# test 1 - Overton, Mays – Midterm 1 – Due: Feb 16 2005,...

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Unformatted text preview: Overton, Mays – 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 μmg . 4 m 2 m 6 m T 1 T 2 F μ Determine the acceleration. 1. a = μg 2. a = 4 μg 3. a = 2 μg 4. a = 6 μg 5. a = 3 μg correct 6. a = 9 μg 7. a = 5 μg 8. a = 8 μg 9. a = 7 μg 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 μmg . 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 ma, where F f = μ ( m 1 + m 2 + m 3 ) g = 12 μmg . 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 μmg- 12 μmg 12 m = 36 μg 12 = 3 μg . 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 μmg = 2 ma 2. T 2 + 2 μmg = 2 ma 3. T 2- T 1- 2 μmg = 2 ma correct 4. T 2- T 1 + 2 μmg = 2 ma 5. T 2- 2 μmg = 2 ma 6. T 2- T 1- 2 μmg- 4 μmg = 2 ma 7. T 2 + T 1- 2 μmg- 4 μmg = 2 ma 8. T 2 + T 1 + 2 μmg- 4 μmg = 2 ma 9. T 2 + T 1 + 2 μmg = 2 ma Overton, Mays – 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- μm 2 g = m 2 a T 2- T 1- 2 μmg = 2 ma , 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- μm 1 g = m 1 a = 4 ma Since F f 1 = μm 1 g = 4 μmg and from Part 1, a = 3 μg , we have T 1- 4 μmg = 12 mμg , so T 1 = (12 + 4) μmg T 1 = 16 μmg Since F = 48 μmg , we have T 1 = 16 μmg µ F 48 μmg ¶ = 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 ....
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## This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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test 1 - Overton, Mays – Midterm 1 – Due: Feb 16 2005,...

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