Give the magnitude and direction of the net force acting on a stone of mass $0.1\; kg$,
$(a)$ just after it is dropped from the window of a stationary train,
$(b)$ just after it is dropped from the window of a train running at a constant velocity of $36\; km/h$,
$(c)$ just after it is dropped from the window of a train accelerating with $1\; m/s^2$,
$(d)$ lying on the floor of a train which is accelerating with $1\; m/s^2$,the stone being at rest relative to the train.
Neglect air resistance throughout.

(A) $1\; N$; vertically downward. Mass of the stone,$m = 0.1\; kg$. Acceleration of the stone,$a = g = 10\; m/s^2$. As per Newton's second law of motion,the net force acting on the stone is $F = ma = mg = 0.1 \times 10 = 1\; N$. Acceleration due to gravity always acts in the downward direction.
$(b)$ $1\; N$; vertically downward. The train is moving with a constant velocity,so its acceleration is zero. Once the stone is dropped,no horizontal force acts on it. The only force acting is gravity,which is $1\; N$ vertically downward.
$(c)$ $1\; N$; vertically downward. Once the stone is dropped,it is no longer in contact with the train. The horizontal force due to the train's acceleration stops acting on the stone. The only force remaining is gravity,which is $1\; N$ vertically downward.
$(d)$ $0.1\; N$; in the direction of motion of the train. The stone is at rest relative to the train,so it shares the train's acceleration $a = 1\; m/s^2$. The net force is $F = ma = 0.1 \times 1 = 0.1\; N$,acting in the direction of the train's motion.