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Unformatted text preview: practice 04 – GADHIA, TEJAS – Due: Feb 6 2007, 10:00 pm 1 Question 1 part 1 of 1 10 points A heavily loaded freight train moves with constant velocity. What is the relationship between the net force on the first car ( F 1 ) and the net force on the last car ( F 2 )? 1. F 1 = F 2 correct 2. F 1 > F 2 3. Unable to determine. 4. F 1 < F 2 Explanation: The net force on each car is zero because the train moves with constant velocity (no acceleration). Question 2 part 1 of 1 10 points Within a book on a table there are bil lions of forces pushing and pulling on all the molecules. Why is it that these forces never by chance add up a net force in one direction, causing the book to accelerate “spontaneously” across the table? 1. In fact, the book is moving across the table spontaneously all the time, which is caused by these forces but the movement is too weak to observe. 2. The billions of force pairs are internal to the book, and exert no net force on the book. correct 3. These forces between molecules are much smaller than the friction between the book and the table. 4. These forces are counteracted by gravi taty. Explanation: The billions of force pairs are internal to the book, and exert no net force on the book. An external net farce is necessary to accelerate the book. Question 3 part 1 of 1 10 points An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1 . 7 s. A passenger in the elevator is holding a 3 . 3 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: 34 . 6237 N (tolerance ± 1 %). 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 parenleftbigg g + 2 h t 2 parenrightbigg = (3 . 3 kg) bracketleftbigg 9 . 8 m / s 2 + 2 (1 m) (1 . 7 s) 2 bracketrightbigg = 34 . 6237 N . Question 4 part 1 of 1 10 points practice 04 – GADHIA, TEJAS – Due: Feb 6 2007, 10:00 pm 2 A 1258 . 4 kg car is traveling at 29 . 1 m / s when the driver takes his foot off the gas pedal. It takes 5 . 5 s for the car to slow down to 20 m / s. How large is the net force slowing the car? Correct answer: 2082 . 08 N (tolerance ± 1 %). Explanation: Basic Concept: Newton’s Second Law: F = M a Solution: From the initial velocity, the final velocity and the time of slowing down, the acceleration of the car is given by a = v − v t = 20 m / s − 29 . 1 m / s 5 . 5 s = − 1 . 65455 m / s 2 Applying Newton’s second law yields the force slowing the car bardbl vector F bardbl =  M a  = vextendsingle vextendsingle (1258 . 4 kg)( − 1 . 65455 m / s 2 ) vextendsingle vextendsingle = 2082 . 08 N ....
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This note was uploaded on 03/09/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Force

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