Question

The arrangement of pulley-block system is released from rest at $t=0$ as shown in figure. The velocity of block $B$ is $v_B=45\text{m/s}$ upwards at some instant. Then, find the magnitude of velocity of block $A$ in $\text{m/s}$. Consider the string to be massless and inextensible, all pulleys are light and smooth.

• A. $9\text{m/s}$
• B. $22.5\text{m/s}$
• C. $1.5\text{m/s}$
• D. $3\text{m/s}$

Answer 1

The correct option is D $3\text{m/s}$

Since tension will be an internal force for the system of (blocks+pulleys+strings), the sum of work done by all internal forces on the system will be zero.
$\sum W_{\text{int}}=\mathrm{0...}\left(i\right)$

Let the displacement of block $B$ be $x_B$ upwards and displacement of block $A$ be $x_A$ downwards at an instant. Distributing tension in the strings as shown below.


For block $A$, work done by net tension force $15T$ will be $-ve$ due to velocity and displacement being in opposite directions.
Simillarly for block $B$, work done by net tension force $T$ will be $+ve$
$\sum W_{\text{int}}=0$
$\Rightarrow (-15T\times x_A)+(T\times x_B)=0$
$\Rightarrow x_B=15x_A..\left(ii\right)$

Differentiating Eq. $\left(ii\right)$ w.r.t time, we get the relation between velocities of the blocks.
$v_B=15v_A$
$\therefore v_A=\frac{v_B}{15}=\frac{45}{15}=3\text{m/s}$ (downward)