This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: sela (ds39632) HW05 janow (11196) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A block is released from rest on an inclined plane and moves 3 . 5 m during the next 1 . 7 s. The acceleration of gravity is 9 . 8 m / s 2 . 8 k g k 24 What is the coefficient of kinetic friction k for the incline? Correct answer: 0 . 174681. Explanation: Given : m = 8 kg , = 3 . 5 m , = 24 , and t = 1 . 7 s . Consider the free body diagram for the block m g s i n N = m g c o s N a m g 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) Applying Newtons Second Law of Motion summationdisplay F i = m a and Eq. 1, the sine component of the weight acts down the plane and friction acts up the plane. The block slides down the plane, so m a = m g sin  k m g cos 2 t 2 = g parenleftBig sin  k cos parenrightBig 2 = g t 2 parenleftBig sin  k cos parenrightBig k = g t 2 sin  2 g t 2 cos (2) = tan  2 g t 2 cos = tan24  2 (3 . 5 m) (9 . 8 m / s 2 ) (1 . 7 s) 2 cos 24 = . 174681 . 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 . 96 kg mass has fallen through 0 . 415 m, its downward speed is 1 . 25 m / s. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 96 kg 3 . 53 kg a What is the frictional force between the 3 . 53 kg mass and the table? Correct answer: 8 . 87291 N. Explanation: Given : m 1 = 1 . 96 kg , m 2 = 3 . 53 kg , v = 0 m / s , and v = 1 . 25 m / s . Basic Concept: Newtons Second Law F = M a sela (ds39632) HW05 janow (11196) 2 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 . 415 m) = 1 . 88253 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 . 96 kg) (9 . 8 m / s 2 ) (1 . 96 kg + 3 . 53 kg) (1 . 88253 m / s 2 ) = 8 . 87291 N ....
View Full
Document
 Spring '08
 moro

Click to edit the document details