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11 October, 17:06

Suppose a 63-kg boy and a 40-kg girl use a massless rope in a tug-of-war on an icy, resistance-free surface. If the acceleration of the girl toward the boy is 3.0 m/s2, find the magnitude of the acceleration of the boy toward the girl.

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  1. 11 October, 17:32
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    a₂ = 1.9 m/s²

    Explanation:

    Newton's third law or principle of action and reaction:

    Force of the boy on the girl = - Force of the girl on the boy

    Fbg = - Fgb Formula (1)

    Newton's second law

    ∑F = m*a Formula (2)

    ∑F : algebraic sum of the forces in Newton (N)

    m : mass in kilograms (kg)

    a : acceleration in meters over second square (m/s²)

    Newton's second law to the girl

    m₁ : girl mass

    a₁ : acceleration of the girl toward the boy

    ∑F = m₁*a₁

    Fbg = (40kg) * (3 m/s²)

    Fbg = 120 N

    Newton's second law to the boy

    m₂: boy mass

    a₂ : acceleration of the boy toward the girl

    ∑F = m₂*a₂

    Fgb = (63) * a₂

    120 = (63) * a₂

    a₂ = 120 / (63)

    a₂ = 1.9 m/s²
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