Two Blocks On A Rough Incline

Consider both blocks as one system linked by an inextensible string over a frictionless pulley, so they share a common acceleration magnitude. The incline is frictionless, so the only driving and opposing agents are the gravity component of the incline block along the slope, $m_{1}g\sin \theta$, and the full weight of the hanging block, $m_{2}g$. The 5 kg block's pull down the slope is $5\times 10\times 0.6=30\text{ N}$, exceeding the hanging block's weight $4\times 10=40\text{ N}$? Here $m_{2}g=40\text{ N}$ is larger, so the hanging block descends. Net force $=40-30=10\text{ N}$ acting on total mass $9\text{ kg}$, giving $a=\frac{m_{2}g-m_{1}g\sin \theta }{m_1+m_2}=\frac{40-30}{9}=1.11{\text{ m/s}}^2$. The 2.22 m/s^2 value halves the denominator mistakenly. The 0.55 m/s^2 value doubles it. The 0.22 m/s^2 value misuses cosine instead of sine when resolving the incline weight. Solving the two blocks separately would give a string tension of $T=m_2(g-a)=4(10-1.11)\approx 35.6\text{ N}$, comfortably between the hanging weight and zero, which validates the system result. This is the standard NCERT incline-pulley setup, where the frictionless surface means the only slope-direction force from the incline block is its gravity component. As a check, the small net 10 N over 9 kg gives a gentle acceleration, consistent with the near-balanced weights.