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Unformatted text preview: morris (gmm643) – HW 6 – Coker – (56625) 1 This printout should have 20 questions. Multiplechoice 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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 Spring '10
 COKER
 Force, Potential Energy, Work, Correct Answer, Coker

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