College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 3 - Problems - Page 118: 109


(a) The acceleration of the elevator is $1.8~m/s^2$ (downward) (b) The elevator's speed is $8.7~m/s$

Work Step by Step

(a) We can find the person's acceleration: $\sum F = ma$ $mg-F_N = ma$ $a = \frac{mg-F_N}{m}$ $a = \frac{(51~kg)(9.80~m/s^2)-408~N}{51~kg}$ $a = 1.8~m/s^2$ (downward) The acceleration of the elevator is $1.8~m/s^2$ (downward) (b) We can find the elevator's speed: $v_f=v_0+at$ $v_f = 1.5~m/s+(1.8~m/s^2)(4.0~s)$ $v_f = 8.7~m/s$ The elevator's speed is $8.7~m/s$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.