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Turner Exam 1 Spring 2005

# Turner Exam 1 Spring 2005 - Trevino Denise Midterm 1 Due...

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Trevino, Denise – Midterm 1 – Due: Feb 16 2005, 10:00 pm – Inst: Turner 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points Consider a force pulling 3 blocks along a rough horizontal surface, where the masses are a multiple of a given mass m , as shown in the figure below. The coefficient of the kinetic friction is μ . The blocks are pulled by a force of 48 μ m g . 4 m 2 m 6 m T 1 T 2 F μ Determine the acceleration. 1. a = μ g 2. a = 4 μ g 3. a = 2 μ g 4. a = 6 μ g 5. a = 3 μ g correct 6. a = 9 μ g 7. a = 5 μ g 8. a = 8 μ g 9. a = 7 μ g Explanation: m 1 m 2 m 3 T 1 T 2 F μ Given : m 1 = 4 m , m 2 = 2 m , m 3 = 6 m , m 1 + m 2 + m 3 = 12 m , and F = 48 μ m g . Basic Concept: Newton’s 2nd law. Solution: Since all three blocks must have the same acceleration (no block can “speed ahead” or “lag behind”) Newton’s 2nd Law applies to the whole system of three blocks. (Take right to be positive.) m 1 + m 2 + m 3 F μ ( m 1 + m 2 + m 3 ) g F f F - F f = ( m 1 + m 2 + m 3 ) a = 12 m a , where F f = μ ( m 1 + m 2 + m 3 ) g = 12 μ m g . a = F - F f ( m 1 + m 2 + m 3 ) = F - μ ( m 1 + m 2 + m 3 ) g ( m 1 + m 2 + m 3 ) = 48 μ m g - 12 μ m g 12 m = 36 μ g 12 = 3 μ g . 002 (part 2 of 3) 10 points The equation of motion for the central mass 2 m is given by 1. T 2 + T 1 - 2 μ m g = 2 m a 2. T 2 + 2 μ m g = 2 m a 3. T 2 - T 1 - 2 μ m g = 2 m a correct 4. T 2 - T 1 + 2 μ m g = 2 m a 5. T 2 - 2 μ m g = 2 m a 6. T 2 - T 1 - 2 μ m g - 4 μ m g = 2 m a 7. T 2 + T 1 - 2 μ m g - 4 μ m g = 2 m a 8. T 2 + T 1 + 2 μ m g - 4 μ m g = 2 m a 9. T 2 + T 1 + 2 μ m g = 2 m a

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Trevino, Denise – Midterm 1 – Due: Feb 16 2005, 10:00 pm – Inst: Turner 2 Explanation: The acceleration is to the right. m 2 T 1 T 2 μ m 2 g The free body diagram for m 2 shows T 2 - T 1 - μ m 2 g = m 2 a T 2 - T 1 - 2 μ m g = 2 m a , since m 2 = 2 m . 003 (part 3 of 3) 10 points Find the tension T 1 in the rope between the masses 4 m and 2 m in terms of F . 1. T 1 = F 2 2. T 1 = F 4 3. T 1 = F 6 4. T 1 = F 7 5. T 1 = F 9 6. T 1 = F 3 correct 7. T 1 = F 5 8. T 1 = F 8 9. T 1 = F Explanation: m 1 T 1 μ m 1 g The free body diagram for m 1 shows T 1 - μ m 1 g = m 1 a = 4 m a Since F f 1 = μ m 1 g = 4 μ m g and from Part 1, a = 3 μ g , we have T 1 - 4 μ m g = 12 m μ g , so T 1 = (12 + 4) μ m g T 1 = 16 μ m g Since F = 48 μ m g , we have T 1 = 16 μ m g F 48 μ m g = 16 48 F = F 3 . 004 (part 1 of 1) 10 points At the beginning of a new school term, a student moves a box of books by attaching a rope to the box and pulling with a force of F = 83 N at an angle of 64 , as shown in the figure. The acceleration of gravity is 9 . 8 m / s 2 . The box of books has a mass of 15 kg and the coefficient of kinetic friction between the bottom of the box and the floor is 0 . 25. 15 kg F 64 μ = 0 . 25 What is the acceleration of the box? Correct answer: 1 . 21898 m / s 2 . Explanation: Given : m = 15 kg , μ k = 0 . 25 , F = 83 N , and θ = 64 .
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Turner Exam 1 Spring 2005 - Trevino Denise Midterm 1 Due...

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