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Unformatted text preview: kx W , , 2 2 1 2 2 1 − = − = θ θ x i x f x=0 x=0 Given: m=1.7 kg, k=310 N/m, θ =30°, x i =0.31 m, x f =0.14 m, frictionless Method: no friction, so we can use conservation of energy Initially θ sin , v v i 2 2 1 2 2 1 i i x h kx mgh m E = = + + = 2 2 1 sin i i i kx mgx E + = Finally 2 2 1 2 2 1 sin v v find , sin f f f f f f f kx mgx m E x h + + = = θ ) ( sin ) ( 2 v ) ( sin ) ( v sin sin v 2 2 2 2 2 1 2 2 1 2 2 1 2 2 1 2 2 1 f i f i f f i f i f i i f f f i f x x m k x x g x x k x x mg m kx mgx kx mgx m E E − + − = − + − = + = + + = s m 2 2 95 . 3 95 . 13 666 . 1 v ) 14 . 31 . ( 7 . 1 310 30 sin ) 14 . 31 . )( 8 . 9 ( 2 v = + = − + − = f f Interesting to plot the potential energies x f x f x f x i x i x i Energy U g U s U total E K...
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This note was uploaded on 12/10/2011 for the course PHY 1111 taught by Professor Stencil during the Fall '11 term at UGA.
 Fall '11
 Stencil
 Force, Work

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