Friction And Connected Blocks

Treat the two blocks as one system sharing a single inextensible string, so they move with a common acceleration. The hanging 2 kg block's weight $m_{2}g$ drives the motion, while kinetic friction $\mu m_{1}g$ on the 4 kg table block opposes it. Applying Newton's Second Law to the system, $m_{2}g-\mu m_{1}g=(m_1+m_2)a$. Substituting, $2(10)-0.25\times 4\times 10=(4+2)a$, giving $20-10=6a$, so $a=10/6=1.67{\text{ m/s}}^2$. The 3.33 m/s^2 value forgets friction and uses only the hanging weight. The 2.5 m/s^2 value omits the table block from the total mass. The 1.25 m/s^2 value overestimates the friction by misusing the normal force. Treating the blocks as a system conveniently sidesteps the internal tension, but one could equally solve the two Newton equations separately and find $T=m_2(g-a)=2(10-1.67)\approx 16.7\text{ N}$, a value between zero and the hanging weight as expected. This is the NCERT horizontal-pulley-with-friction problem, where the normal force on the table block is simply its own weight $m_{1}g$ since the surface is horizontal. As a check, the result is positive yet below the frictionless 3.33 m/s^2, confirming that friction slows but does not halt the descent of the hanging mass.