Circular motion & friction

Circular motion & friction - adamkiewicz(ja28724...

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Unformatted text preview: adamkiewicz (ja28724) – Circular motion friction – Morozova – (11102) 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 m / s 2 down a plane in- clined at angle 26.0 ◦ . The acceleration of gravity is 9 . 81 m / s 2 . m μ k 3 m / s 2 26 ◦ Find μ k between the block and the inclined plane. Correct answer: 0 . 147487. Explanation: Given : a = 3 m / s 2 , θ = 26 ◦ , 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 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 μ 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: 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 . 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 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 = (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 μ s = 0 . 7 and μ k = 0 . 59, respec- tively....
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Circular motion & friction - adamkiewicz(ja28724...

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