Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 4 - Forces and Newton's Laws of Motion - Problems - Page 115: 46


The maximum acceleration that the plane can have is $2.94m/s^2$.

Work Step by Step

The maximum force that the plane can act upon the cup without making it slide backward cannot surpass the maximum static frictional force: $$F_{max}=f_s^{max}=\mu_sF_N=\mu_smg$$ $\mu_s$: coefficient of static friction $m$: mass of the cup $g$: gravitational acceleration According to Newton's 2nd Law, we also have $$F_{max}=ma_{max}$$ $a_{max}$: maximum acceleration of the plane so that the cup does not slide Therefore, $$ma_{max}=\mu_smg$$ $$a_{max}=\mu_sg=0.3\times9.8=2.94m/s^2$$
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