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: 5.36

Answer

(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. $v^2 = v_0^2+2ax = 0 + 2ax$ $v = \sqrt{(2)(0.93~m/s^2)(5.0~m)}$ $v = 3.0~m/s$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.