Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 5 - Circular Motion; Gravitation - General Problems: 83

Answer

The speed of the train is 25.0 m/s.

Work Step by Step

Let $F_T$ be the tension in the cord holding the lamp. $m\frac{v^2}{r} = F_T ~sin(\theta)$ $mg = F_T ~cos(\theta)$ We can divide the first equation by the second equation. $\frac{v^2}{gr} = tan(\theta)$ $v = \sqrt{gr ~tan(\theta)} = \sqrt{(9.80 ~m/s^2)(215 ~m)(tan(16.5^{\circ}))}$ $v = 25.0 ~m/s$ The speed of the train is 25.0 m/s.
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