Two-body system with friction — block on table and hanging mass: A 6.6 kg block A rests on a horizontal platform and is connected over a smooth pulley to a hanging 3.2 kg mass. The coefficient of friction between block A and the platform is 1/3. What is the acceleration of the system (take g ≈ 9.8 m/s^2)?

Introduction / Context:
This is a standard two-mass system: a block on a horizontal surface connected to a hanging mass over a frictionless pulley. The acceleration depends on the net driving force (weight of the hanging mass minus frictional resistance on the block) divided by the total mass of the system.

Given Data / Assumptions:

• Mass on table m1 = 6.6 kg; hanging mass m2 = 3.2 kg.
• Coefficient of friction μ = 1/3 between m1 and table.
• Pulley and string are light and frictionless; motion is impending in the direction pulled by m2.
• Take g = 9.8 m/s^2.

Concept / Approach:

Friction force opposing motion on the table is F_f = μ * N = μ * m1 * g. The driving force is m2 * g − F_f. The acceleration of the connected system is a = (driving force) / (m1 + m2), provided the hanging weight exceeds friction so that motion occurs.

Step-by-Step Solution:

Verification / Alternative check:

Check motion condition: m2 * g > μ * m1 * g → 3.2 > 2.2 (true). Therefore system accelerates with the computed value.

Why Other Options Are Wrong:

(a) and (b) underestimate net driving force; (d) and (e) overestimate by ignoring friction magnitude or miscalculating total mass.

Common Pitfalls:

Using mass difference directly without subtracting friction; forgetting that friction depends on normal reaction m1 * g; rounding g too crudely.

Final Answer:

1.00 m/s^2