A 64-kg boy and a 49-kg girl use an elastic rope while engaged in a tug-of-war on an icy frictionless surface. If the acceleration of the girl toward the boy is 3.8 m/s2, determine the magnitude of the acceleration of the boy toward the girl.
well, F=ma
Since the forces are equal and opposite, plug in the numbers.
To determine the magnitude of the acceleration of the boy toward the girl, we can use Newton's second law of motion. This law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Let's denote the acceleration of the boy toward the girl as a_boy and the acceleration of the girl toward the boy as a_girl.
The net force acting on the boy is equal to the product of the boy's mass (m_boy) and his acceleration (a_boy):
Net force on the boy = m_boy * a_boy
Similarly, the net force acting on the girl is equal to the product of the girl's mass (m_girl) and her acceleration (a_girl):
Net force on the girl = m_girl * a_girl
Since the forces in a tug-of-war are equal and opposite, the net force on one person is equal in magnitude but opposite in direction to the net force on the other person.
Therefore, the net force on the girl is equal in magnitude to the net force on the boy:
m_boy * a_boy = m_girl * a_girl
Given that the boy's mass (m_boy) is 64 kg, the girl's mass (m_girl) is 49 kg, and the girl's acceleration (a_girl) is 3.8 m/s^2, we can substitute these values into the equation and solve for the boy's acceleration (a_boy):
64 kg * a_boy = 49 kg * 3.8 m/s^2
Dividing both sides of the equation by 64 kg:
a_boy = (49 kg * 3.8 m/s^2) / 64 kg
a_boy simplifies to:
a_boy = 2.9 m/s^2
Therefore, the magnitude of the acceleration of the boy toward the girl is 2.9 m/s^2.
To determine the magnitude of the acceleration of the boy toward the girl, you can use Newton's second law of motion, which states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration (F = ma).
In this case, the girl is accelerating towards the boy with an acceleration of 3.8 m/s². The mass of the girl is 49 kg, so we can calculate the force exerted on the girl using F = ma:
Force on the girl = mass of girl * acceleration of girl
Fgirl = 49 kg * 3.8 m/s²
Fgirl = 186.2 N
Since the girl and the boy are engaged in a tug-of-war using an elastic rope, the tension in the rope is the same for both of them. Therefore, this force value (186.2 N) also represents the force exerted on the boy.
Now, we can use this force value and the mass of the boy (64 kg) to calculate the acceleration of the boy toward the girl using Newton's second law:
Force on the boy = mass of boy * acceleration of boy
186.2 N = 64 kg * acceleration of boy
To find the acceleration of the boy, rearrange the equation:
acceleration of boy = Force on the boy / mass of boy
acceleration of boy = 186.2 N / 64 kg
acceleration of boy ≈ 2.91 m/s²
Therefore, the magnitude of the acceleration of the boy toward the girl is approximately 2.91 m/s².