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# HW 6 - johnson(mjj622 HW 6 Coker(56625 This print-out...

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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 ℓ 8 2. - M g ℓ 4 3. - 7 M g ℓ 8 correct 4. - M g ℓ 5. M g ℓ 4 6. M g ℓ 8 7. - M g ℓ 8 8. M g ℓ Explanation: Using Newton’s Second Law, F net = T + M g - ˆ j M a = T - ˆ j M g T = ˆ j parenleftbigg M g - M g 8 parenrightbigg = ˆ j parenleftbigg 7 M g 8 parenrightbigg , so the work done when Δ x = - ˆ j is W = F · Δ x = parenleftbigg 7 M g 8 parenrightbigg ( - ) = - 7 M g ℓ 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 μ = 0 16 N 27 Find the work done by the applied force. Correct answer: 20 . 6713 J. Explanation: Consider the force diagram F θ m g n f k Given : m = 1 . 12 kg , Δ x = 1 . 45 m , f = 16 N , and θ = 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 Δ x = f cos θ Δ 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:

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johnson (mjj622) – HW 6 – Coker – (56625) 2 Vertically, Δ 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, Δ 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 θ so the work done by F net is W = f cos θ Δ x = (16 N) (cos 27 ) (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 . which adds up all the little vector F · vectors parts along the path, taking into account the changing force. Here all the motion is in the ˆ ı direction so ds = dx .
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HW 6 - johnson(mjj622 HW 6 Coker(56625 This print-out...

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