Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 11 - Rotational Dynamics and Static Equilibrium - Problems and Conceptual Exercises - Page 370: 66

Answer

$K.E_i=133J$ $K.E_f=34.8J$

Work Step by Step

We can find the required initial and final kinetic energies as follows: $K.E_i=\frac{1}{2}mv_f^2$ We plug in the known values to obtain: $K.E_i=\frac{1}{2}(34.0)(2.80)$ $K.E_i=133J$ Now $K.E_f=\frac{1}{2}I\omega^2=\frac{1}{2}(I+mr^2)\omega^2$ We plug in the known values to obtain: $K.E_f=\frac{1}{2}[512+(34.0)(2.31)^2](0.317)$ $K.E_f=34.8J$
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