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Question

Physics Question on Newton's Laws of Motion

A wooden block of mass 5kg 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

D

196 N

Answer

291 N

Explanation

Solution

Given:

Total mass = 5kg+25kg=30kg5 \, \text{kg} + 25 \, \text{kg} = 30 \, \text{kg}

Acceleration due to gravity (gg) is taken as 9.8m/s29.8 \, \text{m/s}^2.

The weight of the system is:

W=30×9.8=294NW = 30 \times 9.8 = 294 \, \text{N}

Considering the downward acceleration of the system:

Net force = WN=30×0.1W - N = 30 \times 0.1

Rearranging:

294N=3    N=291N294 - N = 3 \implies N = 291 \, \text{N}