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Physics / University Physics 13 / Chapter 5 / Problem 34E

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

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

Chapter 8, Problem 34E

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QUESTION:

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)?

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QUESTION:

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)?

ANSWER:

Step 1 of 2

a)

Weight of block A = 45 N

Tension on the rope due to block B = 25 N

We know that, Force $\times{}$ coefficient of kinetic friction = Resultant tension on the rope That is, $F\times{}\mu{}=\ T$

This is applicable only if, the other block is moving with constant speed. That is, no acceleration.

Therefore, $45\ N\ \times{}\mu{}\ =\ 25\ N$

Rearranging the equation, $\mu{}\ =\ 25\ N\ /\ 45\ N\ =\ 5/9\ =\ 0.544$

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