Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 7 - Exercises and Problems - Page 132: 48


Please see the work below.

Work Step by Step

We know that according to law of conservation of energy $(\frac{1}{2})mv^2=mg(L-Lcos(\theta))$ This simplifies to: $(\frac{1}{2})v^2=gL(1-cos(\theta))$ $L=\frac{v^2}{2g(1-cos(\theta))}$ We plug in the known values to obtain: $L=\frac{(0.55)^2}{(2)(9.8)(1-cos(8.0))}$ $L=1.6m$
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