University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 5 - Applying Newton's Laws - Problems - Exercises - Page 163: 5.36


(a) $\alpha = 19.3^{\circ}$ (b) $a = 0.93~m/s^2$ (c) $v = 3.0~m/s$

Work Step by Step

(a) The minimum angle is when the force of static friction directed up the ramp is equal in magnitude to the component of the weight directed down the ramp. $F_f = mg~sin(\alpha)$ $mg~cos(\alpha)~\mu_s = mg~sin(\alpha)$ $tan(\alpha) = \mu_s$ $\alpha = arctan(\mu_s) = arctan(0.35)$ $\alpha = 19.3^{\circ}$ (b) $\sum F = ma$ $mg~sin(\alpha) - mg~cos(\alpha)~\mu_k = ma$ $a = g~sin(\alpha) - g~cos(\alpha)~\mu_k$ $a = (9.80~m/s^2)~sin(19.3^{\circ}) - (9.80~m/s^2)~cos(19.3^{\circ})(0.25)$ $a = 0.93~m/s^2$ (c) We can find the speed $v$ after sliding 5.0 meters. $v^2 = v_0^2+2ax = 0 + 2ax$ $v = \sqrt{(2)(0.93~m/s^2)(5.0~m)}$ $v = 3.0~m/s$
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