Problem 5: a) The 2 kg wood box in the Figure below slides up a vertical wood wall while you push on it at a 45° angle. The coefficient of kinetic friction for wood on wood is μ = 0.2. What magnitude force Fpush should you apply to cause the box to slide up at a constant speed? Recall that constant speed means a = 0 m/sec². Answer: 34.6 N. b) Let's re-do part (a), but with the box now sliding down the vertical wood wall while you push on it at a 45° angle. Hk is still μ = 0.2. What magnitude force Fpush should you apply to cause the box to slide down at a constant speed? Recall that constant speed means a = 0 m/sec². Answer: 23.1 N. 2.0 kg 45° push

Problem 5: a) The 2 kg wood box in the Figure below slides up a vertical wood wall while you push on it at a 45° angle. The coefficient of kinetic friction for wood on wood is μ = 0.2. What magnitude force Fpush should you apply to cause the box to slide up at a constant speed? Recall that constant speed means a = 0 m/sec². Answer: 34.6 N. b) Let's re-do part (a), but with the box now sliding down the vertical wood wall while you push on it at a 45° angle. Hk is still μ = 0.2. What magnitude force Fpush should you apply to cause the box to slide down at a constant speed? Recall that constant speed means a = 0 m/sec². Answer: 23.1 N. 2.0 kg 45° push

University Physics Volume 1

18th Edition

ISBN:9781938168277

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

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

Chapter1: Units And Measurement

Section: Chapter Questions

Problem 81AP: Consider the equation s=s0+v0t+a0t2/2+j0t3/6+s0t4/24+ct5/120 , were s is a length and t is a time....

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