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Unformatted text preview: tokarski (at23678) – Friction Circular motion – Morozova – (11104) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A block accelerates 3 . 7 m / s 2 down a plane inclined at angle 25.0 ◦ . The acceleration of gravity is 9 . 81 m / s 2 . m μ k 3 . 7 m / s 2 25 ◦ Find μ k between the block and the inclined plane. Correct answer: 0 . 0501509. Explanation: Given : a = 3 . 7 m / s 2 , θ = 25 ◦ , and g = 9 . 81 m / s 2 . Consider the free body diagram for the block m g s i n θ N = m g c o s θ μ N a mg Basic Concepts: vector F net = mvectora Parallel to the ramp: F x,net = ma x = F g,x F k F g,x = mg sin θ F k = μ k F n Perpendicular to the ramp: F y,net = F n F g,y = 0 F g,y = mg cos θ Solution: Consider the forces parallel to the ramp: F k = F g,x ma x = mg sin θ ma x Consider the forces perpendicular to the ramp: F n = F g,y = mg cos θ Thus the coefficient of friction is μ k = F k F n = mg sin θ ma mg cos θ = g sin θ a g cos θ = 9 . 81 m / s 2 sin 25 ◦ 3 . 7 m / s 2 9 . 81 m / s 2 cos 25 ◦ = . 0501509 . 002 10.0 points Two blocks are arranged at the ends of a mass less string as shown in the figure. The system starts from rest. When the 1 . 47 kg mass has fallen through 0 . 326 m, its downward speed is 1 . 25 m / s. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 47 kg 3 . 02 kg μ a What is the frictional force between the 3 . 02 kg mass and the table? Correct answer: 3 . 64584 N. tokarski (at23678) – Friction Circular motion – Morozova – (11104) 2 Explanation: Given : m 1 = 1 . 47 kg , m 2 = 3 . 02 kg , v = 0 m / s , and v = 1 . 25 m / s . Basic Concept: Newton’s Second Law F = M a Solution: The acceleration of m 1 is obtained from the equation v 2 v 2 = 2 a ( s s ) a = v 2 v 2 2 h = (1 . 25 m / s) 2 (0 m / s) 2 2 (0 . 326 m) = 2 . 39647 m / s 2 . Consider free body diagrams for the two masses T m 1 g a T N μ N a m 2 g Because m 1 and m 2 are tied together with string, they have same the speed and the same acceleration, so the net force exerted on m 2 is F 2 = m 2 a The net force on m 1 is m 1 a = m 1 g T , so that T = m 1 g m 1 a . Thus F 2 = T f k , f k = T F 2 = m 1 g ( m 1 + m 2 ) a = (1 . 47 kg) (9 . 8 m / s 2 ) (1 . 47 kg + 3 . 02 kg) × (2 . 39647 m / s 2 ) = 3 . 64584 N . 003 (part 1 of 3) 10.0 points A block is at rest on the incline shown in the figure. The coefficients of static and ki netic friction are μ s = 0 . 78 and μ k = 0 . 66, respectively....
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 Spring '08
 Circular Motion, Force, Friction, Sin, Cos, µk, Friction Circular motion

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