a = \frac{0-10~m/s}{3.0~s}$ $a = -3.3~m/s^2$ The magnitude of the acceleration is $3.3~m/s^2$ and it is directed in the upward direction. We can use the acceleration to find the apparent weight. $\sum F = ma$ $F_N-mg = ma$ $F_N = m(g+a)$ $F_N = (80~kg)(9.80~m/s^2+3.3~m/s^2)$ $F_N = 1050~N$ The person's apparent weight is 1050 N

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