A steel block of 10 kg rests on a horizontal floor as shown. When three iron cylinders are placed on it as shown, the block and cylinders go down with an acceleration 0.2 m/s2. The normal reaction R' by the floor if mass of the iron cylinders are equal and of 20 kg each, is \underline\hspace2cm N. [Take g = 10 m/s2 and \mu s = 0.2]

Laws of Motion PYQ — JEE Main 2021 · 20 Jul · Shift 1 · Q55 with Solution

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Correct answer: A

A $10\,\text{kg}$ steel block is on a horizontal floor. Three iron cylinders, each of mass $20\,\text{kg}$ , are placed on it.

So, total mass on the floor is

M = 10 + 3\times 20 = 70\,\text{kg}.

The system moves downward with acceleration

a = 0.2\,\text{m/s}^2.

Since the problem asks for the normal reaction $R'$ by the floor, we need the total downward force transmitted to the floor.

If everything were at rest, the floor reaction would be

R = Mg = 70\times 10 = 700\,\text{N}.

But the cylinders and block are accelerating downward with $0.2\,\text{m/s}^2$ , so the apparent load on the floor is reduced.

For the whole set of masses resting on the floor and moving downward with acceleration $a$ , the normal reaction is

R' = M(g-a).

R' = 70(10-0.2) = 70\times 9.8 = 686\,\text{N}.

The obtained value is

686\,\text{N},

which matches Option A.

Stored correct answer: A

Our derived answer: A

So they agree.