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Unformatted text preview: practicework 02 – BAUTISTA, ALDO – Due: Mar 6 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 3. 10 points. The suspended 2 . 2 kg mass on the right is moving up, the 1 . 6 kg mass slides down the ramp, and the suspended 7 . 4 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 17 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 1 . 6 k g μ = . 1 7 34 ◦ 7 . 4 kg 2 . 2 kg What is the acceleration of the three block system? Correct answer: 5 . 13556 m / s 2 (tolerance ± 1 %). Explanation: Let : m 1 = 2 . 2 kg , m 2 = 1 . 6 kg , m 3 = 7 . 4 kg , and θ = 34 ◦ . 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 3 the tension in the left string. Consider the free body diagrams for each mass T 3 m 3 g a T 1 m 1 g a T 3 T 1 N μ N a m 2 g For the mass m 1 , T 1 acts up and the weight m 1 g acts down, with the acceleration a di rected upward F net 1 = m 1 a = T 1 m 1 g (1) For the mass on the table, the parallel compo nent of its weight is m g sin θ and the perpen dicular component of its weight is m g cos θ . ( N = m g cos θ from equilibrium). The accel eration a is directed down the table, T 3 and the parallel weight component m 2 g sin θ act down the table, and T 1 and the frictional force μ N = μ m 2 g cos θ act up the table F net 2 = m 2 a (2) = T 3 + m 2 g sin θ T 1 μ m 2 g cos θ . For the mass m 3 , T 3 acts up and the weight m 3 g acts down, with the acceleration a di rected downward F net 3 = m 3 a = m 3 g T 3 . (3) Adding Eqs. (1), (2), & (3) yields ( m 1 + m 2 + m 3 ) a = m 3 g + m 2 g sin θ μ m 2 g cos θ m 1 g . Solving for a , we have a = [ m 2 sin θ μ m 2 cos θ + ( m 3 m 1 )] g m 1 + m 2 + m 3 = (1 . 6 kg) (9 . 8 m / s 2 ) sin 34 ◦ 2 . 2 kg + 1 . 6 kg + 7 . 4 kg (0 . 17) (1 . 6 kg) (9 . 8 m / s 2 ) cos 34 ◦ 2 . 2 kg + 1 . 6 kg + 7 . 4 kg + (7 . 4 kg 2 . 2 kg) (9 . 8 m / s 2 ) 2 . 2 kg + 1 . 6 kg + 7 . 4 kg = 5 . 13556 m / s 2 . practicework 02 – BAUTISTA, ALDO – Due: Mar 6 2006, 4:00 am 2 Question 2 Part 2 of 3. 10 points. What is the tension in the cord connected to the 2 . 2 kg block? Correct answer: 32 . 8582 N (tolerance ± 1 %). Explanation: Using Eq. 1, we have T 1 = m 1 g + m 1 a = (2 . 2 kg) (9 . 8 m / s 2 + 5 . 13556 m / s 2 ) = 32 . 8582 N . Question 3 Part 3 of 3. 10 points. What is the tension in the cord connected to the 7 . 4 kg block? Correct answer: 34 . 5169 N (tolerance ± 1 %). Explanation: Using Eq. 3, we have T 3 = m 3 g m 3 a = (7 . 4 kg) (9 . 8 m / s 2 5 . 13556 m / s 2 ) = 34 . 5169 N ....
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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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