HW 17 - Morby Grant Homework 17 Due Mar 3 2006 noon Inst...

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Morby, Grant – Homework 17 – Due: Mar 3 2006, noon – Inst: Drummond 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A bead slides without friction around a loop- the-loop. The bead is released from a height h from the bottom of the loop-the-loop which has a radius r . h r Which of the following diagrams best rep- resents the kinetic energy of the bead versus time? 1. t K 2. t K 3. t K 4. t K 5. t K 6. t K correct Explanation: The bead reaches a maximum kinetic en- ergy each time it reaches the bottom of the loop. Consequently, the only curve which fits this criteria is one which has two equally high maxima. Also, the kinetic energy must be zero at t = 0. 002 (part 2 of 2) 10 points Which of the following could represent the gravitational potential energy of the bead ver- sus time? 1. t U correct 2. t U 3. t U 4. t U 5. t U 6. t U Explanation: Since the kinetic energy plus potential en- ergy is constant, this curve must be the inverse of the curve in Part 1.

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Morby, Grant – Homework 17 – Due: Mar 3 2006, noon – Inst: Drummond 2 003 (part 1 of 1) 10 points A particle of mass 9 . 03 kg is attached to two identical springs on a horizontal frictionless tabletop as shown. The springs have spring constant 38 . 9 N / m and equilibrium length L = 1 . 74 m. L L x x k k m Top View If the mass is pulled 0 . 665 m to the right and then released, what is its speed when it reaches the equilibrium point x = 0? Correct answer: 0 . 360294 m / s. Explanation: First calculate the potential energy of the springs. When the mass moves a distance x , the length of each spring changes from L to p L 2 + x 2 , so each exerts a force F = k p L 2 + x 2 - L · toward its fixed end. The y -components can- cel out and the x -components add to F x = 2 F x L 2 + x 2 = - 2 k x + 2 k L x L 2 + x 2 .
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