m. m2 A mass m1 = 12.0 kg sits on a level tabletop, and is attached to a second hanging mass m2 by a rope draped over a frictionless pulley. The coefficient of static friction between the 12.0-kg block and the tabletop is 0.450. What is the minimum mass m2 needed so that the 12.0-kg block will just begin to slide? 5.40 kg 10.8 kg 12.0 kg 14.5 kg

College Physics
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Author:Raymond A. Serway, Chris Vuille
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Chapter4: Newton's Laws Of Motion
Section: Chapter Questions
Problem 74AP: (a) What is the minimum force of friction required to hold the system of Figure P4.74 in...
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A mass m1 = 12.0 kg sits on a level tabletop, and is attached to a second hanging mass m2 by a rope draped over a frictionless pulley.  The coefficient of static friction between the 12.0-kg block and the tabletop is 0.450.   What is the minimum mass m2needed so that the 12.0-kg block will just begin to slide?

m
m2
A mass m1 = 12.0 kg sits on a level tabletop, and is attached to a second hanging mass m2 by a rope draped over a frictionless
pulley. The coefficient of static friction between the 12.0-kg block and the tabletop is 0.450. What is the minimum mass m2
needed so that the 12.0-kg block will just begin to slide?
5.40 kg
10.8 kg
12.0 kg
14.5 kg
Transcribed Image Text:m m2 A mass m1 = 12.0 kg sits on a level tabletop, and is attached to a second hanging mass m2 by a rope draped over a frictionless pulley. The coefficient of static friction between the 12.0-kg block and the tabletop is 0.450. What is the minimum mass m2 needed so that the 12.0-kg block will just begin to slide? 5.40 kg 10.8 kg 12.0 kg 14.5 kg
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