Question

A wooden block of mass 2Kg rests on a soft horizontal floor. When an iron cylinder of mass 25Kg 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 ms-2

. What is the action of the block on the floor (i) before and (ii) after the floor yields?

(Take g = 10 ms-2

)​

Answer 1

★Given:-

• A wooden block rests on a soft horizontal floor.
• Mass of wooden block = 2kg.
• Mass of cylinder= 25kg.
• When an iron cylinder  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

★To find:-

• The action of the block on the floor ,(i) before and (ii) after the floor yields?.

★Solution:-

Here, the block and the cylinder together go down.

So,the action is the normal force.

We know:

✦N=mg

(i)Before the floor yields:-

Before the floor yields,the mass = mass of the block.

=> M = 2kg

Therefore,

=> N=mg

=> N = 2(10) = 20N.

Hence, the action of the block on the floor before  it yields is 20N.

--------------------------

(ii)After the floor yields:-

Given that,the floor yields steadily and the block and the cylinder together go down with an  acceleration of 0.1 m/s².

So,

Mass = Mass of block +cylinder

         = 2+25

          = 27kg.

We can observe that in a system of 27kg going down,

Normal force (N) - Acting upward

Acceleration - Acting downward

Mg - Acting downward

As normal force is acting upward,it has -ve sign.

We know:

✦Force = ma

So,

⇒ Mg - N = ma

⇒ 27(10) - N = 27(0.1)

⇒ 270 - N = 2.7

⇒ N = 270 - 2.7

⇒ N = 267.3N.

Hence,the action of the block after the floor yields is 267.3N.

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