A body of mass `m` is placed over a smooth inclined plane of inclination theta. Which iis placed over a lift which is moving up with an acceleration `a_(0)` . Base length o f the inclined plane is `L` . Calculate the velocity of the block with respect to lift at the bottom, if it is allowed to slide down from the top o fthe plane from rest.

Options

A. `sqrt(2(a_(0)+g)Lsintheta)`
B. `sqrt(2(a_(0)+g)Lcostheta)`
C. `sqrt(2(a_(0)+g)Ltantheta)`
D. `sqrt(2(a_(0)+g)Lcottheta)`

Answer

C

Explanation

Acceleration along the plane with respect to lift is,
`a=(a_(0)+g)sintheta`
Initial velocity=0
`v^(2)=u^(2)+2as`
`v=sqrt(2(a_(0)+g)sintheta(l)/(costheta))` .