Suppose a 65-kg boy and a 45-kg girl use a massless rope in a tug-of-war on an icy, resistance-free surface. If the acceleration of the girl toward the boy is 3.0 m/s2, find the magnitude of the acceleration of the boy toward the girl.

i want to know them

girl: 135 N

boy mass: 65 kg

135/65 = 2.0769 =

2.077 m/s^2

Tension of rope = T N

Mass of girl, m = 45 kg
Acceleration = 3.0 m/s²
T = ma = 135 N
Mass of boy, M = 65 kg
T = MA,
solve for acceleration A.

i want to know them too

To find the magnitude of the acceleration of the boy toward the girl, we can use Newton's second law of motion, which states:

F = m * a

where F is the force, m is the mass, and a is the acceleration.

In this case, the girl is experiencing a force towards the boy due to the tug-of-war. The magnitude of this force can be calculated as:

F = m * a

Taking the girl's mass as 45 kg and her acceleration as 3.0 m/s², we can calculate her force as:

F_girl = 45 kg * 3.0 m/s² = 135 N

According to Newton's third law of motion, for every action, there is an equal and opposite reaction. Therefore, the boy will also experience a force of the same magnitude but in the opposite direction. So, the boy's force is:

F_boy = -135 N

Now, we can calculate the boy's acceleration using Newton's second law:

F_boy = m_boy * a_boy

Rearranging the equation, we get:

a_boy = F_boy / m_boy

Given the boy's mass as 65 kg, substituting the values, we can find the boy's acceleration:

a_boy = -135 N / 65 kg

Simplifying the expression gives us:

a_boy = -2.08 m/s²

Therefore, the magnitude of the acceleration of the boy toward the girl is 2.08 m/s².

-2.1

Mass of the boy is

m
1
=
65

k
g

Mass of the girl is
m
2
=
45

k
g

Acceleration of the girl towards the boy is:

a
1
=
3

m
/
s
2

For the force applied by the boy on the girl :

F
1
=
m
2
×
a
1
F
1
=
45
×
3
F
1
=
135

N
Now since the action and reaction pair will be same (
F
1
=
F
2
)now for the acceleration of the girl:

a
2
=
F
2
m
1
a
2
=
135
65
a
2
=
2.07

m
/
s
2
Thus, the acceleration of the girl towards the boy is
2.07

m
/
s
2

135 Newtons and 2.077 m/s2

no pb m8

The forces are the same

massman*aman=massgirl*agirl

solve for the man's acceleration

um lol but who wants to know these answers?