Given M = 5m, what is the magnitude of the acceleration of the system shown? Assume a frictionless surface and an ideal pulley.

Question

BlackTom AI · Accepted Answer

The problem describes a classic Atwood-like system with one mass on a frictionless horizontal surface connected by a rope over a light pulley to a hanging mass. We are given M = 5m and asked for the magnitude of the acceleration.

First, restate the setup: Mass M sits on a horizontal, frictionless surface, attached by a rope over a pulley to a hanging mass m. The acceleration of both masses is the same in magnitude due to the rope constraint.

Analyze each option by writing down the dynamics equations for the two masses:
- For the block on the table (mass M): The only horizontal force is the tension T, pulling it toward the pulley. Newton’s second law gives M a = T.
- For the hanging mass (mass m): The weight m g acts downward, while the tension T acts upward. The net force is m g − T, which produces acceleration a: m g − T = m a.

Now solve the system:
From the first equation, T = M a. Substitute into the second equation: m g − M a = m a.
This leads to m g = (M + m) a, so a = (m g) / (M + m).

Plug in M = 5m:
a = (m g) / (5m + m) = (m g) / (6m) = g/6.

Option-by-option analysis:
- g/2: This would require a larger acceleration, which would imply a smaller total mass in the denominator or a larger driving force than present. It does not satisfy the balance of forces derived above.
- g: An acceleration of g would mean the hanging mass is in free fall, which cannot happen here because the string and pulley transmit tension and the mass on the table resists motion; the system cannot free-fall with the table mass attached.
- g/6: This matches the derived result a = g/6; it correctly accounts for both masses (5m on the table and m hanging) in the combined mass M + m = 6m.
- g/4: This would correspond to a different distribution of mass or a misapplication of the equations, and would not satisfy the dynamics M a = T and m g − T = m a simultaneously.
- g/5: This would imply a = g divided by 5, which would correspond to an effective total mass of 5m in the denominator, but we have 6m as the total in the expression, so this is incorrect.

In summary, the correct acceleration magnitude is g/6, achieved by balancing the horizontal and vertical forces through the two-branch system and solving for a = m g / (M + m). The other options fail to satisfy the two equations simultaneously or misrepresent the system dynamics.

Given M = 5m, what is the magnitude of the acceleration of the system shown? Assume a frictionless surface and an ideal pulley. 单项选择题

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