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

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems: 40

Answer

The velocity at t = 6.0 seconds is $v = 3.18 ~m/s$.

Work Step by Step

Let $F_N$ be the normal force of the elevator pushing up on the person. Note that this force $F_N$ will be equal to the reading on the scale. We can find the acceleration from t = 0 to t = 3.0 seconds. $\sum F = ma$ $mg - F_N = ma$ $a = \frac{mg-F_N}{m}$ $a = \frac{(95~kg)(9.80~m/s^2)-830~N}{95~kg}$ $a = 1.06~m/s^2$ We can find the acceleration from t = 3.0 seconds to t = 6.0 seconds. $\sum F = ma$ $mg - F_N = ma$ $a = \frac{mg-F_N}{m}$ $a = \frac{(95~kg)(9.80~m/s^2)-930~N}{95~kg}$ $a = 0$ For the first 3.0 seconds, the elevator accelerates downward at a rate of $1.06~m/s^2$. The elevator travels at a constant speed after that. We can find the velocity at t = 3.0 seconds. $v = a~t = (1.06~m/s^2)(3.0~s)$ $v = 3.18~m/s$ Since the elevator does not accelerate after t = 3.0 seconds, the velocity at t = 6.0 seconds is $v = 3.18 ~m/s$
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