Figure shows an ideal pulley block of mass m = 1 kg. resting on a rough ground with friction coefficient `mu = 1.5`. Another block of mass `M = 11 kg` is hanging as shown. When system is released it is found that the magnitude of acceleration of point P on string is a. Find value of 4a in `m//s^(2)` (Use `g = 10 m//s^(2)`)

Answer

13

Explanation

If the point P has an acceleration a upwards then the acceleration of point R will be a downwards.

The point R has an acceleration a downwards so the block will also have an acceleration a downwards.

The point P has an acceleration on a upwards the block as an acceleration a downwards so the acceleration of S will be 3a downwards (because `(vec_(a)_(S)+vec(a)_(P))/(2) = vec(a)_("block")`)
The point Q will also have an acceleration 3a towards right.
`((vec(a)_(S)+vec(a)_(P))/(2)=vec(a)_("block"))`
The F.B.D of 11 kg block

Using FBD of 11 kg block, which will have acceleration a downwards.
`110 - 3T = 11a`...(1)
(in downwards direction)
for 1 kg block, which will have acceleration 3a,
`T - 15 = 3a` (in horizontal direction)
or `3T - 45 = 9a` ....(2)
on adding equation (1) & (2) we get
`20 a = 65`
`rArr 4a = 13 m//s_(2)`