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 162: 5.30

Answer

(a) As the rock slides up the hill, the acceleration is $a = 9.33~m/s^2$ directed down the hill. (b) The rock will slide back down the hill. The acceleration is $a = 2.19~m/s^2$ as it slides down the hill.

Work Step by Step

(a) $\sum F = ma$ $mg~sin(\theta) + mg~cos(\theta)~\mu_k = ma$ $a = g~sin(\theta) + g~cos(\theta)~\mu_k $ $a = (9.80~m/s^2)~sin(36^{\circ})+(9.80~m/s^2)~cos(36^{\circ})(0.45)$ $a = 9.33~m/s^2$ The acceleration is $a = 9.33~m/s^2$ directed down the hill. (b) We can find the maximum possible force of static friction: $F_f = (m~kg)(9.80~m/s^2)~cos(36^{\circ})(0.65)$ $F_f = 5.15m~N$ We can find the component of the weight directed down the hill. $mg~sin(\theta) = (m~kg)(9.80~m/s^2)~sin(36^{\circ}) = 5.76m~N$ Since the component of the weight directed down the hill is greater than the maximum possible force of static friction, the rock will slide back down the hill. We can find the acceleration as the rock slides down. $\sum F = ma$ $mg~sin(\theta) - mg~cos(\theta)~\mu_k = ma$ $a = g~sin(\theta) - g~cos(\theta)~\mu_k$ $a = (9.80~m/s^2)~sin(36^{\circ})-(9.80~m/s^2)~cos(36^{\circ})(0.45)$ $a = 2.19~m/s^2$ The acceleration is $a = 2.19~m/s^2$ as it slides down the hill.
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