With what acceleration 'a' should the lift descend so that the block of mass M exerts a force Mg4 on the floor of the box?
In lift's frame, the lift is moving down with acceleration a. this means the pseudo force on the block will be Ma in upward direction.
Free Body Diagram of block
If normal by block on box is Mg4 then its reaction will also be Mg4
⇒ Ma+N=Mg
Ma+Mg4=Mg
Ma=3Mg4
a=3g4
Ground Frame
The block is at rest with respect to the box which is accelerated with respect to the ground. Hence, the acceleration of the block with respect to the ground is 'a' downward. The forces on the block are
(i) Mg downward (by the earth) and
(ii) N upward (by the floor)
The equation of motion of the block is, therefore
Mg−N=Ma
If N=Mg4, the above equation gives a = 3g4. The block and hence the box should descend with an acceleration 3g4.