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Consider the system shown in Fig. E5.34. Block A weighs

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 34E Chapter 5

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 34E

Consider the system shown in ?Fig. E5.34.? Block A weighs 45.0 N, and block B weighs 25.0 N. Once block B is set into downward motion, it descends at a constant speed. (a) Calculate the coefficient of kinetic friction between block A and the tabletop. (b) A cat, also of weight 45.0 N, falls asleep on top of block A . If block B is now set into downward motion, what is its acceleration (magnitude and direction)?

Step-by-Step Solution:

Solution 34E Step 1: a) eight of block A = 45 N Tension on the rope due to block B = 25 N We know that, Force × coefficient of kinetic friction = Resultant tension on the rope That is, F× = T This is applicable only if, the other block is moving with constant speed. That is, no acceleration. Therefore, 45 N × = 25 N Rearranging the equation, = 25 N / 45 N = 5/9 = 0.544

Step 2 of 2

Chapter 5, Problem 34E is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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Consider the system shown in Fig. E5.34. Block A weighs