317K Final

# 317K Final - Bozanich Katherine Final 1 Due 5:00 pm Inst W...

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Bozanich, Katherine – Final 1 – Due: May 10 2006, 5:00 pm – Inst: W E Drummond 1 This print-out should have 28 questions. Multiple-choice 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 θ with respect to the horizontal. The block starts from rest and the coefficient of kinetic friction is μ k . m D μ k F θ What is the final speed of the block? 1. v = r 2 m ( F sin θ - μ k N ) D 2. v = r 2 m ( F sin θ + μ k N ) D 3. v = r 2 m ( F cos θ + m g sin θ - μ k N ) D 4. v = r 2 m ( F cos θ - m g sin θ ) D 5. v = r 2 m ( F cos θ - m g sin θ - μ k N ) D correct 6. v = r 2 m ( F cos θ + m g sin θ ) D 7. v = r 2 m ( F cos θ - μ k N ) D 8. v = r 2 m ( F cos θ - m g sin θ + μ k N ) D Explanation: The work done by gravity is W grav = m g D cos(90 + θ ) = - m g D sin θ . The work done by the force F is W F = F D cos θ . From the work-energy theorem we know that W net = Δ K , W F + W grav + W friction = 1 2 m v 2 f . Thus v f = r 2 m ( F cos θ - m g sin θ - μ 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 π 2 m 2 2. r = m 2 g T 2 4 π 2 m 1 correct 3. r = m 2 g T 2 2 π 2 m 1 4. r = m 2 g T 2 π 2 m 1 5. r = m 1 g T 2 4 π 2 m 2

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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 π 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 X F x = m a x to m 1 , we obtain T = m 1 a c = m 1 v 2 r and X F y = m a 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 π 2 r T 2 r = m 2 g T 2 4 π 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. The acceleration of gravity is 9 . 8 m / s 2 . P 1 2 . 38 m / s 0 . 12 m P 2 v 2 0 . 05 m 1 . 2 atm 6 . 52 m What is the velocity v 2 of the liquid flowing out of the exit end of the pipe? Correct answer: 13 . 7088 m / s.
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