Two masses are suspended by a cord that passes over a pulley with negligible mass. The cord also has negligible mass. One of the masses, m₁, has a mass of 5.0 kg and the other mass, m₂, has a mass of 3.0 kg. The acceleration of m₁ is A) 3.3 m/s² upward. B) 3.3 m/s² downward. C) 2.5 m/s² upward. D) 2.5 m/s² downward. E) 1.1 m/s² upward.

Question

Quizplus · Accepted Answer

Since the acceleration is acting downwards, the system will move in the direction of the lighter mass (m₂). Therefore, m₂ will experience a net force downwards, while m₁ will experience a net force upwards. We can use Newton's second law to find the acceleration of the system: $\sum F = ma$ $\left(m_{S1} - m_{S2} ight)g = \left(m_{S1} + m_{S2} ight)a$ $a = \frac{\left(m_{S1} - m_{S2} ight)g}{\left(m_{S1} + m_{S2} ight)} = \frac{(5.0 - 3.0) ext{ kg} \cdot 9.81 ext{ m/s}^2}{(5.0 + 3.0) ext{ kg}} = 3.27 ext{ m/s}^2$ Since the acceleration is acting downwards, the correct answer is 3.3 m/s² downward.

Two Masses Are Suspended by a Cord That Passes Over