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15 February, 19:20

When a 4.10-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.90 cm. If the 4.10-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it?

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  1. 15 February, 19:57
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    Using Hooke's law,

    F = - kx

    Where,

    F = force of the mass

    k = spring constant/stiffness

    x = length of the spring

    Given:

    F1 = m1*a

    = 4.1 * 9.81

    = 40.221 N

    F2 = m2*a

    = 1.5 * 9.81

    = 14.715 N

    x1 = 2.9 cm

    F1/x1 = F2/x2

    Therefore,

    x2 = (14.715 * 2.9) / 40.221

    = 1.06 cm.
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