In a system with weights suspended by a light rope on either side of a frictionless pulley, the acceleration of each weight can be determined using Newton's second law of motion.
Given: Weight on one side of the pulley (W1) = 60 N Weight on the other side of the pulley (W2) = 80 N
Let's assume the acceleration of the system is "a," and we'll consider the direction of motion as positive for one weight and negative for the other weight.
For the weight with 60 N (W1): Using Newton's second law: F = m * a Applying it to W1: 60 = m * a1 -- (Equation 1)
For the weight with 80 N (W2): Using Newton's second law: F = m * a Applying it to W2: 80 = m * (-a2) -- (Equation 2)
Since both sides of the rope are connected and have the same acceleration, a1 = -a2 = a.
Combining Equation 1 and Equation 2, we get: 60 = m * a 80 = m * (-a)
Dividing Equation 2 by Equation 1, we can eliminate the mass (m): 80 / 60 = (-a) / a 4/3 = -1
Since the equation 4/3 = -1 is not valid, it means there is no consistent solution that satisfies the given conditions. This indicates that the system is not in equilibrium, and the weights will have different accelerations in different directions.
Therefore, in this scenario, the weights will not have the same acceleration.