College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 4 - Force and Motion - Learning Path Questions and Exercises - Exercises - Page 137: 39


(a) (4) all of the preceding. If the elevator is moving with an acceleration equal to g downwards, then Measured weight$=m(g-g)=m\times0=0$ If the elevator is moving with acceleration in between 0 and $9.8\,m/s^{2}$ downwards, then the measured weight will be in between 0 and 4900 N. If the elevator is accelerating upwards, i.e., moving with some positive acceleration a, then the measured weight will be m(g+a) which is more than 4900 N. Therefore, the correct option is (4). (b) Elevator accelerates downwards at 1.8 $m/s^{2}$.

Work Step by Step

(b) Measured weight= m(g+a)= 4000 N $mg+ma=4000\,N$ $\implies 4900\,N+ma=4000\,N$ which gives $500\,kg\times a=-900\,N$ Or $a=\frac{-900\,N}{500\,kg}=-1.8\,m/s^{2}$ The negative sign shows that the acceleration is downwards.
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