Exam 2-solutions

Exam 2-solutions - Version 014 – Exam 2 – chelikowsky...

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Unformatted text preview: Version 014 – 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 72 N directed at an angle of 47 ◦ to the horizontal. What is the x-component of this force? 1. 21.5535 2. 49.8124 3. 60.0159 4. 61.6695 5. 52.8459 6. 67.7689 7. 72.9553 8. 27.0259 9. 51.423 10. 49.1039 Correct answer: 49 . 1039 N. Explanation: Let : F = 72 N and θ = 47 ◦ . 7 2 N Scale: 10 N 47 ◦ F x = F cos θ = (72 N) cos47 ◦ = 49 . 1039 N . 002 (part 2 of 2) 10.0 points What is the y-component of this force? 1. 71.8772 2. 56.8503 3. 25.2374 4. 38.2917 5. 44.5132 6. 52.6575 7. 35.3861 8. 64.3591 9. 22.7953 10. 49.1868 Correct answer: 52 . 6575 N. Explanation: F y = F sin θ = (72 N) sin47 ◦ = 52 . 6575 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. 100 N 2. 200 N 3. 10 N 4. All are wrong. 5. 0 N correct 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 1 . 6 s. A passenger in the elevator is holding a 8 . 7 kg bundle at the end of a vertical cord. Version 014 – Exam 2 – chelikowsky – (59005) 2 The acceleration of gravity is 9 . 8 m / s 2 . What is the tension in the cord as the ele- vator accelerates? 1. 40.17 2. 100.476 3. 35.02 4. 43.4869 5. 68.7541 6. 66.0999 7. 47.0311 8. 80.071 9. 92.0569 10. 56.2533 Correct answer: 92 . 0569 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 = (8 . 7 kg) bracketleftbigg 9 . 8 m / s 2 + 2 (1 m) (1 . 6 s) 2 bracketrightbigg = 92 . 0569 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 5.- g 3 6. 0 m / s 2 7. g 2 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 07/07/2010 for the course PHY 302 taught by Professor Staff during the Summer '08 term at University of Texas.

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Exam 2-solutions - Version 014 – Exam 2 – chelikowsky...

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