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: 5.26

Answer

(a) As the pull increases from 0 to 75.0 N, the graph shows the force of static friction. As the pull passes 75.0 N, the force of maximum static friction is overcome, and the object starts moving. Then the graph shows the force of kinetic friction when the force of the pull is greater than 75.0 N. (b) $\mu_s = 0.556$ $\mu_k = 0.370$ (c) The graph slants upward at first because the force of static friction continues to increase to oppose the increasing force of the pull. When the pull passes 75.0 N, the force of static friction is overcome, and the graph then shows the force of kinetic friction, which is a constant force directed against the motion. (d) The coefficients of friction would be the same because the coefficients of friction only depend on the types of surfaces that are interacting. The coefficients of friction do not depend on the normal force. If the normal force is doubled, then the force of friction would also double. Then the maximum force of static friction would be 150.0 N and the force of kinetic friction would be 100.0 N. Therefore, the graph would slant upward until the force of the pull was 150.0 N. The force of static friction would also be 150.0 N at this point. Then the graph would jump down to F = 100.0 N (on the y-axis) and stay at a constant level as the force of the pull increased beyond 150.0 N.

Work Step by Step

(a) As the pull increases from 0 to 75.0 N, the graph shows the force of static friction. As the pull passes 75.0 N, the force of maximum static friction is overcome, and the object starts moving. Then the graph shows the force of kinetic friction when the force of the pull is greater than 75.0 N. (b) $F_f = F_N~\mu_s = 75.0~N$ $\mu_s = \frac{75.0~N}{135~N} = 0.556$ $F_f = F_N~\mu_k = 50.0~N$ $\mu_k = \frac{50.0~N}{135~N} = 0.370$ (c) The graph slants upward at first because the force of static friction continues to increase to oppose the increasing force of the pull. When the pull passes 75.0 N, the force of static friction is overcome, and the graph then shows the force of kinetic friction, which is a constant force directed against the motion. (d) The coefficients of friction would be the same because the coefficients of friction only depend on the types of surfaces that are interacting. The coefficients of friction do not depend on the normal force. If the normal force is doubled, then the force of friction would also double. Then the maximum force of static friction would be 150.0 N and the force of kinetic friction would be 100.0 N. Therefore, the graph would slant upward until the force of the pull was 150.0 N. The force of static friction would also be 150.0 N at this point. Then the graph would jump down to F = 100.0 N (on the y-axis) and stay at a constant level as the force of the pull increased beyond 150.0 N.
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