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 ms⁻². The action force of the system on the floor is equal to :

Solution

Physics sir · 3 steps

The system is sinking with some acceleration, so the floor pushes up less than the full weight.

1. One system. The block and the cylinder move down together, so treat them as one body of mass 5 + 25 = 30 kg. Two forces act on it: the weight mg downward and the floor's normal force N upward.
2. Second law. Take downward as positive, since the acceleration a = 0.1 m s⁻² points down. The net force mg - N must equal ma.

   N = m(g - a) = 30 × (9.8 - 0.1) = 30 × 9.7 = 291 N

3. So. By Newton's third law, the system presses on the floor with a force equal to N. The action force is 291 N, option (b).

The idea to keep. Contact force on an accelerating body is m(g ± a). Accelerating down means less than mg, accelerating up means more. The floor force and the force on the floor are an action-reaction pair.

Answer from the NTA final answer key. Solution by PrepWiser.