Suppose a 69-kg boy and a 42-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.
the forces are the same, opposite direction.
42*3=69*ab
solve for the acceleration ab .
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 the mass of the object multiplied by its acceleration (F = m * a).
In this situation, the force acting on the girl (Fgirl) is given by the product of the girl's mass (mgirl) and her acceleration (agirl):
Fgirl = mgirl * agirl
Similarly, the force acting on the boy (Fboy) is given by the product of the boy's mass (mboy) and his acceleration (aboy):
Fboy = mboy * aboy
Since the rope is massless, the tension in the rope is the same for both the girl and the boy. Therefore, the magnitudes of the forces acting on the girl and the boy are equal:
Fgirl = Fboy
Substituting the expressions for the forces, we can write the equation as:
mgirl * agirl = mboy * aboy
Now, we can plug in the given values:
mgirl = 42 kg
agirl = 3.0 m/s^2
mboy = 69 kg
Substituting these values into the equation, we get:
42 kg * 3.0 m/s^2 = 69 kg * aboy
Simplifying the equation, we can solve for aboy:
aboy = (42 kg * 3.0 m/s^2) / 69 kg
Calculating this, we find:
aboy ≈ 1.826 m/s^2
Therefore, the magnitude of the acceleration of the boy toward the girl is approximately 1.826 m/s^2.