1. A 45kg box is being pushed up an incline (theta 27 degrees) at constant speed
ID: 1368329 • Letter: 1
Question
1. A 45kg box is being pushed up an incline (theta 27 degrees) at constant speed. What is the magnitude of force F? Assume ms=0.36 and mk=0.312. A hanging weight (m1=37kg) is connected by a very strong, massless string over a frictionless pulley to a block (m2=21kg) sliding on a flat table. If the coefficient of static friction between the table and the second block is ms=0.33 and the coefficient of kinetic friction is mk=0.28, what is the tension in the string? 1. A 45kg box is being pushed up an incline (theta 27 degrees) at constant speed. What is the magnitude of force F? Assume ms=0.36 and mk=0.31
2. A hanging weight (m1=37kg) is connected by a very strong, massless string over a frictionless pulley to a block (m2=21kg) sliding on a flat table. If the coefficient of static friction between the table and the second block is ms=0.33 and the coefficient of kinetic friction is mk=0.28, what is the tension in the string? 1. A 45kg box is being pushed up an incline (theta 27 degrees) at constant speed. What is the magnitude of force F? Assume ms=0.36 and mk=0.31
2. A hanging weight (m1=37kg) is connected by a very strong, massless string over a frictionless pulley to a block (m2=21kg) sliding on a flat table. If the coefficient of static friction between the table and the second block is ms=0.33 and the coefficient of kinetic friction is mk=0.28, what is the tension in the string?
Explanation / Answer
1) Since the block is moving at constant speed, kinetic friction will be the frictional force applied by the ground.
Foce balance:
F = mg(sin 27o) + µkmg(cos 27o) = mg[(sin 27o) + µk*(cos 27o)]
=> F = 45 * 9.8 * [(sin 27o) + 0.31*(cos 27o)] = 322 N
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