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Unformatted text preview: honea (cth632) H5Forces2 Avram (1955) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Two blocks m 1 (11 kg) and m 2 (17 kg) are connected by strings, as shown in the dia gram. String 2 breaks when T 2 = 4 . 9179 N. Find T 1 when string 2 breaks. m 1 m 2 T 2 T 1 Correct answer: 12 . 5183 N. Explanation: Find T 1 when T 2 breaks at 4 . 9179 N. summationdisplay F x on m 1 = T 2 = m 1 a a = T 2 m 1 = 4 . 9179 N 11 kg = 0 . 447082 m / s 2 summationdisplay F x on the whole system = T 1 = ( m 1 + m 2 ) a = ( m 1 + m 2 ) T 2 m 1 = ( m 1 + m 2 m 1 ) T 2 = (1 + m 2 m 1 ) T 2 = (1 + 1 . 54545) 4 . 9179 N = (2 . 54545) (4 . 9179 N) = 12 . 5183 N 002 (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) 23 kg mass on the left and a(n) 3 . 1 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 1 . 9 m apart. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 9 m 14 cm 23 kg 3 . 1 kg At what rate are the two masses accelerat ing when they pass each other? Correct answer: 7 . 47203 m / s 2 . Explanation: Let : R = 14 cm , m 1 = 3 . 1 kg , m 2 = 23 kg , h = 1 . 9 m , and v = R. Consider the free body diagrams 23 kg 3 . 1 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) honea (cth632) H5Forces2 Avram (1955) 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 = 23 kg 3 . 1 kg 23 kg + 3 . 1 kg (9 . 8 m / s 2 ) = 7 . 47203 m / s 2 . 003 (part 2 of 2) 10.0 points What is the tension in the cord when they pass each other? Correct answer: 53 . 5433 N. Explanation: T = m 1 ( g + a ) = (3 . 1 kg) (9 . 8 m / s 2 + 7 . 47203 m / s 2 ) = 53 . 5433 N . 004 10.0 points A block is pushed up a frictionless incline by an applied horizontal force as shown. The acceleration of gravity is 9 . 8 m / s 2 . 5 k g F = 38 N 38 What is the magnitude of the resulting ac celeration of the block? Correct answer: 0 . 0446044 m / s 2 . Explanation: Basic Concepts: summationdisplay vector F = mvectora Solution: Consider the forces acting on the block and their components along the incline. The horizontal pushing force vector F has upthe incline component + F cos while the weight force vector W = mvectorg has uptheincline component mg sin . Consequently, Ma = F net = F cos  M g sin , a = F cos M g sin = (38 N) cos38 5 kg (9 . 8 m / s 2 ) sin38 = 0 . 0446044 m / s 2 ....
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This note was uploaded on 04/08/2011 for the course PHYS 1402 taught by Professor Avram during the Spring '10 term at Austin Community College.
 Spring '10
 Avram
 Physics, Force

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