A 58 kg boy and a 38 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 3.1 m/s
2
, determine the magnitude of
the acceleration of the boy toward the girl.
Answer in units of m/s
2
The forces will be equal( but opposite).
Force equals mass times acceleration.
Let m = 38kg (the girl's mass)
Let a = 3.1m/s^2 (the girl's acceleration)
Let n = 58kg (the boy's mass)
Find b (the boy's acceleration).
Such that: m*a = -n*b
To solve this problem, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration.
Let's assume that the acceleration of the boy toward the girl is "a" m/s^2.
Using this information, we can set up the following equations:
For the girl:
Force_girl = mass_girl × acceleration_girl
Force_girl = 38 kg × 3.1 m/s^2
For the boy:
Force_boy = mass_boy × acceleration_boy
Force_boy = 58 kg × (-a) m/s^2 (negative sign indicates it is in the opposite direction of the girl's acceleration)
Since the two forces are equal in magnitude (but opposite in direction) when the rope is in tension, we can set up an equation:
Force_girl = Force_boy
38 kg × 3.1 m/s^2 = 58 kg × (-a) m/s^2
Now, let's solve for "a":
38 kg × 3.1 m/s^2 = 58 kg × (-a) m/s^2
118.8 kg*m/s^2 = -58 kg × a m/s^2
Dividing both sides by -58 kg:
a = -118.8 kg*m/s^2 / -58 kg
a ≈ 2.05 m/s^2
Therefore, the magnitude of the acceleration of the boy toward the girl is approximately 2.05 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 force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, we have two objects - the boy and the girl - connected by an elastic rope.
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. We know that the mass of the boy (m_boy) is 58 kg and the mass of the girl (m_girl) is 38 kg. We also know that the acceleration of the girl toward the boy (a_girl) is 3.1 m/s^2.
Now, the force acting on the girl is given by:
F_girl = m_girl * a_girl
And the force acting on the boy is given by:
F_boy = m_boy * a_boy
Since the rope is elastic and the two objects are connected, the forces acting on them are equal in magnitude but opposite in direction. Therefore:
F_girl = - F_boy
Substituting the expressions for the forces:
m_girl * a_girl = - m_boy * a_boy
Rearranging the equation to solve for a_boy:
a_boy = - (m_girl / m_boy) * a_girl
Substituting the given values:
a_boy = - (38 kg / 58 kg) * 3.1 m/s^2
Simplifying the expression:
a_boy = - 2.517 m/s^2
The magnitude of the acceleration of the boy toward the girl is 2.517 m/s^2.