a 55 kh boy and 40 kg girl usa a rope while engaged in a tug-of-war on an icy frictionless surface. If the girl accelerates towards the boy at 5.1 m/s/s, what is the acceleration of the boy towards the girl.
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To determine the acceleration of the boy towards the girl, we can apply Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F = m * a). In this case, the girl exerts a force on the rope, causing her own acceleration. Since the surface is frictionless, we can assume that all the forces cancel out and that the rope is the only force acting between the boy and the girl.
First, we need to find the force (F) that the girl exerts on the rope. We can use her mass (m) and her acceleration (a) for this calculation: F = m * a.
F = (40 kg) * (5.1 m/s^2)
F = 204 N
Since the rope is the only force acting on the boy, the force that the girl exerts on the rope is also the force that the boy exerts on the rope (according to Newton's third law of motion). Therefore, we can use this force to find the boy's acceleration.
Now, we need to find the boy's acceleration (a_boy). We can rearrange Newton's second law (F = m * a) to solve for acceleration:
a = F/m
a_boy = F/m_boy
Since the force (F) is 204 N and the boy's mass (m_boy) is 55 kg:
a_boy = 204 N / 55 kg
a_boy ≈ 3.71 m/s^2
Therefore, the acceleration of the boy towards the girl is approximately 3.71 m/s^2.