problem04_49

University Physics with Modern Physics with Mastering Physics (11th Edition)

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
4.49: a) b) The net force on the system is N 0 . 53 ) s / m 80 . 9 )( kg 00 . 15 ( N 200 2 = - (keeping three figures), and so the acceleration is , s / m 53 . 3 ) kg 0 . 15 /( ) N 0 . 53 ( 2 = up. c) The net force on the 6-kg block is N 2 . 21 ) s / m 53 . 3 )( kg 00 . 6 ( 2 = , so the tension is found from N 2 . 21 = - - mg T F , or N 120 N 2 . 21 ) s / m 80 . 9 )( kg 00 . 6 ( ) N 200 ( 2 = - - = T . Equivalently, the tension at the top of the rope causes the upward acceleration of the rope and the bottom block, so
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a g T ) kg 00 . 9 ( ) kg 00 . 9 ( =-, which also gives N 120 = T . d) The same analysis of part (c) is applicable, but using kg 00 . 2 kg 00 . 6 + instead of the mass of the top block, or 7.00 kg instead of the mass of the bottom block. Either way gives N 3 . 93 = T ....
View Full Document

Ask a homework question - tutors are online