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Unformatted text preview: johnson (mjj622) HW 6 Coker (56625) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Homework on Chapter 7, which introduces the concepts of work, kinetic energy, gravi- tational potential energy, conservation of en- ergy, and inclusion of nonconservative forces such as friction in using energy concepts. A lot of stuff! 001 10.0 points An object with mass M is attached to the end of a string and is being lowered vertically at a constant acceleration of g 8 . If it has been lowered a distance from rest, how much work has been done by the tension in the string? 1. 7 M g 8 2.- M g 4 3.- 7 M g 8 correct 4.- M g 5. M g 4 6. M g 8 7.- M g 8 8. M g Explanation: Using Newtons Second Law, F net = T + M g- j M a = T- j M g T = j parenleftbigg M g- M g 8 parenrightbigg = j parenleftbigg 7 M g 8 parenrightbigg , so the work done when x =- j is W = F x = parenleftbigg 7 M g 8 parenrightbigg (- ) =- 7 M g 8 . 002 (part 1 of 4) 10.0 points A block of mass 1 . 12 kg is pushed 1 . 45 m along a frictionless horizontal table by a constant 16 N force directed 27 below the horizontal. 1 . 12 kg = 0 1 6 N 2 7 Find the work done by the applied force. Correct answer: 20 . 6713 J. Explanation: Consider the force diagram F mg n f k Given : m = 1 . 12 kg , x = 1 . 45 m , f = 16 N , and = 27 . Only the horizontal component of the ap- plied force is used to move the block. The total work due to the applied force is W x = f x x = f cos x = (16 N) (cos27 ) (1 . 45 m) = 20 . 6713 J . 003 (part 2 of 4) 10.0 points Find the work done by the normal force ex- erted by the table. Correct answer: 0 J. Explanation: johnson (mjj622) HW 6 Coker (56625) 2 Vertically, y = 0 , so W g = 0 J . 004 (part 3 of 4) 10.0 points Find the work done by the force of gravity. Correct answer: 0 J. Explanation: Vertically, y = 0 , so W g = 0 J . 005 (part 4 of 4) 10.0 points Find the work done by the net force on the block. Correct answer: 20 . 6713 J. Explanation: The net force is F net = f cos so the work done by F net is W = f cos x = (16 N) (cos27 ) (1 . 45 m) = 20 . 6713 J . 006 10.0 points An applied force varies with position ac- cording to F = k 1 x n- k 2 , where n = 3, k 1 = 6 . 8 N / m 3 , and k 2 = 63 N. How much work is done by this force on an object that moves from x i = 5 . 26 m to x f = 18 . 7 m? Correct answer: 205 . 733 kJ. Explanation: Basic Concepts: W = integraldisplay vector F dvectors Solution: The work done by a varying force is W = integraldisplay x 2 x 1 vector F dvectors....
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