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O is a point at the bottom of a rough pl...

`O` is a point at the bottom of a rough plane inclined at an angle `alpha` to the horizontal. Coefficient of friction between `AB` is `(Tanalpha)/(2)` and between `BO` is `(3Tanalpha)/(2)` . `B` is the mid–point of `AO`. A block is released from rest at A. Then identify which graphs are correct during motion of block from point `A` to `O` taking direction down the incline plane as positive `(sinalpha=1//5)` :

A

B

C

D

Text Solution

Verified by Experts

The correct Answer is:
C

For motion between `AB`
`ma=mg sin alpha-(tanalpha)/(2) mg cos alpha`
`a=(g sinalpha)/(2)` (downward)
For motion between `BO`
`ma=(3tan alpha)/(2) mg cos alpha-mg sin alpha`
`a=(gsinalpha)/(2)` (upward)
The velocity increases from zero to maximum value of `B` and then starts decreasing with same rate and finally becomes zero at `O`.
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