Turner Exam 1 Spring 2005

Turner Exam 1 Spring 2005 - Trevino, Denise Midterm 1 Due:...

Info iconThis preview shows pages 1–3. 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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 . The blocks are pulled by a force of 48 mg . 4 m 2 m 6 m T 1 T 2 F Determine the acceleration. 1. a = g 2. a = 4 g 3. a = 2 g 4. a = 6 g 5. a = 3 g correct 6. a = 9 g 7. a = 5 g 8. a = 8 g 9. a = 7 g Explanation: m 1 m 2 m 3 T 1 T 2 F Given : m 1 = 4 m, m 2 = 2 m, m 3 = 6 m, m 1 + m 2 + m 3 = 12 m, and F = 48 mg . Basic Concept: Newtons 2nd law. Solution: Since all three blocks must have the same acceleration (no block can speed ahead or lag behind) Newtons 2nd Law applies to the whole system of three blocks. (Take right to be positive.) m 1 + m 2 + m 3 F ( m 1 + m 2 + m 3 ) g F f F- F f = ( m 1 + m 2 + m 3 ) a = 12 ma, where F f = ( m 1 + m 2 + m 3 ) g = 12 mg . a = F- F f ( m 1 + m 2 + m 3 ) = F- ( m 1 + m 2 + m 3 ) g ( m 1 + m 2 + m 3 ) = 48 mg- 12 mg 12 m = 36 g 12 = 3 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 mg = 2 ma 2. T 2 + 2 mg = 2 ma 3. T 2- T 1- 2 mg = 2 ma correct 4. T 2- T 1 + 2 mg = 2 ma 5. T 2- 2 mg = 2 ma 6. T 2- T 1- 2 mg- 4 mg = 2 ma 7. T 2 + T 1- 2 mg- 4 mg = 2 ma 8. T 2 + T 1 + 2 mg- 4 mg = 2 ma 9. T 2 + T 1 + 2 mg = 2 ma 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 m 2 g The free body diagram for m 2 shows T 2- T 1- m 2 g = m 2 a T 2- T 1- 2 mg = 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 m 1 g The free body diagram for m 1 shows T 1- m 1 g = m 1 a = 4 ma Since F f 1 = m 1 g = 4 mg and from Part 1, a = 3 g , we have T 1- 4 mg = 12 mg , so T 1 = (12 + 4) mg T 1 = 16 mg Since F = 48 mg , we have T 1 = 16 mg F 48 mg = 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 ....
View Full Document

This note was uploaded on 02/09/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 15

Turner Exam 1 Spring 2005 - Trevino, Denise Midterm 1 Due:...

This preview shows document pages 1 - 3. Sign up to view the full document.

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