A wooden block of mass 5 kg rests on a 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 m s⁻². The action force of the system on the floor is equal to
[Take g=9.8 m s⁻²]

Question

Accepted Answer

(2) 291 N. Idea: the system accelerates downwards, so the floor pushes up with less than the full weight, and by Newton's third law the system presses on the floor with that same reduced force. Total mass =5+25=30 kg, moving down with a=0.1 m s⁻². mg-N=ma, so N=m(g-a). N=30(9.8-0.1)=30×9.7=291 N. The action force on the floor equals this reaction, 291 N.

Step-by-step solution

Why the other options are wrong