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Question -

Givethe 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 constantvelocity of 36 km/h,

(c) just after it is dropped from the windowof a train accelerating with 1 m s–2,

(d) lying on the floor of a train which isaccelerating with 1 m s–2, the stone being at rest relative to thetrain. Neglect air resistance throughout.



Answer -

(a)1 N; vertically downward

Mass of the stone, m = 0.1kg

Acceleration of the stone, a =g = 10 m/s2

Asper Newton’s second law of motion, the net force acting on the stone,

F = ma = mg

=0.1 × 10 = 1 N

Accelerationdue to gravity always acts in the downward direction.

(b)1 N; vertically downward

Thetrain is moving with a constant velocity. Hence, its acceleration is zero inthe direction of its motion, i.e., in the horizontal direction. Hence, no forceis acting on the stone in the horizontal direction.

Thenet force acting on the stone is because of acceleration due to gravity and italways acts vertically downward. The magnitude of this force is 1 N.

(c)1 N; vertically downward

It is given that the train is acceleratingat the rate of 1 m/s2.

Therefore, the net force acting on thestone, F‘ = ma = 0.1 × 1 = 0.1 N

This force is acting in the horizontaldirection. 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 ona body at an instant depends on the situation at that instant and not onearlier situations.

Therefore,the net force acting on the stone is given only by acceleration due to gravity.

F = mg = 1 N

Thisforce acts vertically downward.

(d)0.1 N; in the direction of motion of thetrain

Theweight of the stone is balanced by the normal reaction of the floor. The onlyacceleration is provided by the horizontal motion of the train.

Acceleration of the train, a =0.1 m/s2

Thenet force acting on the stone will be in the direction of motion of the train.Its magnitude is given by:

F = ma

=0.1 × 1 = 0.1 N

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