# Exam 2 - quiz 02 BAUTISTA ALDO Due Mar 8 2006 10:00 pm...

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quiz 02 – BAUTISTA, ALDO – Due: Mar 8 2006, 10:00 pm 1 Version number encoded for clicker entry: V1:1, V2:1, V3:1, V4:5, V5:5. Question 1 Part 1 of 1. 10 points. A(n) 81.2 kg fisherman jumps from a dock into a 126.8 kg rowboat at rest on the west side of the dock. If the velocity of the fisherman is 5.41 m/s to the west as he leaves the dock, what is the final velocity of the fisherman and the boat? 1. - 2 . 18063 m / s 2. - 2 . 11198 m / s correct 3. - 2 . 0481 m / s 4. - 1 . 98539 m / s 5. - 1 . 92412 m / s 6. - 1 . 86544 m / s Explanation: Let west be negative: Let : m 1 = 81 . 2 kg , m 2 = 126 . 8 kg , and v i, 1 = - 5 . 41 m / s . The boat and fisherman have the same final speed, and v i, 2 = 0 m/s, so m 1 ~v i, 1 + m 2 ~v i, 2 = ( m 1 + m 2 ) ~v f m 1 ~v i, 1 = ( m 1 + m 2 ) ~v f v f = m 1 v i m 1 + m 2 = (81 . 2 kg) ( - 5 . 41 m / s) 81 . 2 kg + 126 . 8 kg = - 2 . 11198 m / s , which is 2 . 11198 m / s to the west. Question 2 Part 1 of 1. 10 points. A 2 . 18 kg block is pushed 1 . 58 m up a ver- tical wall with constant speed by a constant force of magnitude F applied at an angle of 56 . 5 with the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 18 kg F 56 . 5 If the coefficient of kinetic friction between the block and wall is 0 . 511, find the work done by F . 1. 46 . 4583 J 2. 47 . 9087 J 3. 49 . 4048 J 4. 51 . 0068 J correct 5. 52 . 6712 J 6. 54 . 3148 J Explanation: Given : m = 2 . 18 kg , μ = 0 . 511 , θ = 56 . 5 , and Δ y = 1 . 58 m . F 56 . 5 v m g f k n The block is in equilibrium horizontally, so X F x = F cos θ - n = 0 , so that n = F cos θ Since the block moves with constant velocity, X F y = F sin θ - m g - f k = 0 F sin θ - m g - μ F cos θ = 0 F (sin θ - μ cos θ ) = m g , so

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quiz 02 – BAUTISTA, ALDO – Due: Mar 8 2006, 10:00 pm 2 F = m g sin θ - μ cos θ = (2 . 18 kg) (9 . 8 m / s 2 ) sin 56 . 5 - 0 . 511 cos 56 . 5 = 38 . 7137 N . Thus W F = ( F sin θ ) (Δ y ) = (38 . 7137 N) (sin 56 . 5 )(1 . 58 m) = 51 . 0068 J , since 1 J = 1 kg · m 2 / s 2 . Question 3 Part 1 of 1. 10 points. An object attached to the end of a string swings in a vertical circle of radius 2 . 4 m as shown. At an instant when θ = 33 , the speed of the object is 14 m / s and the tension in the string is 46 N. The acceleration of gravity is 9 . 8 m / s 2 . 14 m / s 46 N 2 . 4 m 33 9 . 8 m / s 2 What is the mass m of the object? 1. 0 . 602653 kg correct 2. 0 . 623418 kg 3. 0 . 644705 kg 4. 0 . 664931 kg 5. 0 . 68706 kg 6. 0 . 709637 kg Explanation: Let : θ = 33 , v = 14 m / s , r = 2 . 4 m , and g = 9 . 8 m / s 2 . θ v T mg R θ g Basic Concepts: F r = m a r = m v 2 r Solution: The centripetal acceleration is a r = v 2 r . According to Newton’s second law, the total force in the radial direction should be equal to F r = m a r . The forces acting on the mass are the ten- sion T on the string, directed radially inward, and the weight force, whose component in the radial direction is m g sin θ . The radial com- ponent of the resultant force is then F r = m a r = T + m g sin θ , or m v 2 R - g sin θ = T . Finally, m = T v 2 R - g sin θ = (46 N) (14 m / s) 2 (2 . 4 m) - ( 9 . 8 m / s 2 ) sin(33 ) = 0 . 602653 kg . Question 4 Part 1 of 1. 10 points. Vector ~ A has components A x = - 2 . 82 , A y = 6 . 99 , A z = 5 . 32 while vector ~ B has components B x = 5 . 84 , B y = - 6 , B z = 3 . 01 .
quiz 02 – BAUTISTA, ALDO – Due: Mar 8 2006, 10:00 pm 3 What is the angle θ AB between these vectors?

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