A wedge is moving rightwards on which a block of mass 10 kg is placed on it. Friction coefficient between the eedge and the block is 0.8. Select correct alternative (s) among the following options. [take `g=10(m)/(s^(2))`]

Options

A. if wedge is moving with constant velocity then friction acting on block is 64N.
B. if wedge is moving with cosntant velocity then acceleration of block is zero.
C. if wedge is moving with `veca=2(veci)(m)/(s^(2))` then friction acting on block is 44N.
D. if wedge is moving with `veca=10(veci)(m)/(s^(2))` then friction is 20N. Downward on the block along the inclined.

Answer

B, C, D

Explanation

(A,B) if moving with constant velocity then `a=0`
so friction available `=mumgcostheta`
`=(0.8)(10)(10)((4)/(5))=64N`.
But `mgsintheta=60N`
So required friction is 60 N.
So net force is zero.
`(C)a=2hat(i)f=mgsin37^(@)-macos37^(@)=44N`
`(D)f=mgsin37^(@)-macos37^(@)=-20N`