Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 9 - Work and Kinetic Energy - Exercises and Problems: 59

Answer

The box's launch speed is 3.7 m/s

Work Step by Step

The kinetic energy of the box when it leaves the spring will be equal to the initial energy stored in the spring plus the work done on the box by friction. $KE = U_s+W_f$ $\frac{1}{2}mv^2 = \frac{1}{2}kx^2-mg~\mu_k~d$ $v^2 = \frac{kx^2-2mg~\mu_k~d}{m}$ $v = \sqrt{\frac{kx^2-2mg~\mu_k~d}{m}}$ $v = \sqrt{\frac{(250~N/m)(0.12~m)^2-(2)(0.25~kg)(9.80~m/s^2)(0.23)(0.12~m)}{0.25~kg}}$ $v = 3.7~m/s$ The box's launch speed is 3.7 m/s.
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