Question

wooden block of mass 2 kg rests on a soft horizontal floor when an iron cylinder of mass 25 Kg is placed on the top of the block the floor yeilds steadily and the block and the cylinder together go down with an acceleration of 0.1 M per second square what is the action of the block on the floor before and after the floor yeilds

Answer 1

Force = mass × acceleration

Before the floor yields :

Mass = 2 kg + 25 kg = 27kg

g = 9.8

Action force = 9.8 × 27 = 264.6 N

After the floor yields.

Mass = 27kg

a = 0.1

F = 0.1 × 27 = 2.7 N

Answer 2

The total force acting on the block are, therefore the total will be 2x 10 N which is equal to 20N.

Now the acceleration is 0.1 m per second square. The force is 270N.

So the equation goes like this 270-R=27x0.1

267.3N this force will act in vertical downward direction.