A 2.5-kg object is attached to a spring of force constant k = 4.5 kN/m. The spring is stretched 10 cm from equilibrium and released. What is the kinetic energy of the mass-spring system when the mass is 5.0 cm from its equilibrium position? 17 J 11 J 5.6 J 42 J 14 J
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The total energy is the sum of the kinetic and potential energy. When the spring is stretched to its maximum (10 cm), the kinetic energy is zero and all the energy is potential. Potential energy (PE) can be calculated using the formula PE = 0.5 * k * x^2, where Show more…
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