KUmghkxAACC+=++=+01202which yields hkxmg==×=2222212981.35 10 N m 0.055 mkgm s0.174 m4chbgbg ch..Therefore,A°=°=xhsin 300.174 m0.35 m .(b) From this result, we findA=−=0.35 0.0550.29 m , which means that ΔyAsinmθ015.in sliding from point Ato point B. Thus, Eq. 8-18 gives 0120ΔΔΔKUmvmg hB2which yieldsvghB−=298015Δ...bgbg1.7 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 Cis 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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