Question

In the figure shown, the system is released from rest. Find the velocity of block $A$ when block $B$ has fallen a distance ${}^{\prime }{l}^{\prime }$. Assume all pulleys to be massless and frictionless.

• A. $2\sqrt{gl}$ m/s
• B. $\sqrt{gl}$ m/s
• C. $\sqrt{5gl}$ m/s
• D. None of these

Answer 1

The correct option is A $2\sqrt{gl}$ m/s

Let the tension in the string attached with block B be T.

Force balance for block B:

$mg=T$

Therefore,

The tension in the string attached to the block A = 2T

Force balance for block A when it moves down with acceleration a;

$2T-mg=ma$

$a=\frac{2T-mg}{m}=g$

Thus, the acceleration of the block A will, be $g$

As the block is falling from the rest;$u=0$

By the question;

When the block B moves a distance $l$, block A will move distance $2l$.

So, $s=2l$

From the second eqn of motion:

$v^2=u^2+2as$

$v=2g\times 2l$

$v=2\sqrt{gl}m/s$

Thus, when the block A move a distance $l$, the velocity of block B will be $2\sqrt{gl}m/s^2$