Analytical Mech Homework Solutions 16

Analytical Mech Homework Solutions 16 - F2F 1 2F 2 t1 + t1...

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

View Full Document Right Arrow Icon
() 2 2 11 1 12 22 FF F 1 x t ttt tt mm m =+− + DD D At t 1 2: t = 222 111 5 FFF F 2 1 x mmm m =++= DDD D t 2.11 3 2 dv dv dx dv c av dt dx dt dx m ==⋅= ⋅= v 1 2 c v d x m =− vd max 1 2 0 vx v c v d x m D ∫∫ 1 2 max 2 c m −= D 1 2 max 2 mv x c = D 2.12 Going up: sin30 cos30 x Fm g m g µ 2 0.1cos30 5.749 m xg s + ±± vv a t =+ D at the highest point so 0 v = 0.174 up v a = D D tv s 2 1 0.174 .087 0.087 2 up up up 2 x vt at v v v m = = D D Going down: 2 0.087 x v = , v 0 = D , ( ) 9.8 0.5 0.0866 ′=− a 2 2 1 0 0.087 4.0513 2 down down x D == 0.207 down = D s s 0.381 total up down t v = D 2.13 At the top so 0 v = max 2 2 kx g k
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

Ask a homework question - tutors are online