An ideal spring hangs from the ceiling. A 1.95 kg mass is hung from the spring, stretching the spring a distance d=0.0865 m from its origina

Question

An ideal spring hangs from the ceiling. A 1.95 kg mass is hung from the spring, stretching the spring a distance d=0.0865 m from its original length when it reaches equilibrium. The mass is then lifted up a distance L=0.0325 m from the equilibrium position and released. What is the kinetic energy of the mass at the instant it passes back through the equilibrium position?

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Thái Dương 6 months 2021-08-04T12:06:47+00:00 1 Answers 6 views 0

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    2021-08-04T12:08:31+00:00

    Answer:

    kinetic energy = 0.1168 J

    Explanation:

    From Hooke’s law, we know that ;

    F = kx

    k = F/x

    We are given ;

    Mass; m = 1.95 kg

    Spring stretch; d = x = 0.0865

    So, Force = mg = 1.95 × 9.81

    k = 1.95 × 9.81/0.0865 = 221.15 N/m

    Now, initial energy is;

    E1 = mgL + ½k(x – L)²

    Also, final energy; E2 = ½kx² + ½mv²

    From conservation of energy, E1 = E2

    Thus;

    mgL + ½k(x – L)² = ½kx² + ½mv²

    Making the kinetic energy ½mv² the subject, we have;

    ½mv² = mgL + ½k(x – L)² – ½kx²

    We are given L=0.0325 m

    Plugging other relevant values, we have ;

    ½mv² = (1.95 × 9.81 × 0.0325) + (½ × 221.15(0.0865 – 0.0325)² – ½(221.15 × 0.0865²)

    ½mv² = 0.62170875 + 0.3224367 – 0.82734979375

    ½mv² = 0.1168 J

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