HW5-solutions - Agha(sa9896 HW5 bohm(59970 This print-out...

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Agha (sa9896) – HW5 – bohm – (59970) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 5.0 points Two 40 N forces and a 80 N force act on a hanging box as shown. 40 N 40 N 80 N Will the box experience acceleration? 1. Yes; downward. correct 2. Yes; upward. 3. Unable to determine without the angle. 4. No; it is balanced. Explanation: The horizontal components of the two 40 N forces cancel, leaving an upward force that is less than 80 N. Thus, the net force on the box is down, causing it to accelerate downward. 002 10.0 points Two forces are the only forces acting on a 4 . 5 kg object which moves with an accelera- tion of 2 . 6 m / s 2 in the positive y direction. One of the forces acts in the positive x direc- tion and has a magnitude of 11 N. What is the magnitude of the other force f 2 ? Correct answer: 16 . 059 N. Explanation: Basic Concepts: summationdisplay F = m a Solution: f 1 f 2 f vector f 2 is the hypotenuse of a right triangle, so vector f 2 = radicalBig f 2 + f 1 2 003 (part 1 of 2) 10.0 points A 28500 kg sailboat experiences an eastward force 22600 N due to the tide pushing its hull while the wind pushes the sails with a force of 67000 N directed toward the northwest (45 westward of North or 45 northward of West). What is the magnitude of the resultant ac- celeration of the sailboat? Correct answer: 1 . 87592 m / s 2 . Explanation: According to Newton’s Second Law, mvectora = vector F net = vector F wind + vector F tide . To find the magnitude of the net force, we draw the parallelogram for vector addition vector F wind vector F tide vector F net θ α NW W E Note that 135 is the angle between the tide and the wind forces. Use the Law of Cosines: F 2 net = F 2 wind + F 2 tide - 2 F wind F tide cos θ = (67000 N) 2 + (22600 N) 2 - 2 (67000 N)(22600 N) cos 135
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