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Unformatted text preview: Bautista, Aldo – Homework 7 – Due: Oct 18 2005, 4:00 am – Inst: Maxim Tsoi 1 This printout should have 27 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 1) 10 points A particle of mass 7 . 04 kg is attached to two identical springs on a horizontal frictionless tabletop as shown. The springs have spring constant 25 . 6 N / m and equilibrium length L = 0 . 514 m. L L x x k k m Top View If the mass is pulled 0 . 423 m to the right and then released, what is its speed when it reaches the equilibrium point x = 0? Correct answer: 0 . 409041 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 . We choose U = 0 at x = 0. Then at any point U = Z x F x dx = k x 2 + 2 k L ‡ L p L 2 + x 2 · = (25 . 6 N / m)(0 . 423 m) 2 + 2(25 . 6 N / m)(0 . 514 m) × µ . 514 m q (0 . 514 m) 2 + (0 . 423 m) 2 ¶ = 0 . 588946 J . From conservation of energy ( K + U ) i = ( K + U ) f , we obtain 0 + U = mv 2 f 2 + 0 . Therefore v f = r 2 U m = s 2(0 . 588946 J) 7 . 04 kg = . 409041 m / s . 002 (part 1 of 3) 10 points A small box of mass m is at the top of a spherical dome with radius R . Starting from rest after a slight push, the box slides down along the frictionless spherical surface (see figure). Ignore the initial kinetic energy (ob tained from the slight push). θ R m v Apply the principle of conservation of en ergy to select the correct equation that relates the speed v of the box to the polar angle θ and the radius R while the box is sliding down on the surface. 1. v = p g R cot θ 2. v = p g R 3. v = p g R sin θ 4. v = r g R (1 + tan θ ) 2 Bautista, Aldo – Homework 7 – Due: Oct 18 2005, 4:00 am – Inst: Maxim Tsoi 2 5. v = p 2 g R (1 cos θ ) correct 6. v = p g R tan θ 7. v = p 2 g R sin θ 8. v = p 2 g R cos θ 9. v = p 2 g R (1 sin θ ) 10. v = p 2 g R Explanation: Basic Concepts: Potential Energy, Ki netic Energy, Centripetal Force θ R m v θ mg Since the surface is frictionless, the me chanical energy of the box is conserved, so the initial mechanical energy is the same as the final mechanical energy mg R + 1 2 mv 2 i = mg R cos θ + 1 2 mv 2 . It starts from rest (the push is “slight”) so v i = 0, and we find mg R (1 cos θ ) = 1 2 mv 2 v = p 2 g R (1 cos θ ) . 003 (part 2 of 3) 10 points Select the inequality relation which corre sponds to the condition that the small box will stay on the surface....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
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