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solution-hw8

# solution-hw8 - lai(yl8859 HW08 Ross(15200 This print-out...

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lai (yl8859) – HW08 – Ross – (15200) 1 This print-out should have 5 questions. Multiple-choice 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 μ = 0 . 276 240 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 , α = 35 , μ = 0 . 276 , m = 20 kg , and g = 9 . 8 m / s 2 , m F α N m g F μ N The block is in equilibrium vertically, so summationdisplay F up = summationdisplay F down F sin α + N = m g N = m g - F sin α The block is accelerating horizontally, so F net = m a = F + F cos α - μ N F net = m a = F (1 + cos α + μ sin α ) - μ m g a = F m parenleftBig 1 + cos α + μ sin α parenrightBig - μ g = 240 N 20 kg × bracketleftBig 1 + cos 35 + (0 . 276) sin 35 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. There is a frictional force between the block on the incline and the incline.

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solution-hw8 - lai(yl8859 HW08 Ross(15200 This print-out...

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