Question

Figure shows three blocks that are being pulled along a smooth horizontal surface. The masses of the blocks are shown in figure and tension force in cord connecting A and B = 120 N. The pulling force F is:

• A. 50 N
• B. 100 N
• C. 125 N
• D. 200 N

Answer 1

Answer: (D)

200 N

Answer 2

The tension $T_{AB}$ (120 N) accelerates the combined mass of blocks B and C ($4 \text{ kg} + 2 \text{ kg} = 6 \text{ kg}$), giving an acceleration $a = 120/6 = 20 \text{ m/s}^2$. This acceleration is also imparted to block A. The pulling force $F$ must overcome the inertia of block A and also provide the tension $T_{AB}$ to pull block B. Thus, $F = m_A a + T_{AB} = (4 \text{ kg})(20 \text{ m/s}^2) + 120 \text{ N} = 80 \text{ N} + 120 \text{ N} = 200 \text{ N}$. Alternatively, the force $F$ accelerates the total mass of all three blocks ($4+4+2=10$ kg) at $a=20 \text{ m/s}^2$, so $F = (10 \text{ kg})(20 \text{ m/s}^2) = 200 \text{ N}$.