CW8-solutions - kuehner (rjk679) CW8 Mackie (20211) 1 This...

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Unformatted text preview: kuehner (rjk679) CW8 Mackie (20211) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. DEADLINE IS CENTRAL TIME 001 (part 1 of 2) 10.0 points A light, inextensible cord passes over a light, frictionless pulley with a radius of 11 cm. It has a(n) 14 kg mass on the left and a(n) 7 . 9 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 2 . 3 m apart. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 3 m 11 cm 14 kg 7 . 9 kg At what rate are the two masses accelerat- ing when they pass each other? Correct answer: 2 . 72968 m / s 2 . Explanation: Let : R = 11 cm , m 1 = 7 . 9 kg , m 2 = 14 kg , h = 2 . 3 m , and v = R . Consider the free body diagrams 14 kg 7 . 9 kg T T m 2 g m 1 g a a Since the larger mass will move down and the smaller mass up, we can take motion downward as positive for m 2 and motion up- ward as positive for m 1 . Apply Newtons second law to m 1 and m 2 respectively and then combine the results: For mass 1: summationdisplay F 1 : T- m 1 g = m 1 a (1) For mass 2: summationdisplay F 2 : m 2 g- T = m 2 a (2) We can add Eqs. (1) and (2) above and obtain: m 2 g- m 1 g = m 1 a + m 2 a a = m 2- m 1 m 1 + m 2 g = 14 kg- 7 . 9 kg 14 kg + 7 . 9 kg (9 . 8 m / s 2 ) = 2 . 72968 m / s 2 ....
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CW8-solutions - kuehner (rjk679) CW8 Mackie (20211) 1 This...

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