JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Laws of Motion

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A wooden block of mass 5 kg rests on soft horizontal floor. When an iron cylinder of mass 25 kg is placed on the top of the block, the floor yields and the block and the cylinder together go down with an acceleration of

0.1 ms^{- 2}

. The action force of the system on the floor is equal to:

(A)297 N
(B)294 N
(C)291 N✓ Correct
(D)196 N

Answer: (C)

Solution

Total mass

m = 5 + 25 = 30

kg descends with acceleration

a = 0.1 m s^{- 2}

(downward). Taking

g = 9.8 m s^{- 2}

and applying Newton's second law with the floor's normal reaction

N

upward:

m g - N = ma \Rightarrow N = m (g - a) = 30 (9.8 - 0.1) = 291 N .

By Newton's third law the action force of the system on the floor is

291

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Single correct option

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