The system shown above is released from rest.
3) The system shown above is released from rest. If friction is negligible, the acceleration of the 4.0 kg block sliding on the table shown above is most nearly
4) A 50-N crate sits on the horizontal floor where the coefficient of static friction between the crate and the floor is 0.50. A 20-N force is applied to the crate acting to the right. What is the resulting static frictional force acting on the crate.
(a) 20 N to the right
(b) 20 N to the left
(c) 25 N to the right
(d) 25 N to the left
(e) none of the above; the crate starts to move
Homework Answers
5) An object is moving to the left at constant velocity. A constant force to the left is now applied to the object...
5) An object is moving to the left at constant velocity. A constant force to the left is now applied to the object. Which of the following is correct? A) The object continues to move to the left, and its speed increases. B) The object continues to move to the left with a constant speed. C) The object continues to move to the left, and its speed decreases. D) The object immediately stops and then begins moving to the right...
In the figure above, block A has mass mA = 25 kg and block B has...
In the figure above, block A has mass mA = 25 kg and block B has mass mB = 10 kg. Both blocks move with constant acceleration a = 2 m/s2 to the right, and the coefficient of static friction between the two blocks is = 0.8. The static frictional force acting between the blocks is (A) 20 N (B) 50 N (C) 78 N (D) 196 N E) 274 N
Chapter 3, Practice Problem 3/022 The system is released from rest in the position shown. Calculate...
Chapter 3, Practice Problem 3/022 The system is released from rest in the position shown. Calculate the tension T in the cord and the acceleration a positive if to the right, negative if to the left) of the 21-kg block. The small pulley attached to the block has negligible mass and friction. (Suggestion: First establish the kinematic relationship between the accelerations of the two bodies.) 21 kg Hs = H = = 0.50 23 kg Answers: N Jm/s²
Why is the friction equal to the force? A crate of mass 180 kg sits on...
Why is the friction equal to the force?
A crate of mass 180 kg sits on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.4, and the coefficient of kinetic friction is 0.36. You push the crate with a horizontal force of 600 N. What is the magnitude of the frictional force on the crate by the floor in this situation? [Use g = 9.81 m/s2] Incorrect 635.69 N. Incorrect 706.32 N Incorrect...
Suppose you have to move a heavy crate of weight 875 N by sliding it along...
Suppose you have to move a heavy crate of weight 875 N by sliding it along a horizontal concrete floor. You push the crate to the right with a horizontal force of magnitude 300 N, but friction prevents the crate from sliding. What is the magnitude Fp of the minimum force you need to exert on the crate to make it start sliding along the floor? Let the coefficient of static friction μs between the crate and the floor be...
10b.4 A 5-kg block is at rest o 60 N to the right, F2 the left...
10b.4 A 5-kg block is at rest o 60 N to the right, F2 the left is at rest on a frictional floor. The coefficient of static friction - 15 N to the left. What is the frictional force? lAnswer: 40 N be block and the floor is 0.8. Two horizontal forces are acted on the block: Fi
Suppose you have to move a heavy crate of weight 875 N by sliding it along a horizontal concrete floor. You push the cr...
Suppose you have to move a heavy crate of weight 875 N by sliding it along a horizontal concrete floor. You push the crate to the right with a horizontal force of magnitude 300 N, but friction prevents the crate from sliding. Part A Draw a free-body diagram of the crate. Part B What is the magnitude Fp of the minimum force you need to exert on the crate to make it start sliding along the floor? Let the coefficient...
A 50-N crate sits on a horizontal floor where the coefficient of static friction between the...
A 50-N crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is .50 and that of kinetic is .40 gradually increasing horizontal force to the right is applied to the crate how would you describe the motion of the box at the moment forces reached to 20 N (a) Not moving (b) not moving bit anootbto move (c) moving with a constant speed (d) Accelerating
Although friction is a complicated force, many aspects of it can be described with a simple model.
Tactics Box 5.1 Working with Friction ForcesLearning Goal:To practice Tactics Box 5.1 Working with friction forces.Although friction is a complicated force, many aspects of it can be described with a simple model. This Tactics Box presents the conditions under which these models are valid and outlines the essential steps in solving friction problems.TACTICS BOX 5.1 Working with friction forces1. If the object is not moving relative to the surface it's in contact with, the friction force is static friction. Draw...
A large crate is at rest on a horizontal floor. The coefficient of static friction between...
A large crate is at rest on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.500. A force F⃗ is applied to the crate in a direction 30.0∘ above the horizontal. The minimum value of F required to get the crate to start sliding is 410 N. Part A What is the mass of the crate? Express your answer with the appropriate units.
5) An object is moving to the left at constant velocity. A constant force to the left is now applied to the object. Which of the following is correct? A) The object continues to move to the left, and its speed increases. B) The object continues to move to the left with a constant speed. C) The object continues to move to the left, and its speed decreases. D) The object immediately stops and then begins moving to the right...
In the figure above, block A has mass mA = 25 kg and block B has mass mB = 10 kg. Both blocks move with constant acceleration a = 2 m/s2 to the right, and the coefficient of static friction between the two blocks is = 0.8. The static frictional force acting between the blocks is (A) 20 N (B) 50 N (C) 78 N (D) 196 N E) 274 N
Chapter 3, Practice Problem 3/022 The system is released from rest in the position shown. Calculate the tension T in the cord and the acceleration a positive if to the right, negative if to the left) of the 21-kg block. The small pulley attached to the block has negligible mass and friction. (Suggestion: First establish the kinematic relationship between the accelerations of the two bodies.) 21 kg Hs = H = = 0.50 23 kg Answers: N Jm/s²
Why is the friction equal to the force?
A crate of mass 180 kg sits on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.4, and the coefficient of kinetic friction is 0.36. You push the crate with a horizontal force of 600 N. What is the magnitude of the frictional force on the crate by the floor in this situation? [Use g = 9.81 m/s2] Incorrect 635.69 N. Incorrect 706.32 N Incorrect...
10b.4 A 5-kg block is at rest o 60 N to the right, F2 the left is at rest on a frictional floor. The coefficient of static friction - 15 N to the left. What is the frictional force? lAnswer: 40 N be block and the floor is 0.8. Two horizontal forces are acted on the block: Fi
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