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) We have here,
Mass, m = 0.1 kg
Acceleration due to gravity, a = +g = 10 m/s2
Net force, F = ma = 0.1 × 10 = 1.0 N, acting in vertically downward direction.
b) When the train is running at a constant velocity, its acceleration = 0.
Due to this motion, no force is acting on the stone.
Force on the stone, F = weight of stone = mg
= 0.1 × 10 = 1.0 N
This force acts in the vertically downward direction.
c) Acceleration of the train, a = 1 m s-2
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.
Net force on the stone, F = mg = 0.1 × 10
= 1.0 N, acting vertically downwards.
d) When the stone is lying on the train, its acceleration is same as that of the train.
Therefore, the force acting on stone, F = ma
= 0.1 × 1 = 0.1 N
This force is along the horizontal direction of motion of the train.
In each case, the weight of the stone is being balanced by the normal reaction.