Given the magnitude and direction of the 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//hr`
(c) just after it is droped from the window of a train accelerating with `1 ms^(-2)`
(d) lying on the floor of a trin which is accelerating with `1 ms^(-2)` the stone being at rest relative to the train .
Neglect the resistance of air throuhout .

Explanation

(a) Here ,`m = 0 .1 kg , a = + g = 10 m//s^(2)`
Net force `F = ma = 0. 1 xx 10 = 1.0 N`
This force acts vertically downwards .
(b) When the train is runing at a constant velocity its acc = 0 No force acts on the stone due to this motion Therefore force on the stone F = mg ` = 0.1 xx 10 = 1.0 N` This force also acts vertically downwards
(c) When the train is accelerating with `1 ms^(-2)` an additional Force `F = ma = 0.1 xx 1 = 0. 1` N acts on the stone in the horizontal direction But once the stone is dropped from tha train F becomes Zero and the net force on the stone is `F = mg = 0.1 xx 10 = 0.1 N` acting vertically downwards
(d) As the stone is lying on the floor of the train its acceleration is same as that of the train `:.` force acting on stone ,`F = ma = 0.1 xx 1 = 0.1 N`
This force is along the horizontal direction of motion of the train Note that weight of the stone in this case is being balanced by the normal reaction.