Q.A block A of mass $4\ \text{kg}$ rests on a frictionless horizontal table and is connected by a light inextensible string, passing over a frictionless pulley fixed at the edge of the table, to a block B of mass $2\ \text{kg}$ hanging freely. Find

Given: $m_A = 4\ \text{kg}$ (on the frictionless table), $m_B = 2\ \text{kg}$ (hanging), $g = 9.8\ \text{m/s}^2$.

Acceleration. Treating both blocks together as a single system of total mass $m_A + m_B$, driven by the weight of the hanging block $B$ (block $A$'s table is frictionless, so it contributes no resistance), $a = \frac{m_B g}{m_A + m_B} = \frac{2 \times 9.8}{4 + 2} = \frac{19.6}{6} \approx 3.27\ \text{m/s}^2$

Tension. Applying Newton's second law to block $A$ alone (the only horizontal force on it is the string tension, pulling it toward the pulley): $T = m_A a = 4 \times 3.27 \approx 13.07\ \text{N}$ This can be checked using block $B$: $m_B g - T = m_B a \Rightarrow T = m_B(g - a) = 2 \times (9.8 - 3.27) = 2 \times 6.53 \approx 13.07\ \text{N}$, in agreement.

Find the common acceleration by treating both blocks as one system driven by the hanging block's weight, $a = m_B g/(m_A+m_B)$, then find tension by applying Newton's second law to either block alone and checking the two agree.

• Using the full weight of the hanging block, $m_B g$, as the tension itself, rather than solving the connected-system equations.
• Forgetting that block $A$'s table is frictionless, and including a friction term that is not present.