TOzad55Dh8oy_1210016089_jwk572

TOzad55Dh8oy_1210016089_jwk572 - homework 04 KIM, JI Due:...

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Unformatted text preview: 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 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 m / s , and v = 1 . 33 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 ) homework 04 KIM, JI Due: Sep 26 2007, 4:00 am 2 a = v 2- v 2 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 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 . 1 4 k g 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 , = 33 , and t = 3 . 7 s . Consider the free body diagram for the block m g s i n N = m g c o s N a mg The acceleration can be obtained through kinematics. Since v = 0, = v t + 1 2 a t 2 = 1 2 a t 2 a = 2 t 2 (1) = 2 (2 . 7 m) (3 . 7 s) 2 = . 394449 m / s 2 ....
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This note was uploaded on 05/05/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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TOzad55Dh8oy_1210016089_jwk572 - homework 04 KIM, JI Due:...

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