Question

For the system shown in figure, all the surfaces are smooth. What should be the force F to be applied so that block A has acceleration of 8 m/s²? (strings and pulleys are light)

• A. 4 N
• B. 8 N
• C. 16 N
• D. 32 N

Answer 1

Answer: (B)

8 N

Answer 2

Let $T$ be the tension in the string. Since the force $F$ is applied to the string, we have $F = T$.

For block A, the pulley attached to it is a movable pulley. The string wraps around this pulley, and there are two segments of the string pulling block A to the right. Therefore, the net force exerted by the string on block A is $2T$.

The equation of motion for block A is given by Newton's second law: $F_{net, A} = m_A a_A$ $2T = m_A a_A$

We are given: Mass of block A, $m_A = 2$ kg Acceleration of block A, $a_A = 8$ m/s²

Substituting these values into the equation: $2T = (2 \text{ kg}) \times (8 \text{ m/s}^2)$ $2T = 16 \text{ N}$

Solving for the tension $T$: $T = \frac{16 \text{ N}}{2}$ $T = 8 \text{ N}$

Since $F = T$, the force $F$ that needs to be applied is 8 N.

For block B, the string is attached to the wall, goes around the pulley on B, and then to the pulley on A. The string segment from the wall pulls on B to the right with tension $T$. The string segment from pulley B to pulley A pulls on B to the left with tension $T$. Thus, the net force on block B is $T - T = 0$, which means block B has zero acceleration ($a_B = 0$). This information is not required to solve for $F$ but confirms the consistency of the system's analysis.