A 57 kg boy and a 42 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 3.4 m/s2, determine the magnitude of
the acceleration of the boy toward the girl. Answer in units of m/s2.
The tension on the elastic rope can be calculated by Newton's second law:
F=ma
From the girl's acceleration, F can be found, which can then be applied to the boy to find his acceleration.
2.505263158 m/s2
To find the magnitude of acceleration of the boy toward the girl, we can apply 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.
First, let's calculate the net force acting on the system. We know that the girl experiences an acceleration of 3.4 m/s^2 toward the boy. So, the net force acting on the girl is given by:
Net force on the girl = mass of the girl * acceleration of the girl
= 42 kg * 3.4 m/s^2
= 142.8 N
According to Newton's third law of motion, when the girl exerts a force on the boy, the boy exerts an equal and opposite force on the girl. Therefore, the net force on the system is zero.
Since the net force acting on the system is zero, we can say that the net force on the boy is:
Net force on the boy = -Net force on the girl
= -142.8 N
Now, we can use Newton's second law to find the acceleration of the boy. Rearranging the equation, we have:
Net force on the boy = mass of the boy * acceleration of the boy
Substituting the values we know, we get:
-142.8 N = 57 kg * acceleration of the boy
Now, we can solve for the acceleration of the boy:
acceleration of the boy = -142.8 N / 57 kg
= -2.51 m/s^2
Therefore, the magnitude of the acceleration of the boy toward the girl is 2.51 m/s^2.