Question

multiple-object accelerating systems: a 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. if the masses are gently released, what is the resulting acceleration of the masses?

Answer 1

When two objects are connected by a rope over an ideal pulley, the acceleration of the system can be calculated using Newton's second law of motion. The tension in the rope is equal to the difference in weights between the two masses, and the acceleration of the system is equal to the tension divided by the sum of the masses.

Step-by-step explanation:

When two objects are connected by a rope over an ideal pulley, the acceleration of the system can be calculated using Newton's second law of motion. The net force acting on the system is the difference between the weights of the two masses. Since the pulley is frictionless and the rope is light, the tension in the rope is equal to the force required to accelerate the masses:

Tension = (m₁ - m₂) * g

where m₁ is the mass of the larger object, m₂ is the mass of the smaller object, and g is the acceleration due to gravity (approximately 9.8 m/s²).

In this case, with a 5.0-kg mass and a 3.0-kg mass, the tension in the rope is:

Tension = (5.0 kg - 3.0 kg)* 9.8 m/s² = 19.6 N

Since both masses are accelerating together as a system, the acceleration of the masses is the same and can be calculated using the following formula:

Acceleration = Tension / (m₁ + m₂)

Substituting the values:

Acceleration = 19.6 N / (5.0 kg + 3.0 kg) = 2.45 m/s²