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-1,
(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) Mass of stone = 0.1 kg
Net force, F = mg = 0.1 x 10 = 1.0 N. (vertically downwards).
(b) When the train is running at a constant velocity, its acceleration is zero. No force acts on the stone due to this motion. Therefore, the force on the stone is the same (1.0 N.).
(c) The stone will experience an additional force F’ (along horizontal) i.e.,F = ma = 0.1 x 1= 0.1 N
As the stone is dropped, the force F’ no longer acts and the net force acting on the stone F = mg = 0.1 x 10 = 1.0 N. (vertically downwards).
(d) As the stone is lying on the floor of the train, its acceleration is the same as that of the train.
.So, force acting on the stone, F = ma = 0.1 x 1 = 0.1 N.
It acts along the direction of motion of the train.

 

Question 5.4: