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Figure shows an ideal pulley block of ma...

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)`)

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The correct Answer is:
13

Sol.
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 a upwards, the block has 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.
The `F.B.D.` of `11kg` block
The `F.B.D.` of `1kg` block

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