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: 42

Answer

(a) $F = 6750~N$ (b) $F = 1.35\times 10^6~N$

Work Step by Step

(a) We can find the rate of deceleration as; $a = \frac{0-v_0^2}{2x}$ $a = \frac{-(15~m/s)^2}{(2)(1~m)}$ $a = -112.5~m/s^2$ We can use the magnitude of deceleration to find the net force on the person as; $F = ma$ $F = (60~kg)(112.5~m/s^2)$ $F = 6750~N$ (b) We can find the rate of deceleration as; $a = \frac{0-v_0^2}{2x}$ $a = \frac{-(15~m/s)^2}{(2)(0.005~m)}$ $a = -22500~m/s^2$ We can use the magnitude of deceleration to find the net force on the person as; $F = ma$ $F = (60~kg)(22500~m/s^2)$ $F = 1.35\times 10^6~N$
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