College Physics (4th Edition)

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

Chapter 3 - Problems - Page 115: 63


(a) $a = 3.48~m/s^2$ (b) The elevator's velocity 4.00 seconds later is $15.1~m/s$

Work Step by Step

(a) We can find the net force on the elevator: $\sum F = F_t-mg$ $\sum F = 33,600~N-(2530~kg)(9.80~m/s^2)$ $\sum F = 8806~N$ We can find the acceleration of the elevator: $\sum F = ma$ $a = \frac{\sum F}{m}$ $a = \frac{8806~N}{2530~kg}$ $a = 3.48~m/s^2$ (b) We can find the velocity 4.00 seconds later: $v_f = v_0+at$ $v_f = (1.20~m/s)+(3.48~m/s^2)(4.00~s)$ $v_f = 15.1~m/s$ The elevator's velocity 4.00 seconds later is $15.1~m/s$
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