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L18_phys131_29oct07

# L18_phys131_29oct07 - 29 29 29 θ θ μ-θ =-θ = θ θ μ...

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Summary for Lecture 18 on October 29, 2007 Mechanical energy only f 2 f 2 f i 2 i 2 i gf sf f gi si i 2 g 2 s mgy kx 2 1 mv 2 1 mgy kx 2 1 mv 2 1 U U K U U K mv 2 1 K mgy U kx 2 1 U + + = + + + + = + + = = = Spring potential energy Gravitational potential energy Kinetic energy Conservation of energy

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Add thermal and internal energy ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 f INT f TH f MEC i INT i TH i MEC INT TH 2 2 s g MEC E E E E E E : energy of on conservati Extended energy ernal int of forms other All E energy Thermal E kx 2 1 mgy mv 2 1 U U K E + + = + + = = + + = + + =
A 10 kg block begins sliding at 20 m/s up a plane inclined at 30 0 with respect to horizontal. After sliding up 10 meters, it reaches a relaxed spring with weak spring constant 100 n/m. How far is the spring compressed when the block stops sliding? The block-plane coefficient of friction is 0.4 10 m ( 29 ( 29 ( 29 ( 29

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Unformatted text preview: ( 29 ( 29 ( 29 θ θ μ + -θ + =-θ = θ θ μ-+ + = + + = μ + + + = + +-+ = sin y cos mg 10 sin y k 2 1 mgy mv 2 1 10 sin y x : is n compressio spring final the and y is height vertical final The sin y cos mg kx 2 1 mgy ) mv 2 1 hill up ce tan dis final L NL U U K U U K E E E E f 2 f f 2 i f f f 2 f 2 i sf gf f si gi i i TH f TH f MEC i MEC 30 The solution, plus a little extra y f = 7.040 m Spring compression= 4.08 m E TH =478 J U gf = 690 J U sf =832 J These sum to the total initial energy K i =2000 J...
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L18_phys131_29oct07 - 29 29 29 θ θ μ-θ =-θ = θ θ μ...

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