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Unformatted text preview: hernandez (ejh742) oldhomework 11 Turner (58120) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points The suspended 2 . 4 kg mass on the right is moving up, the 1 . 4 kg mass slides down the ramp, and the suspended 7 . 7 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 12 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 1 . 4 k g = . 1 2 37 7 . 7 kg 2 . 4 kg What is the acceleration of the three block system? Correct answer: 5 . 12018 m / s 2 . Explanation: Let : m 1 = 2 . 4 kg , m 2 = 1 . 4 kg , m 3 = 7 . 7 kg , and = 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 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 and the perpen dicular component of its weight is mg cos . ( N = mg cos from equilibrium). The accel eration a is directed down the table, T 3 and the parallel weight component m 2 g sin act down the table, and T 1 and the frictional force N = m 2 g cos act up the table F net 2 = m 2 a (2) = T 3 + m 2 g sin T 1 m 2 g cos . For the mass m 3 , T 3 acts up and the weight m 3 g acts down, with the acceleration a di rected downward F net 3 = m 3 a = m 3 g T 3 . (3) Adding Eqs. (1), (2), & (3) yields ( m 1 + m 2 + m 3 ) a = m 3 g + m 2 g sin m 2 g cos m 1 g . Solving for a , we have a = [ m 2 sin m 2 cos + ( m 3 m 1 )] g m 1 + m 2 + m 3 = (1 . 4 kg) (9 . 8 m / s 2 ) sin 37 2 . 4 kg + 1 . 4 kg + 7 . 7 kg (0 . 12) (1 . 4 kg) (9 . 8 m / s 2 ) cos 37 2 . 4 kg + 1 . 4 kg + 7 . 7 kg + (7 . 7 kg 2 . 4 kg) (9 . 8 m / s 2 ) 2 . 4 kg + 1 . 4 kg + 7 . 7 kg = 5 . 12018 m / s 2 . hernandez (ejh742) oldhomework 11 Turner (58120) 2 002 (part 2 of 3) 10.0 points(part 2 of 3) 10....
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 Spring '08
 Turner
 Physics, Mass, Work

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