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Unformatted text preview: adamkiewicz (ja28724) Circular motion friction Morozova (11102) 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 m / s 2 down a plane in clined at angle 26.0 . The acceleration of gravity is 9 . 81 m / s 2 . m k 3 m / s 2 26 Find k between the block and the inclined plane. Correct answer: 0 . 147487. Explanation: Given : a = 3 m / s 2 , = 26 , 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 26  3 m / s 2 9 . 81 m / s 2 cos 26 = . 147487 . 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 3 . 26 kg mass has fallen through 0 . 396 m, its downward speed is 1 . 29 m / s. The acceleration of gravity is 9 . 8 m / s 2 . 3 . 26 kg 4 . 78 kg a What is the frictional force between the 4 . 78 kg mass and the table? Correct answer: 15 . 0549 N. adamkiewicz (ja28724) Circular motion friction Morozova (11102) 2 Explanation: Given : m 1 = 3 . 26 kg , m 2 = 4 . 78 kg , v = 0 m / s , and v = 1 . 29 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 . 29 m / s) 2 (0 m / s) 2 2 (0 . 396 m) = 2 . 10114 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 = (3 . 26 kg) (9 . 8 m / s 2 ) (3 . 26 kg + 4 . 78 kg) (2 . 10114 m / s 2 ) = 15 . 0549 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 kinetic friction are s = 0 . 7 and k = 0 . 59, respec tively....
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This note was uploaded on 04/14/2009 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.
 Spring '08
 moro
 Physics, Circular Motion, Friction

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