## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 211: 14

#### Answer

The speed at the bottom of the dip is 12.1 m/s

#### Work Step by Step

Let $F_N$ be the normal force exerted on the passengers. Note that $F_N$ is equal to the apparent weight of the passengers. We can let $F_N = 1.5~mg$ to find the speed at the bottom of the dip. $\sum F = \frac{mv^2}{r}$ $F_N-mg = \frac{mv^2}{r}$ $1.5~mg-mg = \frac{mv^2}{r}$ $0.5~g = \frac{v^2}{r}$ $v = \sqrt{0.5~g~r}$ $v = \sqrt{(0.5)(9.80~m/s^2)(30~m)}$ $v = 12.1~m/s$ The speed at the bottom of the dip is 12.1 m/s.

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