Question

The acceleration of 2 kg mass in the given figure is (all surfaces are smooth and pulleys are massless and frictionless)

3 kg 2 kg 1 kg © Darwin by MCC

Answer 1

a = g/6 ≈ 1.63 m/s² (downward)

Answer 2

Solution Explanation We choose a coordinate system so that the 3 kg block accelerates to the left with magnitude a. Then, because of the inextensible strings:

• The 2 kg (left) block moves downward with acceleration a.
• The 1 kg (right) block moves upward with acceleration a.

For the 2 kg block (taking downward as positive):   Weight – Tension = mass × acceleration     2g – T₁ = 2a      …(1)   ⇒ T₁ = 2g – 2a

For the 1 kg block (taking upward as positive):   Tension – Weight = mass × acceleration     T₂ – g = 1a      …(2)   ⇒ T₂ = g + a

For the 3 kg block on the table, the only horizontal forces are due to the two tensions. The left string pulls toward the left with T₁, and the right string pulls toward the right with T₂; since the block accelerates to the left, the net force is   T₁ – T₂ = 3a      …(3)

Substitute T₁ and T₂ from (1) and (2) into (3):   (2g – 2a) – (g + a) = 3a   2g – 2a – g – a = 3a   g – 3a = 3a   g = 6a Thus,   a = g/6

Taking g = 9.8 m/s²,   a ≈ 9.8/6 ≈ 1.63 m/s²

So, the 2 kg mass (on the left) accelerates downward with an acceleration of g/6 (≈1.63 m/s²).

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