# HW4 - boland(kbb544 – HW4 – mackie –(10611 1 This...

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Unformatted text preview: boland (kbb544) – HW4 – mackie – (10611) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. NOTE DEADLINE is 11pm (central) tues 9/28 BTW, 6/10 of these were done (or will be done by tues) in class. 001 (part 1 of 2) 10.0 points A block of mass 7 . 13 kg lies on a frictionless horizontal surface. The block is connected by a cord passing over a pulley to another block of mass 6 . 26 kg which hangs in the air, as shown on the following picture. Assume the cord to be light (massless and weightless) and unstretchable and the pulley to have no friction and no rotational inertia. The acceleration of gravity is 9 . 8 m / s 2 . 7 . 13 kg 6 . 26 kg Calculate the acceleration of the first block. Correct answer: 4 . 58163 m / s 2 . Explanation: Given : m 1 = 7 . 13 kg , and m 2 = 6 . 26 kg . m 1 m 2 a T N m 1 g a T m 2 g Since the cord is unstretchable, the first block accelerates to the right at exactly the same rate a as the second (hanging) block ac- celerates downward. Also, the cord’s tension pulls the first block to the right with exactly the same tension T as it pulls the second block upward. The only horizontal force acting on the first block is the cord’s tension T , hence by New- ton’s Second Law m 1 a = F net → 1 = T . The second block feels two vertical forces: The cord’s tension T (upward) and the block’s own weight W 2 = m 2 g (downward). Conse- quently, m 2 a = F net ↓ 2 = m 2 g- T . Adding the two equations together, we arrive at ( m 1 + m 2 ) a = m 2 g , and hence a = m 2 m 1 + m 2 g = 6 . 26 kg 7 . 13 kg + 6 . 26 kg (9 . 8 m / s 2 ) = 4 . 58163 m / s 2 . 002 (part 2 of 2) 10.0 points Calculate the tension in the cord. Correct answer: 32 . 667 N. Explanation: T = m 1 a = (7 . 13 kg) (4 . 58163 m / s 2 ) = 32 . 667 N . 003 (part 1 of 2) 10.0 points A light, inextensible cord passes over a light, frictionless pulley with a radius of 14 cm. It has a(n) 22 kg mass on the left and a(n) 2 . 5 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 2 . 8 m apart. The acceleration of gravity is 9 . 8 m / s 2 . boland (kbb544) – HW4 – mackie – (10611) 2 2 . 8 m 14 cm ω 22 kg 2 . 5 kg At what rate are the two masses accelerat- ing when they pass each other? Correct answer: 7 . 8 m / s 2 . Explanation: Let : R = 14 cm , m 1 = 2 . 5 kg , m 2 = 22 kg , h = 2 . 8 m , and v = ω R . Consider the free body diagrams 22 kg 2 . 5 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 =...
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## This note was uploaded on 02/20/2011 for the course PHYS 1062 taught by Professor Mackie during the Spring '11 term at Temple.

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HW4 - boland(kbb544 – HW4 – mackie –(10611 1 This...

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