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Unformatted text preview: Pickering, Tracey Homework 11 Due: Feb 15 2006, noon Inst: Drummond 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The force required to stretch a spring varies directly with the amount the spring is stretched. A force of 17 pounds is needed to stretch a spring 65 inches, as shown in the righthand figure below. 54 00 65 00 W 17 lbs How much force is required to stretch the spring 54 inches? Correct answer: 14 . 1231 lbs. Explanation: Given : F 2 = 17 lbs , x 2 = 65 inches , and x 1 = 54 inches . F x , so F = k x = k = F x . Thus F 1 x 1 = F 2 x 2 = F 2 x 1 x 2 = (17 lbs)(54 inches) 65 inches = 14 . 1231 lbs . 002 (part 1 of 3) 10 points The suspended 2 . 4 kg mass on the right is moving up, the 2 . 4 kg mass slides down the ramp, and the suspended 7 . 3 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 19 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 2 . 4 k g = . 1 9 32 7 . 3 kg 2 . 4 kg What is the acceleration of the three block system? Correct answer: 4 . 68545 m / s 2 . Explanation: Let : m 1 = 2 . 4 kg , m 2 = 2 . 4 kg , m 3 = 7 . 3 kg , and = 32 . Basic Concept: F net = ma 6 = 0 Solution: The acceleration a of each mass is the same, but the tensions in the two strings will be different. Let T 1 be the tension in the right string and T 3 the tension in the left string. Consider the free body diagrams for each mass Pickering, Tracey Homework 11 Due: Feb 15 2006, noon Inst: Drummond 2 T 3 m 3 g a T 1 m 1 g a T 3 T 1 N N a m 2 g For the mass m 1 , T 1 acts up and the weight m 1 g acts down, with the acceleration a di rected upward F net 1 = m 1 a = T 1 m 1 g (1) For the mass on the table, the parallel compo nent of its weight is mg sin and the perpen dicular component of its weight is mg cos . ( N = mg cos from equilibrium). The accel eration a is directed down the table, T 3 and the parallel weight component m 2 g sin act down the table, and T 1 and the frictional force N = m 2 g cos act up the table F net 2 = m 2 a (2) = T 3 + m 2 g sin  T 1 m 2 g cos . For the mass m 3 , T 3 acts up and the weight m 3 g acts down, with the acceleration a di rected downward F net 3 = m 3 a = m 3 g T 3 . (3) Adding Eqs. (1), (2), & (3) yields ( m 1 + m 2 + m 3 ) a = m 3 g + m 2 g sin  m 2 g cos  m 1 g . Solving for a , we have a = [ m 2 sin  m 2 cos + ( m 3 m 1 )] g m 1 + m 2 + m 3 = (2 . 4 kg)(9 . 8 m / s 2 ) sin32 2 . 4 kg + 2 . 4 kg + 7 . 3 kg (0 . 19)(2 . 4 kg)(9 . 8 m / s 2 ) cos32 2 . 4 kg + 2 . 4 kg + 7 . 3 kg + (7 . 3 kg 2 . 4 kg)(9 . 8 m / s 2 ) 2 . 4 kg + 2 . 4 kg + 7 . 3 kg = 4 . 68545 m / s 2 ....
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This note was uploaded on 06/23/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
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