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h5forces2_solution

# h5forces2_solution - honea(cth632 H5Forces2 Avram(1955 This...

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honea (cth632) – H5Forces2 – Avram – (1955) 1 This print-out should have 18 questions. Multiple-choice 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 ω 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 = ω 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 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)

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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 kg 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 up-the- incline component + F cos θ while the weight force vector W = mvectorg has up-the-incline component - mg sin θ . Consequently, Ma = F net = F cos θ - M g sin θ, a = F cos θ M - g sin θ = (38 N) cos 38 5 kg - (9 . 8 m / s 2 ) sin 38 = 0 . 0446044 m / s 2 . 005 10.0 points A 4 . 3 kg object hangs at one end of a rope that is attached to a support on a railroad boxcar. When the car accelerates to the right, the rope makes an angle of 36 with the vertical The acceleration of gravity is 9 . 8 m / s 2 .
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