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Unformatted text preview: homework 10 – BAUTISTA, ALDO – Due: Feb 13 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 1. 10 points. Two blocks are arranged at the ends of a massless string as shown in the figure. The system starts from rest. When the 2 . 34 kg mass has fallen through 0 . 448 m, its down ward speed is 1 . 26 m / s. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 34 kg 3 . 7 kg μ a What is the frictional force between the 3 . 7 kg mass and the table? Correct answer: 12 . 2299 N (tolerance ± 1 %). Explanation: Given : m 1 = 2 . 34 kg , m 2 = 3 . 7 kg , v = 0 m / s , and v = 1 . 26 m / s . Basic Concept: Newton’s Second Law F = M a Solution: The acceleration of m 1 is obtained from the equation v 2 v 2 = 2 a ( s s ) a = v 2 v 2 2 h = (1 . 26 m / s) 2 (0 m / s) 2 2 (0 . 448 m) = 1 . 77188 m / s 2 . Consider free body diagrams for the two masses T m 1 g a T N μ N a m 2 g Because m 1 and m 2 are tied together with string, they have same the speed and the same acceleration, so the net force exerted on m 2 is F 2 = m 2 a The net force on m 1 is m 1 a = m 1 g T , so that T = m 1 g m 1 a . Thus F 2 = T f k , f k = T F 2 = m 1 g ( m 1 + m 2 ) a = (2 . 34 kg) (9 . 8 m / s 2 ) (2 . 34 kg + 3 . 7 kg) × (1 . 77188 m / s 2 ) = 12 . 2299 N . Question 2 Part 1 of 1. 10 points. The suspended m 1 mass on the right is moving up, the m 2 mass slides down the ramp, and the suspended m 3 mass on the left is moving down. There is friction between the block and the ramp, with the coefficient of the kinetic friction μ . The acceleration of gravity is g and the acceleration of the three block system is a . The pulleys are massless and frictionless. homework 10 – BAUTISTA, ALDO – Due: Feb 13 2006, 4:00 am 2 m 2 μ θ m 3 m 1 What is the equation of motion of the three block system? 1. m 3 g + m 2 g (cos θ μ sin θ ) m 1 g = ( m 1 + m 2 + m 3 ) a 2. m 3 g + m 2 g (sin θ μ sin θ ) m 1 g = ( m 1 + m 2 + m 3 ) a 3. m 3 g + m 2 g (1 μ ) m 1 g = ( m 1 + m 3 ) a 4. m 3 g + m 2 g (sin θ μ cos θ ) m 1 g = ( m 1 + m 2 + m 3 ) a correct 5. m 3 g + m 2 g (cos θ μ cos θ ) m 1 g = ( m 1 + m 2 + m 3 ) a 6. m 3 g + m 2 g (cos θ μ sin θ ) m 1 g = ( m 1 + m 3 ) a 7. m 3 g + m 2 g (cos θ μ cos θ ) m 1 g = ( m 1 + m 3 ) a 8. m 3 g + m 2 g (sin θ μ sin θ ) m 1 g = ( m 1 + m 3 ) a 9. m 3 g + m 2 g (sin θ μ cos θ ) m 1 g = ( m 1 + m 3 ) a 10. m 3 g + m 2 g (1 μ ) m 1 g = ( m 1 + m 2 + m 3 ) a Explanation: Basic Concept: F net = m a 6 = 0 Solution: The acceleration a of each mass is the same, but the tensions in the two strings will be different. Let T 1 be the tension in the right string and T 2 the tension in the left string....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Mass, Work

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