{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

problem07_23

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

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

7.23: a) In this case, J 000 , 625 1 = K as before, J 000 , 17 other - = W and J. 900 , 50 ) 00 . 1 ( ) s m (9.80 kg) 2000 ( m) 00 . 1 ( m) N 10 41 . 1 )( 2 1 ( ) 2 1 ( 2 2 5 2 2 2 2 = - + - × = + = mgy ky U The kinetic energy is then J 557,100 J 000 , 17 J 900 , 50 J 000 , 625 2 = - - = K , corresponding to a speed s. m 6 . 23 2 = v b) The elevator is moving down, so the friction force is up (tending to stop the elevator, which is the idea). The net upward force is then
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: , kx f mg N 400 , 138 ) m 00 . 1 ( ) m N 10 41 . 1 ( N 000 , 17 ) s m 80 . 9 )( kg 2000 ( 5 2 =-×-+-=-+-for an upward acceleration of . s m 2 . 69 2...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online