## 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. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-5 Laws Of Motion

### 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.

(a) Here, m = 0.1 Kg, a = + g = 10 m/s2
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.