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HW 6-solutions

HW 6-solutions - morris(gmm643 – HW 6 – Coker –(56625...

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Unformatted text preview: morris (gmm643) – 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 6 . If it has been lowered a distance ℓ from rest, how much work has been done by the tension in the string? 1. M g ℓ 6 2. M g ℓ 3.- M g ℓ 6 4.- M g ℓ 5.- M g ℓ 3 6. M g ℓ 3 7. 5 M g ℓ 6 8.- 5 M g ℓ 6 correct 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 6 parenrightbigg = ˆ j parenleftbigg 5 M g 6 parenrightbigg , so the work done when Δ x =- ˆ j ℓ is W = F · Δ x = parenleftbigg 5 M g 6 parenrightbigg (- ℓ ) =- 5 M g ℓ 6 . 002 (part 1 of 4) 10.0 points A block of mass 3 . 66 kg is pushed 3 . 05 m along a frictionless horizontal table by a constant 16 N force directed 27 ◦ below the horizontal. 3 . 66 kg μ = 0 1 6 N 2 7 ◦ Find the work done by the applied force. Correct answer: 43 . 4811 J. Explanation: Consider the force diagram F θ mg n f k Given : m = 3 . 66 kg , Δ x = 3 . 05 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 ◦ ) (3 . 05 m) = 43 . 4811 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: morris (gmm643) – 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: 43 . 4811 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) (cos27 ◦ ) (3 . 05 m) = 43 . 4811 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 = 5 N / m 3 , and k 2 = 32 N. How much work is done by this force on an object that moves from x i = 4 . 33 m to x f = 11 . 2 m? Correct answer: 19 . 0098 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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HW 6-solutions - morris(gmm643 – HW 6 – Coker –(56625...

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