This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Muraj, Hamza – Homework 18 – Due: Mar 6 2006, 4:00 am – Inst: Florin 1 This printout should have 11 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 5 . 6 g mass is released from rest at C which has a height of 1 m above the base of a loop theloop and a radius of 0 . 27 m . The acceleration of gravity is 9 . 8 m / s 2 . 1 m . 27 m B D C A 5 . 6 g Find the normal force pressing on the track at A , where A is at the same level as the center of the loop. Correct answer: 0 . 296759 N. Explanation: The normal force at A (where the mass presses the track) is the same as the force which the track presses the mass. The later is the centripetal force, which is given by F c = mv 2 R . The quantity mv 2 at A may be obtained through the conservation of energy relation K = 1 2 mv 2 = mg ( h R ) . So the normal force is N = F c = 2 mg ( h R ) R = 2(5 . 6 g)(9 . 8 m / s 2 )(1 m . 27 m) . 27 m = 0 . 296759 N . 002 (part 2 of 2) 10 points Consider a different situation when the initial height at C has not yet been specified. What is the minimum kinetic energy of the block at B , which is located at the top of the loop, so that the block can pass by this point without falling off from the track? Correct answer: 0 . 0074088 J. Explanation: At B , the equation of motion is given by mg + N = ma c . The stayon track condition is when N ≥ 0. The limiting case is when N = 0. so mg = ma c = mv 2 R , or K min = 1 2 mv 2 = 1 2 mg R = 1 2 (5 . 6 g)(9 . 8 m / s 2 )(0 . 27 m) = 0 . 0074088 J . 003 (part 1 of 1) 10 points An ore car of mass 41000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 14 m lower vertically, is a horizontally situated spring with constant 420000 N / m. The acceleration of gravity is 9 . 8 m / s 2 . Ignore friction. How much is the spring compressed in stop ping the ore car? Correct answer: 5 . 17558 m. Explanation: Energy is conserved, so the change of po tential energy from when the car is at rest to when it just hits the spring is mg h = 1 2 mv 2 . The kinetic energy is then converted to po tential energy in the spring as the cart comes to rest. When the spring is fully compressed by an amount d , all of the kinetic energy has been converted to potential energy so 1 2 mv 2 = 1 2 k d 2 . Muraj, Hamza – Homework 18 – Due: Mar 6 2006, 4:00 am – Inst: Florin 2 Thus, 1 2 k d 2 = mg h, and solving for d we have d = r 2 mg h k = s 2(41000 kg)(9 . 8 m / s 2 )(14 m) (420000 N /...
View
Full
Document
This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas.
 Spring '09
 Turner

Click to edit the document details