Exam2-solutions - Version 084 Exam2 Chiu (60180) 1 This...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 B o o k 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. 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 . 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 3. S d = 3 5 S 4. S d = 1 5 S 5. The scale reading is decreasing with time. 6. S d = 2 5 S 7. S d = 6 5 S 8. S d = 4 5 S correct 9. S d = 2 S 10. S d = 7 5 S Explanation: Looking at the left-hand free body diagram and using Newtons second law, we have S scale mg = 1 5 mg S scale = 4 5 mg = 4 5 S . 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: Newtons second law. F net = ma 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 ....
View Full Document

This note was uploaded on 03/11/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 11

Exam2-solutions - Version 084 Exam2 Chiu (60180) 1 This...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online