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 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 156: 46


(a) $F = 6670~N$ (b) $t = 0.600 ~ms$

Work Step by Step

(a) We can find the rate of deceleration after the bullet strikes the block. $a = \frac{0-v_0^2}{2x}$ $a = \frac{-(400~m/s)^2}{(2)(0.12~m)}$ $a = -6.67\times 10^5~m/s^2$ We can use the magnitude of acceleration to find the force. $F = ma$ $F = (0.010~kg)(6.67\times 10^5~m/s^2)$ $F = 6670~N$ (b) We can find the time it takes the bullet to stop. $v_f= v_0 + at$ $t = \frac{v_f-v_0}{a}$ $t = \frac{0-400~m/s}{-6.67\times 10^5~m/s^2}$ $t = 0.600~ms$
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