flyingpig
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Why do ideal springs stop at xf = 0? We have friction in this problem, does that mean it really doesn't stop at xf = 0, but xf < 0?
The discussion revolves around a physics problem involving a 12 kg box pressed by a spring with a hardness coefficient of 800 N/m. The participants analyze the box's velocity at point B (0.5m from the starting point) and the distance it travels before stopping, factoring in kinetic friction of 0.15. Key formulas discussed include the kinetic energy equation (Ek = mv²/2) and the elastic potential energy from the spring (Eel = c * OA² / 2). The conversation emphasizes the conversion of elastic energy to kinetic energy while accounting for energy lost due to friction.
PREREQUISITESStudents studying physics, particularly those focusing on mechanics, as well as educators seeking to enhance their understanding of energy transformations in dynamic systems.
flyingpig said:Why do ideal springs stop at xf = 0? We have friction in this problem, does that mean it really doesn't stop at xf = 0, but xf < 0?