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Unformatted text preview: lai (yl8859) HW08 Ross (15200) 1 This printout should have 5 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points The magnitude of each force is 240 N, the force on the right is applied at an angle 35 and the mass of the block is 20 kg. The coefficient of friction is 0 . 276. The acceleration of gravity is 9 . 8 m / s 2 . 20 kg = 0 . 276 2 4 N 35 240 N What is the magnitude of the resulting ac celeration? Correct answer: 21 . 0247 m / s 2 . Explanation: Given : F = 240 N , = 35 , = 0 . 276 , m = 20 kg , and g = 9 . 8 m / s 2 , m F N m g F N The block is in equilibrium vertically, so summationdisplay F up = summationdisplay F down F sin + N = m g N = m g F sin The block is accelerating horizontally, so F net = m a = F + F cos  N F net = m a = F (1 + cos + sin ) m g a = F m parenleftBig 1 + cos + sin parenrightBig g = 240 N 20 kg bracketleftBig 1 + cos 35 + (0 . 276) sin35 bracketrightBig (0 . 276) (9 . 8 m / s 2 ) = 21 . 0247 m / s 2 . 002 (part 1 of 2) 10.0 points Two blocks are attached by a thin inextensi ble string over a frictionless, massless pulley....
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This note was uploaded on 02/24/2011 for the course PHYS 152 taught by Professor Button during the Winter '08 term at IUPUI.
 Winter '08
 BUTTON
 mechanics, Force

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