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Unformatted text preview: burgin (lcb626) – WH04 – markert – (56480) 1 This printout should have 15 questions. Multiplechoice 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 15 cm. It has a(n) 17 kg mass on the left and a(n) 2 . 7 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 4 . 6 m apart. The acceleration of gravity is 9 . 8 m / s 2 . 4 . 6 m 15 cm ω 17 kg 2 . 7 kg At what rate are the two masses accelerat ing when they pass each other? Correct answer: 7 . 11371 m / s 2 . Explanation: Let : R = 15 cm , m 1 = 2 . 7 kg , m 2 = 17 kg , h = 4 . 6 m , and v = ω R. Consider the free body diagrams 17 kg 2 . 7 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 = 17 kg 2 . 7 kg 17 kg + 2 . 7 kg (9 . 8 m / s 2 ) = 7 . 11371 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: 45 . 667 N. Explanation: T = m 1 ( g + a ) = (2 . 7 kg) (9 . 8 m / s 2 + 7 . 11371 m / s 2 ) = 45 . 667 N . 003 10.0 points burgin (lcb626) – WH04 – markert – (56480) 2 A force F = 103 N acts at an angle α = 37 ◦ with respect to the horizontal on a block of mass m = 23 . 8 kg, which is at rest on a horizontal plane. The acceleration of gravity is 9 . 81 m / s 2 . If the static frictional coefficient is μ s = . 82, what is the force of static friction? 23 . 8 kg μ s = 0 . 82 1 3 N 37 ◦ 1. 258 . 905 N 2. 123 . 999 N 3. 67 . 4528 N 4. 0 N 5. 50 . 8293 N 6. 82 . 2595 N correct 7. 242 . 281 N 8. 140 . 623 N 9. 61 . 9869 N 10. 191 . 452 N Explanation: Let : m = 23 . 8 kg , F applied = 103 N , α = 37 ◦ , μ s = 0 . 82 , and g = 9 . 81 m / s 2 m F a p p l i e d α mg N F s Basic Concepts: F applied,x = F applied cos α F applied,y = F applied sin α F y,net = N  F applied,y mg = 0 F x,net = ma x = F applied,x F s = 0 when at rest....
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This note was uploaded on 10/25/2010 for the course PHY PHYSICS taught by Professor Swinney during the Fall '08 term at University of Texas.
 Fall '08
 Swinney

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