The slender 6-kg bar AB is horizontal and at rest and the spring is...?

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  • 2 years ago
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    At the top position, the bar has gravitational potential energy.

    At the bottom of the swing, the spring has elastic potential energy.

    The kinetic energy is 0 at both positions.

    GPE = EPE

    mgh = 1/2 kx²

    The trick is to use geometry to find h and x.

    h is pretty simple to find. At the top of the swing, the center of the bar is at the same height as the hinge. At the bottom of the swing, the center of the bar is 1 m below the hinge. So:

    h = 0 − (-1)

    h = 1 m

    To find x, we first need to find the length of the spring at the bottom position using Pythagorean theorem.

    L² = (2 m)² + (1.5 m + 2 m)²

    L ≈ 4.03 m

    x, the increase from the unstretched length, is:

    x = L − 1.5

    x ≈ 2.53 m

    Therefore:

    mgh = 1/2 kx²

    (6)(9.8)(1) = 1/2 k (2.53)²

    k = 18.4 N/m

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