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: 9

Answer

(a) W = 1.89 J (b) The speed of the coin just as it hits the water is 26.5 m/s

Work Step by Step

(a) We can find the work done by gravity. $W = F\cdot d$ $W = F ~d~cos(\theta)$ $W = (mg) ~d~cos(0^{\circ})$ $W = (0.0055~kg)(9.80~m/s^2)(35~m)(1)$ $W = 1.89~J$ (b) We can use the work-energy theorem to find the speed when the coin hits the water. $KE_2 = KE_1+W$ $\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2+W$ $v_2^2 = \frac{mv_1^2+2W}{m}$ $v_2 = \sqrt{\frac{mv_1^2+2W}{m}}$ $v_2 = \sqrt{\frac{(0.0055~kg)(4.0~m/s)^2+(2)(1.89~J)}{0.0055~kg}}$ $v_2 = 26.5~m/s$ The speed of the coin just as it hits the water is 26.5 m/s
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