A 55 kg boy and a 37 kg girl use an elastic rope
while engaged in a tug-of-war on a frictionless
icy surface.
If the acceleration of the girl toward the
boy is 2.4 m/s^2, determine the magnitude of
the acceleration of the boy toward the girl in m/s^2
To determine the magnitude of the acceleration of the boy toward the girl, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
In this scenario, the boy and the girl are connected by an elastic rope, and they are engaged in a tug-of-war. Since they are on a frictionless icy surface, there are no other external forces acting on them.
Let's assume that the acceleration of the boy toward the girl is represented by a, and the acceleration of the girl toward the boy is given as 2.4 m/s^2. We can write the following equations:
For the boy:
Mass of the boy (m1) = 55 kg
Acceleration of the boy (a) = ?
Net force acting on the boy (F1_net) = m1 * a
For the girl:
Mass of the girl (m2) = 37 kg
Acceleration of the girl (2.4 m/s^2) = 2.4 m/s^2
Net force acting on the girl (F2_net) = m2 * 2.4
Since the rope is elastic, the magnitudes of the net forces acting on the boy and the girl should be equal. Therefore, we can equate the two equations:
F1_net = F2_net
m1 * a = m2 * 2.4
Substituting the given values:
55 kg * a = 37 kg * 2.4 m/s^2
Now, we can solve for a:
a = (37 kg * 2.4 m/s^2) / 55 kg
a ≈ 1.61 m/s^2
Therefore, the magnitude of the acceleration of the boy toward the girl is approximately 1.61 m/s^2.