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17 October, 07:17

A body of mass 1kg suspended from the free end of a spring having force constant 400N/m is executing S. H. M when the total energy of system is 2 joule find the minimum acceleration

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  1. 17 October, 07:29
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    -30 m/s²

    Explanation:

    The total energy is the sum of the gravitational potential energy of the mass, the kinetic energy of the mass, and the elastic energy in the spring.

    ∑E = PE + KE + EE

    ∑E = mgh + ½ mv² + ½ kx²

    At the bottom, h = 0 and v = 0.

    ∑E = ½ kx²

    2 = ½ (400) x²

    2 = 200 x²

    0.01 = x²

    x = 0.1

    At the top, v = 0 and x = 0.1 - h.

    ∑E = mgh + ½ kx²

    2 = (1) (10) h + ½ (400) (0.1 - h) ²

    2 = 10h + 200 (0.01 - 0.2h + h²)

    2 = 10h + 2 - 40h + 200h²

    0 = 200h² - 30h

    0 = 10h (20h - 3)

    h = 0.15

    So x ranges from - 0.05 m to 0.10 m.

    Sum of forces on the mass:

    ∑F = ma

    kx - mg = ma

    The minimum acceleration is when x is a minimum.

    (400 N/m) (-0.05 m) - (1 kg) (10 m/s²) = (1 kg) a

    a = - 30 m/s²

    The maximum acceleration is when x is a maximum.

    (400 N/m) (0.1 m) - (1 kg) (10 m/s²) = (1 kg) a

    a = 30 m/s²

    And of course, acceleration is 0 when kx = mg.

    (400 N/m) x = (1 kg) (10 m/s²)

    x = 0.025 m
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