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    Find the total energy at the equilibrium position

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    A 0.50-kg mass is attached to a spring with a spring constant of 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. What is the total energy of the system?

    © BrainMass Inc. brainmass.com December 24, 2021, 10:23 pm ad1c9bdddf
    https://brainmass.com/physics/energy/finding-total-energy-equilibrium-position-470027

    SOLUTION This solution is FREE courtesy of BrainMass!

    m = 0.5kg
    k = 20 N/m
    v = 1.5m/s

    At the equilibrium place the spring is not stretched or compressed, hence the entire energy in the system is in the form of kinetic energy of the mass.

    E = 1/2 m * v^2
    =0.5*0.5*1.5^2
    =0.5625 J.

    Therefore, the total energy of the system is 0.5625 J.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 10:23 pm ad1c9bdddf>
    https://brainmass.com/physics/energy/finding-total-energy-equilibrium-position-470027

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