Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg.
Neglect air resistance throughout.
(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 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//s2`,
Net force, F = ma `=0.1xx9.8=0.98 N`
(b) When the train is running at a constant velocity its acc. = 0. No force acts on the stone due to this motion.
Therefore, force on the stone F = weight of stone
`= mg = 0.1xx9.8=0.98 N`
This force also acts vertically downwards.
(c ) When the train is accelerating with 1 m/s 2, an additional force `F^(1)=ma = 0.1xx1=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 `= 0.1xx9.8=0.9N`, acting vertically downdards.
(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 he normal rection.