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 10 - Interactions and Potential Energy - Exercises and Problems - Page 258: 43


(a) $k = 21,560~N/m$ (b) $v = 18.6~m/s$

Work Step by Step

(a) We can find the minimum spring constant required to send the car to the top of the 10 meter hill: $U_e = \Delta U_g$ $\frac{1}{2}kx^2 = mgh$ $k = \frac{2mgh}{x^2}$ $k = \frac{(2)(400~kg)(9.8~m/s^2)(10~m)}{(2.0~m)^2}$ $k = 19,600~N/m$ Since the actual spring constant needs to be 10% larger, the spring constant should be $~~k = 21,560~N/m$ (b) The maximum speed will occur at the track's lowest point which is 5 meters below the starting point. We can use conservation of energy to find the maximum speed of a 350-kg car: $U_{g2}+K_2 = U_{g1}+U_e$ $K_2 = U_{g1}-U_{g2}+U_e$ $\frac{1}{2}mv^2 = mgh_1-mgh_2+\frac{1}{2}kx^2$ $v^2 = 2g~(h_1-h_2)+\frac{kx^2}{m}$ $v = \sqrt{2g~(h_1-h_2)+\frac{kx^2}{m}}$ $v = \sqrt{(2)(9.8~m/s^2)~(5~m-0)+\frac{(21,560~N/m)(2.0~m)^2}{350~kg}}$ $v = 18.6~m/s$
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