A smooth wedge of mass M is pushed with an acceleration `a=gtantheta` and a block of mass m is projected down the slant with a velocity v relative to the wedge.
`The horizontal force applied on the wedge is:

Answer

D

Explanation

`(1)/(2)rhov^(2)=rhogh+(Mg)/(A)impliesV=sqrt(2gh+(2Mg)/(rhoA))=sqrt(2xx10xx6+(2xx50xx10)/(10^(3)xx1))=sqrt(120+1)=sqrt(121)=11m//s`