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/s2, find the magnitude of the acceleration of the boy toward the girl.
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Is it possible they don't want the negative sign? Is is possible they only want two significant digits?
My crystal ball is not working. I know physics, but I can't guess what the author supposed was the "right" answer.
okay i changed it they didn't want a negative sign for some reason
if arnold strongman and suzie small each pull very hard on opposite ends of a massless rope in a tug-of-war the greater force on the rope is exerted by ? i think both the same
To find the magnitude of the acceleration of the boy toward the girl, we can use Newton's second law of motion: F = ma.
In this case, the force acting on the girl is the force pulling her toward the boy, and the force acting on the boy is the force pulling him toward the girl. The magnitude of the force experienced by both of them will be the same, as they are connected by a massless 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.
From the given information, we know that the acceleration of the girl toward the boy is 3.0 m/s^2.
We can assume that the force of tension in the rope is T. According to Newton's third law, the force exerted by the boy on the girl is also T in the opposite direction, and the force exerted by the girl on the boy is also T in the opposite direction.
The net force acting on the girl is given by F_net,girl = ma_girl = T.
Similarly, the net force acting on the boy is given by F_net,boy = ma_boy = T.
Since both F_net,girl and F_net,boy are the same, we have:
ma_girl = ma_boy.
Given the mass of the girl (m_girl) is 40 kg, we have:
40 kg * 3.0 m/s^2 = m_boy * a_boy.
Now, we need to find the mass of the boy (m_boy). Given the mass of the boy and girl together (m_total) is 68 kg, we have:
m_total = m_boy + m_girl.
68 kg = m_boy + 40 kg.
Solving this equation for m_boy, we find:
m_boy = 68 kg - 40 kg
m_boy = 28 kg.
Substituting this value of m_boy into the equation, we have:
40 kg * 3.0 m/s^2 = 28 kg * a_boy.
Simplifying this equation, we find:
a_boy = (40 kg * 3.0 m/s^2) / 28 kg.
Evaluating this expression, we get:
a_boy = 4.286 m/s^2.
Therefore, the magnitude of the acceleration of the boy toward the girl is approximately 4.286 m/s^2.