University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 6 - Work and Kinetic Energy - Problems - Exercises - Page 197: 6.42

Answer

(a) W = 0.0625 J (b) v = 0.177 m/s

Work Step by Step

(a) The work done on the block will be equal to the elastic potential energy stored in the spring. $W = \frac{1}{2}kx^2$ $W = \frac{1}{2}(200~N/m)(0.025~m)^2$ $W = 0.0625~J$ (b) The work done by the spring will be equal to the block's kinetic energy as it leaves the spring. $\frac{1}{2}mv^2 = 0.0625~J$ $v^2 = \frac{(2)(0.0625~J)}{m}$ $v = \sqrt{\frac{(2)(0.0625~J)}{4.00~kg}}$ $v = 0.177~m/s$
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