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Exam2-solutions

Exam2-solutions - Version 084 Exam2 Chiu(60180 This...

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Version 084 – Exam2 – Chiu – (60180) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A book is at rest on an incline as shown above. A constant force vertically downward is in contact with the book. F Book The following figures show several attempts at drawing free-body diagrams for the book. Which figure has the correct directions for each force? The magnitudes of the forces are not necessarily drawn to scale. 1. normal force friction weight 2. weight force normal friction 3. weight force normal friction 4. normal friction weight force 5. weight normal friction force 6. weight friction normal force 7. weight friction force normal 8. weight force friction normal correct Explanation: The normal force points perpendicular to the surface of the inclined plane. The weight force points down. The F hand also points down. The friction force keeps the block from sliding and consequently points up the incline. 002 10.0 points Consider a man standing on a scale in an elevator.

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Version 084 – Exam2 – Chiu – (60180) 2 Scale Given: g is the magnitude of the gravita- tional acceleration. When the elevator stays at rest, the scale reads S 0 . Suppose the elevator now acceler- ates downward at a constant rate of 1 5 g . What is the scale reading S d during the downward motion? 1. The scale reading is increasing with time. 2. S d = S 0 3. S d = 3 5 S 0 4. S d = 1 5 S 0 5. The scale reading is decreasing with time. 6. S d = 2 5 S 0 7. S d = 6 5 S 0 8. S d = 4 5 S 0 correct 9. S d = 2 S 0 10. S d = 7 5 S 0 Explanation: Looking at the left-hand free body diagram and using Newton’s second law, we have S scale m g = 1 5 m g S scale = 4 5 m g = 4 5 S 0 . 003 10.0 points The horizontal surface on which the objects slide is frictionless. 1 . 4 kg 2 . 8 kg 4 . 2 kg T 2 T 1 F If the tension in string T 1 = 22 N, deter- mine F . 1. 44.0 2. 58.0 3. 28.0 4. 68.0 5. 32.0 6. 56.0 7. 54.0 8. 48.0 9. 50.0 10. 40.0 Correct answer: 44 N. Explanation: Let : m 1 = 1 . 4 kg , m 2 = 2 . 8 kg , m 3 = 4 . 2 kg , and m 1 + m 2 + m 3 = 8 . 4 kg . m 1 m 2 m 3 T 2 T 1 F Basic Concept: Newton’s second law. F net = m a for each body. m 1 T 2
Version 084 – Exam2 – Chiu – (60180) 3 m 2 T 2 T 1 m 3 T 1 F Solution: For the leftmost body, T 2 acts to the right, so m 1 a = T 2 . For the middle body, T 1 acts to the right and T 2 to the left, so m 2 a = T 1 T 2 . Adding these equations, ( m 1 + m 2 ) a = T 1 , so a = T 1 m 1 + m 2 = 22 N 4 . 2 kg = 5 . 2381 m / s 2 . For the rightmost body, F acts to the right and T 1 to the left, so m 3 a = F T 1 F = T 1 + m 3 a = T 1 + m 3 T 1 ( m 1 + m 2 ) = T 2 parenleftbigg m 1 + m 2 + m 3 m 1 + m 2 parenrightbigg = (22 N) parenleftbigg 1 . 4 kg + 2 . 8 kg + 4 . 2 kg 1 . 4 kg + 2 . 8 kg parenrightbigg = (22 N) parenleftbigg 8 . 4 kg 4 . 2 kg parenrightbigg = 44 N . 004 10.0 points Two blocks are connected by a thin inexten- sible string over a frictionless massless pulley as shown on the picture below The acceleration of gravity is 9 . 8 m / s 2 .

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Exam2-solutions - Version 084 Exam2 Chiu(60180 This...

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