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**Unformatted text preview: **tokarski (at23678) Friction Circular motion Morozova (11104) 1 This print-out should have 13 questions. Multiple-choice 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 k 3 . 7 m / s 2 25 Find k between the block and the inclined plane. Correct answer: 0 . 0501509. Explanation: Given : a = 3 . 7 m / s 2 , = 25 , and g = 9 . 81 m / s 2 . Consider the free body diagram for the block m g s i n N = m g c o s 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 F k = k F n Perpendicular to the ramp: F y,net = F n- F g,y = 0 F g,y = mg cos Solution: Consider the forces parallel to the ramp: F k = F g,x- ma x = mg sin - ma x Consider the forces perpendicular to the ramp: F n = F g,y = mg cos Thus the coefficient of friction is k = F k F n = mg sin - ma mg cos = g sin - a g cos = 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 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: Newtons 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 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 s = 0 . 78 and k = 0 . 66, respectively....

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