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Unformatted text preview: Muraj, Hamza Homework 16 Due: Mar 1 2006, 4:00 am Inst: Florin 1 This printout should have 12 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 4) 10 points A 2 . 97 kg block is pushed 1 . 55 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of 63 . 9 with the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 97 kg F 6 3 . 9 If the coefficient of kinetic friction between the block and wall is 0 . 449, find the work done by F . Correct answer: 57 . 8361 J. Explanation: Given : m = 2 . 97 kg , = 0 . 449 , = 63 . 9 , and y = 1 . 55 m . F 6 3 . 9 v mg f k n The block is in equilibrium horizontally, so X F x = F cos  n = 0 , so that n = F cos Since the block moves with constant velocity, X F y = F sin  mg f k = 0 F sin  mg F cos = 0 F (sin  cos ) = mg F = mg sin  cos = (2 . 97 kg)(9 . 8 m / s 2 ) sin63 . 9  . 449 cos63 . 9 = 41 . 5506 N Thus W F = ( F sin )( y ) = (41 . 5506 N)(sin63 . 9 )(1 . 55 m) = 57 . 8361 J . since 1J = 1kg m 2 / s 2 . 002 (part 2 of 4) 10 points Find the work done by the force of gravity. Correct answer: 45 . 1143 J. Explanation: W g = mg (cos180 ) y = (2 . 97 kg)(9 . 8 m / s 2 )(cos180 )(1 . 55 m) = 45 . 1143 J ....
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