Unformatted text preview: 1 2 mv 2 2 − μ 2 mg ( f + δ ) − 1 2 k δ 2 = 0 Substituting for 1 2 mv 2 2 from above, this equation becomes 1 2 mV 2 + 2 mg f w sin θ − 1 2 cos θ W = μ 2 mg ( f + δ ) + 1 2 k δ 2 We are given δ = 1 2 f and μ 2 = 1 3 , wherefore μ 2 mg ( f + δ ) + 1 2 k δ 2 = 1 3 mg w 3 2 f W + 1 2 k w 1 2 f W 2 = 1 2 mg f + 1 8 k f 2 Combining these two equations yields 1 2 mg f + 1 8 k f 2 = 1 2 mV 2 + 2 mg f w sin θ − 1 2 cos θ W Solving for k , we conclude that k = 4 mg f w 4 sin θ − 2 cos θ − 1 + V 2 g f W...
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 Spring '06
 Shiflett
 Energy, Force, Potential Energy, 1 K, 1 1 mg

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