Question

A block A of mass 7 kg is placed on a frictionless table A thread tied to it passes over a frictionless pulley and carries a body B of mass 3 kg at the other end. The acceleration of the system is (given $g=10m{s}^{-2}$)

• A. $100m{s}^{-2}$
• B. $3m{s}^{-2}$
• C. $10m{s}^{-2}$
• D. $30m{s}^{-2}$

Answer 1

The correct option is B

$3m{s}^{-2}$

Step1: Analyzing the given diagram

1. Let a be the system's acceleration, and T be the tension force acting on the sting.
2. Then block B will begin to move in a downward direction, while block A will begin to move in a forward direction.
3. As illustrated in the diagram. Since tension is the force acting on the string, the forces on block B are $m_{B}g,m_{B}a$, and the forces on block A are $m_{A}a$.

Step2: Given data

Mass of block A, $m_A=7kg$

Mass of block B, $m_B=3kg$

$g=10m{s}^{-2}$

Step3: Formula used

$T=ma(whereT=tension,m=mass,a=acceleration)$

Step4: Calculating the tension

1. For block A, the tension force is the only force that allows block A to move forward
2. So $T=m_{A}a.........\left(i\right)$ is the result of balancing both forces.
3. For block B, we can see that the weight and motion forces $m_{B}g,m_{B}a$ are acting in a downward direction, which is balanced by tension T acting on the string, which can be written as $m_{B}a=m_{B}g-T.......\left(ii\right)$

Step5: Calculating acceleration

From equation (i) put the value of $T=m_{A}a$ we get,

$$\begin{gathered} (m_B+m_A)a=m_{B}g \\ (3+7)a=3\times 10 \\ 10a=30 \\ a=3m{s}^{-2} \end{gathered}$$

Hence, option B is the correct answer.