## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 4 - Problems - Page 151: 63

#### Answer

(a) The person's apparent weight is $567.3~N$ (b) The person's apparent weight is $628.3~N$

#### Work Step by Step

(a) We can find the person's mass: $mg = 598~N$ $m = \frac{598~N}{9.80~m/s^2}$ $m = ~61.0~kg$ Let $F_N$ be the upward force exerted on the person by the elevator. Note that $F_N$ is the person's apparent weight. The elevator is accelerating downward. We can find the person's apparent weight: $\sum F = ma$ $mg-F_N = ma$ $F_N = m~(g-a)$ $F_N = (61.0~kg)(9.80~m/s^2-0.50~m/s^2)$ $F_N = 567.3~N$ The person's apparent weight is $567.3~N$ (b) Let $F_N$ be the upward force exerted on the person by the elevator. Note that $F_N$ is the person's apparent weight. The elevator is accelerating upward. We can find the person's apparent weight: $\sum F = ma$ $F_N-mg = ma$ $F_N = m~(g+a)$ $F_N = (61.0~kg)(9.80~m/s^2+0.50~m/s^2)$ $F_N = 628.3~N$ The person's apparent weight is $628.3~N$

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