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Unformatted text preview: Bozanich, Katherine Final 1 Due: May 10 2006, 5:00 pm Inst: W E Drummond 1 This printout should have 28 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. Answer 24 questions. Do NOT answer more than 24. 001 (part 1 of 1) 10 points A block of mass m is pushed a distance D up an inclined plane by a horizontal force F . The plane is inclined at an angle with respect to the horizontal. The block starts from rest and the coefficient of kinetic friction is k . m D k F What is the final speed of the block? 1. v = 2 m ( F sin  k N ) D 2. v = 2 m ( F sin + k N ) D 3. v = 2 m ( F cos + mg sin  k N ) D 4. v = 2 m ( F cos  mg sin ) D 5. v = 2 m ( F cos  mg sin  k N ) D correct 6. v = 2 m ( F cos + mg sin ) D 7. v = 2 m ( F cos  k N ) D 8. v = 2 m ( F cos  mg sin + k N ) D Explanation: The work done by gravity is W grav = mg D cos(90 + ) = mg D sin . The work done by the force F is W F = F D cos . From the workenergy theorem we know that W net = K , W F + W grav + W friction = 1 2 mv 2 f . Thus v f = 2 m ( F cos  mg sin  k N ) D . 002 (part 1 of 1) 10 points Mass m 1 moves in a circular path of radius r on a frictionless horizontal table. It is at tached to a string that passes through a fric tionless hole in the center of the table. A second mass m 2 is attached to the other end of the string. r v m 1 m 2 Determine an expression for r in terms of m 1 , m 2 , and the period T (the time for one revolution). 1. r = m 1 g T 2 2 2 m 2 2. r = m 2 g T 2 4 2 m 1 correct 3. r = m 2 g T 2 2 2 m 1 4. r = m 2 g T 2 2 m 1 5. r = m 1 g T 2 4 2 m 2 Bozanich, Katherine Final 1 Due: May 10 2006, 5:00 pm Inst: W E Drummond 2 Explanation: The linear velocity v can be expressed in terms of the distance it travels each revolu tion: v = 2 r T . Consider the forces acting on each mass: . N . T m 1 .g . T m 2 .g The tension T in the string provides the centripetal force required to keep m 1 moving in a circular path. Applying F x = ma x to m 1 , we obtain T = m 1 a c = m 1 v 2 r and F y = ma y = 0 to m 2 , so m 2 g T = 0 m 2 g = m 1 v 2 r m 2 g = m 1 4 2 r T 2 r = m 2 g T 2 4 2 m 1 . 003 (part 1 of 2) 10 points Note: P atm = 101300 Pa. The viscosity of the fluid is negligible and the fluid is incompress ible. A liquid of density 1349 kg / m 3 flows with speed 2 . 38 m / s into a pipe of diameter 0 . 12 m. The diameter of the pipe decreases to 0 . 05 m at its exit end. The exit end of the pipe is 6 . 52 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1 . 2 atm....
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This note was uploaded on 11/03/2010 for the course PHYSICS 303 taught by Professor Shih during the Spring '10 term at University of Texas at Austin.
 Spring '10
 SHIH

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