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Unformatted text preview: hernandez (ejh742) – homework 11 – Turner – (58120) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points There is friction between the block and the table. The suspended 3 kg mass on the left is moving up, the 4 kg mass slides to the right on the table, and the suspended mass 8 kg on the right is moving down. The acceleration of gravity is 9 . 8 m / s 2 . 3 kg 4 kg 8 kg μ = 0 . 19 What is the magnitude of the acceleration of the system? Correct answer: 2 . 77013 m / s 2 . Explanation: m 1 m 2 m 3 μ a Let : m 1 = 3 kg , m 2 = 4 kg , m 3 = 8 kg , and μ = 0 . 19 . Basic Concepts: The acceleration a of each mass is the same, but the tensions in the two strings will be different. F net = ma negationslash = 0 Solution: Let T 1 be the tension in the left string and T 2 be the tension in the right string. Consider the free body diagrams for each mass T 1 m 1 g a T 2 m 3 g a T 1 T 2 N μ N a m 2 g For the mass m 1 , T 1 acts up and the weight m 1 g acts down, with the acceleration a di rected upward, so F net 1 = m 1 a = T 1 − m 1 g . (1) For the mass on the table, a is directed to the right, T 2 acts to the right, T 1 acts to the left, and the motion is to the right so that the frictional force μ m 2 g acts to the left and F net 2 = m 2 a = T 2 − T 1 − μ m 2 g . (2) For the mass m 3 , T 2 acts up and the weight m 3 g acts down, with the acceleration a di rected downward, so F net 3 = m 3 a = m 3 g − T 2 . (3) Adding these equations yields ( m 1 + m 2 + m 3 ) a = m 3 g − μ m 2 g − m 1 g a = m 3 − μ m 2 − m 1 m 1 + m 2 + m 3 g = 8 kg − (0 . 19) (4 kg) − 3 kg 3 kg + 4 kg + 8 kg × (9 . 8 m / s 2 ) = 2 . 77013 m / s 2 . 002 10.0 points The suspended 2 . 4 kg mass on the right is moving up, the 2 . 4 kg mass slides down the ramp, and the suspended 7 . 9 kg mass on the hernandez (ejh742) – homework 11 – Turner – (58120) 2 left is moving down. There is friction between the block and the ramp. The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 2 . 4 k g μ = . 1 4 37 ◦ 7 . 9 kg 2 . 4 kg What is the tension in the cord connected to the 7 . 9 kg block? Correct answer: 36 . 7226 N. Explanation: Let : m 1 = 2 . 4 kg , m 2 = 2 . 4 kg , m 3 = 7 . 9 kg , and θ = 37 ◦ . Basic Concept: F net = ma negationslash = 0 Solution: The acceleration a of each mass is the same, but the tensions in the two strings will be different. Let T 1 be the tension in the right string and T 3 the tension in the left string. Consider the free body diagrams for each mass T 3 m 3 g a T 1 m 1 g a T 3 T 1 N μ N a m 2 g For the mass m 1 , T 1 acts up and the weight m 1 g acts down, with the acceleration a di rected upward F net 1 = m 1 a = T 1 − m 1 g (1) For the mass on the table, the parallel compo nent of its weight is mg sin θ and the perpen dicular component of its weight is mg cos θ ....
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This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Friction, Work

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