Exam 2 phy - Version 030 – Exam 2 – chelikowsky...

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Unformatted text preview: Version 030 – Exam 2 – chelikowsky – (59005) 1 This print-out should have 24 questions. Multiple-choice 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 x-component 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 y-component 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.

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Exam 2 phy - Version 030 – Exam 2 – chelikowsky...

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