Consider the two blocks as particles. The free-body diagram of the above situation can be developed as follows:

Looking at string length changes, Block 1 moves 1/2 of the distance moved by Block 2.

Using d = vt and v = at, Then d = at². t² is a constant (same for both blocks), so d ∝ a. The acceleration follows the relationship between lengths.

Block 2 accelerates at twice the rate of block 1.

Therefore, a₂ = 2a₁ --------------------(i)

OR a₁ = (1/2)a₂ .

Applying Newton's second law to block 2, (consider left and down to be positive)

(ΣF_{on 2})_y = ma₂

F_G2 - T = m₂•a₂ --------------------------------------(ii)

Applying Newton's second law to block 1,

2T = m₁•a₁

T = (1/2)m₁•a₁ ------------------------------------------------(iii)

using equation (iii) in (ii),

F_G2 - (1/2)m₁•a₁ = m₂•a₂

from (i),

m₂•g - (1/2)m₁•a₁ = m₂•(2a₁)

m₂•g = (1/2)m₁•a₁ + m₂•(2a₁)

m₂•g = a₁[(1/2)m₁ + 2m₂]

(1.5 kg)(9.8 m/s²) = a₁[(1/2)(5.0 kg) + 2 × 1.5 kg]

a₁ = 2.7 m/s²)

The correct answer is option D.

Forces in Connected Systems of Objects Practice Problems