## 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: 31

#### Answer

$v = 5.5~m/s$

#### Work Step by Step

We can use the equation for centripetal force to find the speed at the bottom as: $\sum F = \frac{mv^2}{r}$ $T-mg = \frac{mv^2}{r}$ $v^2 = \frac{r~(T-mg)}{m}$ $v = \sqrt{\frac{r~(T-mg)}{m}}$ $v = \sqrt{\frac{(1.5~m)[15~N-(0.500~kg)(9.80~m/s^2)]}{0.500~kg}}$ $v = 5.5~m/s$

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