Question

the system is released from rest find the velocity of block A when block B Has Fallen a distance l assume all pulleys to be mass less and friction less

Answer 1

Check attached file for ans

Answer 2

Answer:

Required velocity is

$2 \sqrt{gl} \: m {s}^{ - 1}$

Explanation:

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:

${v}^{2} = {u}^{2} + 2as \\ v = 2g \times 2l \\ v = 2 \sqrt{gl} \: m {s}^{ - 1}$

Thus, when the block A move a distance l, the velocity of block B will be

$2 \sqrt{gl} \: m {s}^{ - 1}$