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Unformatted text preview: Version One Homework 3 savrasov 39822 Oct 15, 2007 1 This printout should have 8 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. Accelerating Elevator 05:04, trigonometry, numeric, &gt; 1 min, nor mal. 001 (part 1 of 1) 2 points An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1 . 4 s. A passenger in the elevator is hold ing a 5 kg bundle at the end of a vertical cord. The acceleration of gravity is 9 . 8 m / s 2 . What is the tension in the cord as the ele vator accelerates? Correct answer: 54 . 102 N. Explanation: T mg a elevator g Let h be the distance traveled and a the acceleration of the elevator. With the initial velocity being zero, we simplify the following expression and solve for acceleration of the elevator: h = v t + 1 2 at 2 = 1 2 at 2 = a = 2 h t 2 . The equation describing the forces acting on the bundle is F net = ma = T mg T = m ( g + a ) = m g + 2 h t 2 = (5 kg) 9 . 8 m / s 2 + 2(1 m) (1 . 4 s) 2 = 54 . 102 N . keywords: Race Track Acceleration 05:04, trigonometry, numeric, &gt; 1 min, nor mal. 002 (part 1 of 1) 2 points After a day of testing race cars, you decide to take your own 1550 kg car onto the test track. While moving down the track at 10 m / s, you suddenly accelerate to 30 m / s in 10 s. What average net force have you applied to the car during that time? Correct answer: 3100 N. Explanation: We can apply the definition of acceleration to Newtons Second Law of Motion: F = ma = m v t = m v f v o t = (1550 kg) 30 m / s 10 m / s 10 s = 3100 N keywords: Serway CP 04 29 05:05, trigonometry, numeric,...
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This note was uploaded on 01/10/2010 for the course PHY na taught by Professor Richardscalettar during the Fall '09 term at UC Davis.
 Fall '09
 RichardScalettar
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