Question

In the shown figure.

`{:(,"Column I",,"Column II",,),((A),"if "mu_(1) = 0"," mu = 0", then for block m",(P),"Work dine by normal force between the wedge and block is always positive in ground frame.",,),((B),"if" mu_(1) = 0"," mu_(2) = 0", then for wedge M.",(Q),"Work done by normal force between the wedge and block is always negative or zero in ground frame.",,),((C ),"if " mu_(1) != 0"," mu_(1) = 0", then for block m.",(R ),"Work done by constact force between the wedge and block is always negative in ground frame.",,),(,,(S),"Work done by friction force betwen the wedge and ground is always negative in ground frame.",,):}`

Answer 1

Correct Answer - (A) = QR, (B) = P, (C ) = Q

(A) Wedge slides towards right. So work of N is `+ve` on wedge.
As total work of N on system is 0, so `W_(N)` on block is `-ve`.
(C ) Case (1) wedge does not slide on ground `W_(N) = 0`
Case (2) Wedge slides on ground towards right.
So `W_(N)` on wedge is `+ve`
Then `W_(N)` on block has to be `-ve`