A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of` 0.1 m s^(-2)`. What is the action of the block on the floor (a) before and (b) after the floor yields ? Take g = `10 m s^(-2)` . Identify the action-reaction pairs in the problem.

Explanation

At floor block is statinary its free body diagram is shown in figure
Gravitational on block are and normal force (R ) By Newtons third law
( b) When system (block + cyclinder moves in downward direction with acceleration of gravitational force (25 + 2 = 27 kg )
and normal reaction by floor R
Neglect internal force between cyclinder d and block
`270 - R = 27 xx 0.1`
`R= 270 - 2.7`
` =267 .3 N`

By third law reaction will be 267.3 N in downward direction
Action reaction pairs
For (a ) : (1) the force of gravity (20 N) on the gravity on the earth teh block the figure (ii) the force one the floor by the floor (reaction ) .
for ( b) (i) the force of gravity (270 N) on the gravity on the earth by the system (reaction) the figure ) . (ii) the force on teh floor by the system (action ) . the force on the system by the the block by the cylinder and the force on the cylinder by the block alos constitute and action reaction pair.