ch08-p031 - 31. We refer to its starting point as A, the...

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KU mgh kx AACC +=+ += + 0 1 2 0 2 which yields h kx mg == × = 2 2 2 2 212 98 1.35 10 N m 0.055 m kg m s 0.174 m 4 c hb g b g c h . . Therefore, A ° = ° = x h sin 30 0.174 m 0.35 m . (b) From this result, we find A =− = 0.35 0.055 0.29 m , which means that Δ y A sin m θ 015 . in sliding from point A to point B . Thus, Eq. 8-18 gives 0 1 2 0 ΔΔ Δ K U mv mg h B 2 which yields vg h B = 29 8 0 1 5 Δ .. . b gb g 1.7 m s 31. We refer to its starting point as A , the point where it first comes into contact with the spring as B , and the point where the spring is compressed | x | = 0.055 m as C . Point C is our reference point for computing gravitational potential energy. Elastic potential energy (of the spring) is zero when the spring is relaxed. Information given in the second
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