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.

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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explanation:

c. magnitude and direction