Question

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/ s2 . What is the action of the block on the floor 9. (a) before (b) after the floor yields ? g = 10 m/s2.

Answer 1

This is a question on Newtons law of motion.

We solve the question as follows:

a) Before the floor yields

Before the floor yields the forces acting on the block and the block are:

1) Gravitational force of the block on the floor.

2) The normal force of the floor on the block.

So the action is the gravitational force which is given by :

2 × 10 = 20 N

b) After the floor yields

The system which is made up of the block and the cylinder accelerates downwards at 0.1m/s²

We also have two forces here that is the normal force and the gravitational force.

Now:

Gravitational force = (25 + 2) × 10 = 270 N

Force due to downward acceleration is :

0.1 × 27 = 2.7N

The action is thus given by :

270 - 2.7 = 267.3N

= 267.3 N