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26 August, 22:42

If you have a 60 g egg dropped from a height of 10 m with cushioning material that delays the impact by 2 seconds what was the force exerted on the egg by the floor?

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  1. 26 August, 23:08
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    The potential energy of the egg when it begins to fall is equal to the kinetic energy of the egg just before touching the ground:

    m * g * h = (1/2) * m * v ^ 2

    Clearing the speed:

    v = root (2 * h * g)

    Substituting the values

    v = root (2 * (10) * (9.8)) = 14

    Then, by definition, we have

    F * deltat = m * deltav

    Clearing F

    F = (m * deltav) / (deltat)

    Substituting the values

    F = ((0.060) * (14)) / (2) = 0.42

    answer

    the force exerted on the egg by the floor was 0.42N
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