This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Pickering, Tracey Homework 17 Due: Mar 3 2006, noon Inst: Drummond 1 This printout should have 8 questions. Multiplechoice 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 theloop. The bead is released from a height h from the bottom of the looptheloop 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 correct 4. t K 5. t K 6. t K 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 2. t U 3. t U 4. t U 5. t U 6. t U correct Explanation: Since the kinetic energy plus potential en ergyisconstant, thiscurvemustbetheinverse of the curve in Part 1. Pickering, Tracey Homework 17 Due: Mar 3 2006, noon Inst: Drummond 2 003 (part 1 of 1) 10 points A particle of mass 9 . 49 kg is attached to two identical springs on a horizontal frictionless tabletop as shown. The springs have spring constant 58 . 5 N / m and equilibrium length L = 0 . 697 m. L L x x k k m Top View If the mass is pulled 0 . 578 m to the right and then released, what is its speed when it reaches the equilibrium point x = 0? Correct answer: 0 . 73202 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 ycomponents can cel out and the xcomponents add to F x = 2 F x L 2 + x 2 = 2 k x + 2 k Lx L 2 + x 2 ....
View
Full
Document
 Fall '08
 Turner
 Work

Click to edit the document details