Suppose a 68-kg boy and a 40-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/s^2, find the magnitude of the acceleration of the boy toward the girl.

Fboy+Fgirl=0

68*a+40*3.0=0
solve for a.

so it should be -1.76 m/s^2

yes, the negative sign indicates direction.

its a web assign and i typed that answer in but it said it was wrong

lol. In web assagain, if you type ::corrrect, it shows you the answer.

To find 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. In this case, the force acting on the boy is equal to the force acting on the girl.

Let's assume that the force exerted by the boy is F_boy and the force exerted by the girl is F_girl. Since the net force on both objects is the same (as they are connected by a massless rope and are moving together), we can write:

F_boy = F_girl

Now, we know that force is equal to mass times acceleration:

F_boy = m_boy * a_boy
F_girl = m_girl * a_girl

Given:
m_boy = 68 kg (mass of the boy)
m_girl = 40 kg (mass of the girl)
a_girl = 3.0 m/s^2 (acceleration of the girl)

The masses of both the boy and girl are given, and we need to find the acceleration of the boy. We can rearrange the equation to solve for the acceleration of the boy:

a_boy = F_boy / m_boy

Since F_boy = F_girl, we can substitute this in the equation:

a_boy = F_girl / m_boy

Now, substituting the values we have, we get:

a_boy = (m_girl * a_girl) / m_boy

Plugging in the given values:

a_boy = (40 kg * 3.0 m/s^2) / 68 kg

Now, performing the calculation:

a_boy = 1.76 m/s^2

Therefore, the magnitude of the acceleration of the boy toward the girl is 1.76 m/s^2.