Question

A 57 kg boy and a 37 kg girl use a elastic rope while engaging in a tug-of-war on a frictionless icy surface. If the acceleration of the girl toward the boy is 3.2 m/s^2, determine the magnitude of the acceleration of the boy towards the girl.

Answer 1

The boy's acceleration toward the girl is 2.08 m/s^2.

Step-by-step explanation:

According the Newton's Third Law, the force causing the girl accelerate towards the boy will be me with equal counter-force causing the boy to accelerate towards the girl. These forces will be of equal magnitude and allows us to determine the boy's acceleration:

`F_(boy) = F_(girl)\\F_(girl) = m_(girl)\cdot a_(girl) = 37kg\cdot 3.2 (m)/(s^2)=118.40N\\\implies F_(boy) = 118.40N\\a_(boy) = (F_(boy))/(m_(boy))=(118.4N)/(57kg)=2.08(m)/(s^2)`

The boy's acceleration toward the girl is 2.08 m/s^2.