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: 54

Answer

(a) $v = 14.1~m/s$ (b) $d = 32.1~m$

Work Step by Step

(a) We can use conservation of energy to find the speed: $K = U_e$ $\frac{1}{2}mv^2 = \frac{1}{2}kx^2$ $v^2 = \frac{kx^2}{m}$ $v = (\sqrt{\frac{80,000~N/m}{100~kg}})~(0.50~m)$ $v = 14.1~m/s$ (b) We can use conservation of energy to find the kinetic energy at the bottom of the incline: $K+U_{g2} = U_e+U_{g1}$ $K+0 = \frac{1}{2}kx^2+mgh$ $K = \frac{1}{2}(80,000~N/m)(0.50~m)^2+(100~kg)(9.8~m/s^2)(10~m)$ $K = 19,800~J$ We can use work and energy to find the distance traveled along the incline: $U_{gf} = K+W_{ext}$ $mgh = K-mg~cos~\theta~\mu_k~d$ $mg~d~sin~\theta + mg~cos~\theta~\mu_k~d= K$ $d = \frac{K}{(mg~sin~\theta + mg~cos~\theta~\mu_k)}$ $d = \frac{19,800~J}{(100~kg)(9.8~m/s^2)(sin~30^{\circ}+0.15~cos~30^{\circ})}$ $d = 32.1~m$
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