Give the magnitude and direction of the net ating on a stone of mass 0.1 kg .
(a) just after it is dropped from the window of stationary train,
(b) just after it is dripped from the window of a train running at a constant velocity of 36 km/h
(c ) just after it is dropped frrom the window of a train accelerating with `1 ms^(-2)`
(d) lying on the floor of a train which is accelerating with `1 ms^(-2)` , the stone being at rest relative to the train .

Explanation

(a) Here , m = 0.1 kg , `a = + g = 9.8 m//s^(2)`
Net force , `F = ma = 0.1 xx 9.8 = 0.98 N `
This force acts vertically downwards
(b) When the train is running at a constant velocity its acc. = 0 .No force acts on the stone due to this motion .
(c ) When the train is accelerating with `1 m//s^(2)` an additional force `F^(1) = ma = 0.1 xx1 = 0.1 N ` acts on the stone in the horizontal direction . But once the stone is dropped from the train `F^(1)` becomes zero and the net force on the stone is `F = mg = 0.1 xx 9.8 = 0.98 N ` ,a cting vertically downwards .
(d) As the stone is lying on the horizontal direction of motion of the train .Note the weight of the stone in this case is being balanced by the normal reaction .