Question

Give the magnitude and direction of the net force acting on a stone of mass 0.1 \(\text {kg}\),

1. just after it is dropped from the window of a stationary train,
2. just after it is dropped from the window of a train running at a constant velocity of 36 \(\text {km}/\text h\),
3. just after it is dropped from the window of a train accelerating with 1 \(\text m \; \text s^{-2}\),
4. lying on the floor of a train which is accelerating with 1 \(\text m \; \text s^{-2}\), the stone being at rest relative to the train.

Neglect air resistance throughout.

Answer 1

(a) 1 \(\text N\); vertically downward
Mass of the stone, \(\text m\) = 0.1 \(\text {kg}\) 
Acceleration of the stone, \(\text a\) = \(\text g\) = 10 \(\text m/ \text s^{2}\) 
As per Newton’s second law of motion, the net force acting on the stone, 
\(\text F\) = \(\text {ma}\) = \(\text {mg}\) = 0.1 × 10 = 1 \(\text N\) 
Acceleration due to gravity always acts in the downward direction. 

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(b) 1 \(\text N\); vertically downward 
The train is moving with a constant velocity. Hence, its acceleration is zero in the direction of its motion, i.e., in the horizontal direction. Hence, no force is acting on the stone in the horizontal direction.
The net force acting on the stone is because of acceleration due to gravity and it always acts vertically downward. The magnitude of this force is 1 \(\text N\).

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(c) 1 \(\text N\); vertically downward
It is given that the train is accelerating at the rate of 1 \(\text m/ \text s^{2}\).
Therefore, the net force acting on the stone, \(\text F'\)' = \(\text {ma}\) = 0.1 × 1 = 0.1 \(\text N\)
This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force F,' stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations.
Therefore, the net force acting on the stone is given only by acceleration due to gravity.
\(\text F\) = \(\text {mg}\) = 1 \(\text N\)
This force acts vertically downward.

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(d) 0.1 \(\text N\); in the direction of motion of the train
The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train.
Acceleration of the train, a = 0.1 \(\text m/ \text s^{2}\)
The net force acting on the stone will be in the direction of motion of the train. Its magnitude is given by:
\(\text F\) = \(\text {ma}\) = 0.1 × 1 = 0.1 \(\text N\)