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# TOzad55Dh8oy_1210016089_jwk572 - homework 04 KIM JI Due...

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homework 04 – KIM, JI – Due: Sep 26 2007, 4:00 am 1 Question 1, chap 5, sect 5. part 1 of 1 10 points The system is in equilibrium, and the pul- leys are weightless and frictionless. The acceleration of gravity is 9 . 8 m / s 2 . 17 N 13 N T 3 T 2 T T 3 T 2 Find the tension T . Correct answer: 47 N (tolerance ± 1 %). Explanation: Let : W 1 = 17 N and W 2 = 13 N . W 1 W 2 1 2 T 3 T 2 T 1 T 3 T 2 At pulley 2, 2 T 2 = W 2 T 2 = W 2 2 . At the weight W 1 , T 3 = T 2 + W 1 = W 2 2 + W 1 . At pulley 1, T 1 = 2 T 3 = W 2 + 2 W 1 = (13 N) + 2 (17 N) = 47 N . Question 2, chap 6, sect 1. part 1 of 1 10 points Two blocks are arranged at the ends of a massless string as shown in the figure. The system starts from rest. When the 4 . 18 kg mass has fallen through 0 . 372 m, its down- ward speed is 1 . 33 m / s. The acceleration of gravity is 9 . 8 m / s 2 . 4 . 18 kg 5 . 66 kg μ a What is the frictional force between the 5 . 66 kg mass and the table? Correct answer: 17 . 5689 N (tolerance ± 1 %). Explanation: Given : m 1 = 4 . 18 kg , m 2 = 5 . 66 kg , v 0 = 0 m / s , and v = 1 . 33 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 0 = 2 a ( s - s 0 )

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homework 04 – KIM, JI – Due: Sep 26 2007, 4:00 am 2 a = v 2 - v 2 0 2 h = (1 . 33 m / s) 2 - (0 m / s) 2 2 (0 . 372 m) = 2 . 37755 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 = (4 . 18 kg) (9 . 8 m / s 2 ) - (4 . 18 kg + 5 . 66 kg) × (2 . 37755 m / s 2 ) = 17 . 5689 N . Question 3, chap 6, sect 1. part 1 of 2 10 points A block is released from rest on an inclined plane and moves 2 . 7 m during the next 3 . 7 s. The acceleration of gravity is 9 . 8 m / s 2 . 14 kg μ k 33 What is the magnitude of the acceleration of the block? Correct answer: 0 . 394449 m / s 2 (tolerance ± 1 %). Explanation: Given : m = 14 kg , = 2 . 7 m , θ = 33 , and t = 3 . 7 s . Consider the free body diagram for the block m g sin θ N = m g cos θ μ N a m g The acceleration can be obtained through kinematics. Since v 0 = 0, = v 0 t + 1 2 a t 2 = 1 2 a t 2 a = 2 t 2 (1) = 2 (2 . 7 m) (3 . 7 s) 2 = 0 . 394449 m / s 2 . Question 4, chap 6, sect 1. part 2 of 2 10 points What is the coefficient of kinetic friction μ k for the incline?
homework 04 – KIM, JI – Due: Sep 26 2007, 4:00 am 3 Correct answer: 0 . 601415 (tolerance ± 1 %).

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TOzad55Dh8oy_1210016089_jwk572 - homework 04 KIM JI Due...

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