D Look at the system shown in the figure above. Block C weighs 450 N and Block D weights 25.0 N. Once block D is set into a downward motion, it descends at a constant speed. (a) What is the coefficient of kinetic friction between block C and the table top? (b) A small dog, also of weight 45.0 N, falls asleep on top of block C. If Block D is now set into a downward motion, what is the acceleration (magnitude and direction)?

D Look at the system shown in the figure above. Block C weighs 450 N and Block D weights 25.0 N. Once block D is set into a downward motion, it descends at a constant speed. (a) What is the coefficient of kinetic friction between block C and the table top? (b) A small dog, also of weight 45.0 N, falls asleep on top of block C. If Block D is now set into a downward motion, what is the acceleration (magnitude and direction)?

Name: DLook at the system shown in the figure above. Block C weighs 450 N a
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Description: DLook at the system shown in the figure above. Block C weighs 450 N and Block D weights 25.0 N. Onceblock D is set into a downward motion, it descends at a constant speed.(a) What is the coefficient of kinetic friction between block C and the table

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter6: Applications Of Newton's Laws

Section: Chapter Questions

Problem 6.7CYU: Check Your Understanding A block of mass 1.0 kg rests on a horizontal surface. The frictional...

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I don't understand why Fs <= Fk. What does the F in (1) and (2) stand for? If it's force applied, wouldn't F = Fs + Fk for box 2, and (F - Fs - Fk) = Fs for box 1? My thinking is that the Force applied on box 2 minus the force of kinetic friction and the force of static friction applied by Box 1 on 2 would equal to the net force. Then the net force of box 2 would cause the same netforce on box 1, if box 1 isn't moving, then it's Fs should be equal to that net force applied on box 1. Therefore Force applied wouldn't be equal to Static Friction Force, and shouldn't it be bigger than the Kinetic Friction Force since it's moving in the direction of Force applied?