In the figure shown, the system is released frrom rest. Find the velocity of block `A` when block `B` has fallen a distance `l`. Assume all pulleys to be massless and frictionless.

Answer

A

Explanation

Evidently , when the block `B` descends by a distance `l`. The pulley `P` , descends by a distance `(l)/(2)` , and consequently , the block `A` , (up). `2v`, would be the speed of block `B` (down).
Net loss in `PE` of the system
= Loss in PE of block `B` - gain in P.E., of block `A`
`= mgl - mg (l)/(2) = mg (l)/(2)`
Total gain in `K.E. = (1)/(2) mv^(2) + (1)/(2) m (2v)^(2) = (5)/(2) mv^(2)`
Now, law of conservation of energy demands
Loss in P.E. = Gain in K.E.
`mg (l)/(2) = (5)/(2) mv^(2) implies v^(3) = (gl)/(5)`
`v = sqrt((gl)/(5))`.