### Q.3:- **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.

**Answer:-**

(a) Here, *m* = 0.1 Kg, a = + g = 10 m/s^{2}

Net force, *F* = *ma* = 0.1 × 10 = 1.0 N

This forcer acts vertically downwards.

(b) When the train is running at a constant velocity, its acceleration = 0, No force acts on the stone due to this motion. Therefore, force on the stone *F* = weight of stone = *mg* = 0.1 × 10 = 1.0 N

This force also acts vertically downwards.

(c) When the train is accelerating with 1 m s^{-2}, an additional force *F’* = *ma* = 0.1 × 1 = 0.1 N acts on the stone in the horizontal direction. But once the stone is dropped from the train, *F’* becomes zero and the net force on the stone is *F* = *mg* = 0.1 × 10 = 1.0 N, acting vertically downwards.

(d) As the stone is lying on the floor of the trin, its acceleration is same as that of the train.

∴ force acting on stone, *F* = *ma* = 0.1 × 1 = 0.1 NThis force is along the horizontal direction of motion of the train.

Note that in each case, the weight of the stone is being balanced by the normal reaction.