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A small block of mass m is released from...

A small block of mass `m` is released from a fixed smooth wedge in figure. Initial point is marked as `A`. Bottom of wedge is marked as `B` and at a point `C` the block stops because the straight part of floor is rough.

The velocity of the block at the midpoint between `B` to `C` will be `:`

A

`(sqrt(2gh))/(2)`

B

`sqrt(2gh)`

C

`sqrt(gh)`

D

`(sqrt(gh))/(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`(` work done by gravity `-` work done by friction `) =` change in `K.E.`
`:. Mgh - mu mg`` (x_(0))/(2)=(1)/(2)mv_(1)^(2)-(1)/(2)mv_(1)^(2)`
`=mgh-(h)/(x_(0))mg``(x_(0))/(2)=(1)/(2) mv_(1)^(2)-0 :' mu=(h)/(x_(0))`
`:. (mgh)/(2)=(1)/(2)mv_(1)^(2)rArr v_(f)=sqrt(gh)`
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