Connected Bodies And Pulleys

In an Atwood machine the heavier block descends while the lighter one rises with the same magnitude of acceleration, because the string is inextensible and the pulley frictionless. Writing Newton's Second Law for each block, the unbalanced driving force is the weight difference $(m_2-m_1)g$ while the total inertia being accelerated is $(m_1+m_2)$. Hence $a=\frac{(m_2-m_1)g}{m_1+m_2}=\frac{(5-3)(10)}{5+3}=\frac{20}{8}=2.5{\text{ m/s}}^2$. The 1.25 m/s^2 value halves this result, perhaps by using only one mass in the denominator incorrectly. The 5 m/s^2 value would require the full weight difference to act on just the 2 kg net mass, ignoring that both blocks share the acceleration. The 1.6 m/s^2 value comes from an arithmetic mismatch of the masses. The same framework also yields the string tension $T=\frac{2m_1{m}_{2}g}{m_1+m_2}$, which always lies between the two individual weights, a useful consistency marker for connected-body problems. This is the standard NCERT pulley result, derived by writing one Newton equation per block and eliminating the common tension. As a sanity check, the acceleration is well below g, which is expected since the lighter block partly counterbalances the heavier one, and were the two masses equal the acceleration would correctly fall to zero.