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

Well, it sounds like this boy and girl really know how to have a cool time on that icy surface! Let's see if we can figure out the magnitude of the acceleration of the boy using some good old physics.

First, we need to understand Newton's second law, which tells us that force equals mass times acceleration. Since the girl is accelerating towards the boy, the force acting on her is given by F = m * a, where m is her mass and a is her acceleration.

Using this equation, we have F = 37 kg * 2.4 m/s^2.

Now, according to Newton's third law, every action has an equal and opposite reaction. So, the force acting on the boy is the same as that on the girl, just in the opposite direction. Therefore, the magnitude of the acceleration of the boy towards the girl can be calculated using the same equation:

F = m * a, where m is the boy's mass (55 kg) and a is his acceleration (unknown).

Since the force experienced by the boy is the same as the force experienced by the girl, we have:

37 kg * 2.4 m/s^2 = 55 kg * a.

Simplifying this equation, we find that the acceleration of the boy towards the girl is approximately 1.6 m/s^2.

So, the magnitude of the acceleration of the boy towards the girl is about 1.6 m/s^2. Keep in mind that this calculation assumes that the elastic rope is massless and that there are no other external forces acting on the system.

Hope that cleared things up in a fun way!

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 force equals mass multiplied by acceleration.

Let's assume the mass of the boy is represented by mb and the mass of the girl by mg. Given that the mass of the boy is 55 kg and the mass of the girl is 37 kg, we have:
mb = 55 kg
mg = 37 kg

We know that the acceleration of the girl toward the boy is 2.4 m/s^2.
Therefore, the force acting on the girl is:
Fg = mg * ag
= 37 kg * 2.4 m/s^2
= 88.8 N

Since the boy and girl are connected by the same rope and experiencing the same tension, the force acting on the boy is also 88.8 N.

Now, we can determine the acceleration of the boy using Newton's second law:
Fb = mb * ab
= 55 kg * ab
= 88.8 N

Dividing both sides of the equation by 55 kg:
ab = 88.8 N / 55 kg
ab ≈ 1.616 m/s^2

Therefore, the magnitude of the acceleration of the boy toward the girl is approximately 1.616 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 its mass multiplied by its acceleration (F = m * a).

Given that the acceleration of the girl toward the boy is 2.4 m/s^2, we can use this information to calculate the force exerted on the girl:

Force (girl) = mass (girl) * acceleration (girl)
F_g = 37 kg * 2.4 m/s^2
F_g = 88.8 N

Now, since the boy and the girl are engaged in a tug-of-war using an elastic rope, the magnitude of the force exerted on the boy (F_b) and the girl (F_g) are the same. Therefore, F_b = F_g = 88.8 N.

To find the acceleration of the boy toward the girl, we can rearrange the equation F = m * a to solve for acceleration:

a = F / m

Substituting the known values, we get:

Acceleration (boy) = Force (boy) / mass (boy)
a_b = 88.8 N / 55 kg
a_b ≈ 1.616 m/s^2

Therefore, the magnitude of the acceleration of the boy toward the girl is approximately 1.616 m/s^2.