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Unformatted text preview: Muraj, Hamza Quiz 1 Due: Feb 15 2006, 10:00 pm Inst: Florin 1 This printout should have 27 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points Three objects can only move along a straight, level path. The graphs below show the po sition d of each of the objects plotted as a function of time t . d t I d t II d t III The magnitude of the velocity k ~v k of the object increases in which of the cases? 1. I only 2. I, II, and III 3. II only 4. I and III only 5. II and III only 6. III only correct 7. I and II only Explanation: Case I: The object moves at constant speed. Case II: The object remains at rest. Case III: The speed of the object increases with time; i.e. , constant acceleration. Thus, the magnitude of the velocity of the object increases only in case III. 002 (part 2 of 2) 10 points The sum of the forces X F i = 0 on the object is zero in which of the cases? 1. I, II, and III 2. III only 3. I and II only correct 4. I only 5. II and III only 6. I and III only 7. II only Explanation: When the sum of the forces on the object is zero, the acceleration of the object is zero by Newtons 2 nd law. In cases I and II, the velocity of the object doesnt change with time, so the sum of the forces on the object is zero. In case III, the object is accelerating, so the sum of the forces on the object is not zero. 003 (part 1 of 2) 10 points Consider the 601 N weight held by two cables shown below. The lefthand cable had tension T 2 and makes an angle of 2 with the ceiling. The righthand cable had tension 390 N and makes an angle of 42 with the ceiling. The righthand cable makes an angle of 42 with the ceiling and has a tension of 390 N . 601 N T 2 3 9 N 4 2 2 a) What is the tension T 2 in the lefthand cable slanted at an angle of 2 with respect to the wall? Correct answer: 446 . 795 N. Explanation: Observe the freebody diagram below. Muraj, Hamza Quiz 1 Due: Feb 15 2006, 10:00 pm Inst: Florin 2 F 2 F 1 1 2 W g Note: The sum of the x and ycomponents of F 1 , F 2 , and W g are equal to zero. Given : W g = 601 N , F 1 = 390 N , 1 = 42 , and 2 = 90  . Basic Concept: Vertically and Horizontally, we have F x net = F x 1 F x 2 = 0 = F 1 cos 1 F 2 cos 2 = 0 (1) F y net = F y 1 + F y 2W g = 0 = F 1 sin 1 + F 2 sin 2W g = 0 (2) Solution: Using Eqs. 1 and 2, we have F x 2 = F 1 cos 1 (1) = (390 N)cos42 = 289 . 826 N , and F y 2 = F 3 F 1 sin 1 (2) = 601 N (390 N)sin42 = 601 N 260 . 961 N = 340 . 039 N , so F 2 = q ( F x 2 ) 2 + ( F y 2 ) 2 = q (289 . 826 N) 2 + (340 . 039 N) 2 = 446 . 795 N ....
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This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Turner

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