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# hw09 - 5.4 cm m2 g a 3.6 kg m1 g 16 kg a 001(part 1 of 2...

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alexander (jra2623) – homework 09 – Turner – (92510) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A light, inextensible cord passes over a light, frictionless pulley with a radius of 5 . 4 cm. It has a(n) 16 kg mass on the left and a(n) 3 . 6 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 1 . 7 m apart. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 7 m 5 . 4 cm ω 16 kg 3 . 6 kg At what rate are the two masses accelerat- ing when they pass each other? Correct answer: 6 . 2 m / s 2 . Explanation: Let : R = 5 . 4 cm , m 1 = 3 . 6 kg , m 2 = 16 kg , h = 1 . 7 m , and v = ω R . Consider the free body diagrams 16 kg 3 . 6 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 Newton’s 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 = 16 kg - 3 . 6 kg 16 kg + 3 . 6 kg (9 . 8 m / s 2 ) = 6 . 2 m / s 2 . 002 (part 2 of 2) 10.0 points What is the tension in the cord when they pass each other? Correct answer: 57 . 6 N. Explanation: T = m 1 ( g + a ) = (3 . 6 kg) (9 . 8 m / s 2 + 6 . 2 m / s 2 ) = 57 . 6 N . 003 10.0 points

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alexander (jra2623) – homework 09 – Turner – (92510) 2 A person weighing 0 . 6 kN rides in an elevator that has a downward acceleration of 2 . 8 m / s 2 .
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