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2 April, 12:30

A glider with mass m = 0.200 kg sits on a frictionless, horizontal air track. It is connected to a spring of negligible mass and force constant k = 5.00 N/m. You pull on the glider to stretch the spring 0.100 m, and then release the glider. The glider begins to move back toward the equilibrium position (x = 0). What is its speed when x = 0.0800 m?

0.400 m/s

0.300 m/s

0.016 m/s

0.025 m/s

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  1. 2 April, 12:56
    0
    0·300 m/s

    Explanation:

    Given

    Mass of the glider = 0·2 kg

    Spring constant = k = 5 N/m

    At x = 0·1 m, the glider is released, it means that initial velocity of the glider is zero

    ∴ Total Mechanical energy = potential energy

    Equilibrium position is at x = 0 cm

    Potential energy in case of spring = 0·5 * k * x²

    where k is the spring constant

    x is the distance from equilibrium position

    Initially potential energy = 0·5 * 5 * (0·1 - 0) ² = 0·025 J

    As there is no dissipative force, therefore mechanical energy of the glider remains constant

    At x = 0·08 m

    Mechanical energy of the glider = Kinetic energy + Potential energy

    Let the velocity of the glider at x = 0·08 m be v m/s

    Kinetic energy at this instant = 0.5 * m * v² = 0·5 * 0·2 * v² J

    Potential energy at this instant = 0·5 * k * x² = 0·5 * 5 * 0·08² = 0·016 J

    Mechanical energy = 0·5 * 0·2 * v² + 0·016

    As mechanical energy is constant

    0·025 = 0·5 * 0·2 * v² + 0·016

    0·009 = 0·1 * v²

    ∴ v = 0·300 m/s

    ∴ Speed when x = 0.0800 m is 0·300 m/s
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