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Analytical Mech Homework Solutions 16

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

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( ) ( ) 2 2 1 1 1 1 2 2 2 F F F 1 x t t t t t t m m m = + + D D D At t 1 2 : t = 2 2 2 1 1 1 5 2 2 F F F F 2 1 x t t t m m m m = + + = D D D D t 2.11 3 2 dv dv dx dv c a v dt dx dt dx m = = = = − v 1 2 c v dx m = − v d max 1 2 0 v x v c v dv dx m = D 1 2 max 2 c v x m = − D 1 2 max 2 mv x c = D 2.12 Going up: sin30 cos30 x F mg mg µ = − D D ( ) 2 sin30 0.1cos30 5.749 m x g s = − + = − D D ±± v v at = + D at the highest point so 0 v = 0.174 up v a = − = D D t v s 2 2 2 1 0.174 .087 0.087 2 up up up 2 x v t at v v v m = + = = D D D D Going down: 2 0.087 x v = D D , v 0 = D , ( ) 9.8 0.5 0.0866 ′ = − a 2 2 1 0 0.087 4.0513 2 down down x v t D = = 0.207 down t v = D s s 0.381 total up down t t t v = + = D
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