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Unformatted text preview: Version 030 – Exam 2 – chelikowsky – (59005) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A man is pulling on a rope with a force of 94 N directed at an angle of 49 ◦ to the horizontal. What is the xcomponent of this force? 1. 60.1947 2. 48.9787 3. 20.7436 4. 58.3893 5. 53.1408 6. 31.5126 7. 21.9638 8. 57.0456 9. 61.6695 10. 72.9553 Correct answer: 61 . 6695 N. Explanation: Let : F = 94 N and θ = 49 ◦ . 9 4 N Scale: 10 N 49 ◦ F x = F cos θ = (94 N) cos49 ◦ = 61 . 6695 N . 002 (part 2 of 2) 10.0 points What is the ycomponent of this force? 1. 70.9427 2. 27.5943 3. 28.1326 4. 59.1445 5. 60.2114 6. 27.0 7. 49.1868 8. 52.9469 9. 17.785 10. 45.7379 Correct answer: 70 . 9427 N. Explanation: F y = F sin θ = (94 N) sin49 ◦ = 70 . 9427 N . 003 10.0 points What is the net force on a Mercedes convert ible traveling along a straight road at a steady speed of 100 km/h? 1. 200 N 2. 0 N correct 3. 10 N 4. 100 N 5. All are wrong. Explanation: The net force is zero because the Mercedes is traveling at constant velocity, which means with zero acceleration. 004 10.0 points An elevator starts from rest with a constant upward acceleration and moves 1 m in the first Version 030 – Exam 2 – chelikowsky – (59005) 2 1 . 8 s. A passenger in the elevator is holding a 5 . 1 kg bundle at the end of a vertical cord. The acceleration of gravity is 9 . 8 m / s 2 . What is the tension in the cord as the ele vator accelerates? 1. 99.4067 2. 57.2951 3. 81.2548 4. 78.6905 5. 95.1311 6. 35.6729 7. 104.958 8. 64.9224 9. 37.7713 10. 53.1281 Correct answer: 53 . 1281 N. Explanation: T mg a elevator g Let h be the distance traveled and a the acceleration of the elevator. With the initial velocity being zero, we simplify the following expression and solve for acceleration of the elevator: h = v t + 1 2 a t 2 = 1 2 a t 2 = ⇒ a = 2 h t 2 . The equation describing the forces acting on the bundle is F net = ma = T mg T = m ( g + a ) = m parenleftbigg g + 2 h t 2 parenrightbigg = (5 . 1 kg) bracketleftbigg 9 . 8 m / s 2 + 2 (1 m) (1 . 8 s) 2 bracketrightbigg = 53 . 1281 N . 005 10.0 points When you jump vertically off the ground, what is your acceleration when you reach your highest point? Up is positive. 1. g 2 2. g correct 3. g 3 4. g 2 5. g 3 6. 0 m / s 2 7. g 8. All are wrong. Explanation: At the top of your jump your acceleration is still g . Let the equation for acceleration via Newton’s second law guide your thinking: a = F m = mg m = g . Gravity does not cease to act at any point of your jump. The acceleration of gravity is directed to wards the center of the Earth....
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This note was uploaded on 07/24/2010 for the course PHY 59005 taught by Professor Chelikowsky during the Fall '09 term at University of Texas.
 Fall '09
 Chelikowsky

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