(a) Here, m = 0.1 kg, a = + g = 10 m s$^{–2}$ Net force, F = ma = 0.1 × 10 = 1.0 N This force acts vertically downwards. (b) When the train is running at a constant velocity, its acceleration = 0. No force acts on the stone due to this motion. Therefore, force on thestone F = weight of stone = mg = 0.1 × 10 = 1.0 N This force also acts vertically downwards. When the train is accelerating with 1 m s$^{–2}$, an additional forceF′ = ma = 0.1 × 1 = 0.1 N acts on the stone in the horizontal direction.But once the stone is dropped from the train, F′ becomes zero and the netforce on the stone isF = mg = 0.1 × 10 = 1.0 N, acting vertically downwards.(d) As the stone is lying on the floor of the train, its acceleration is sameas that of the train.∴ Force acting on stone, F = ma = 0.1 × 1 = 0.1 NThis force is along the horizontal direction of motion of the train. Note thatweight of the stone in this case is being balanced by the normal reaction.