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  • 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.

Give the magnitude and direction of the | Class 11 Physics Chapter Laws of Motion, Laws of Motion NCERT Solutions

Q3. 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)1 N; vertically downward

Mass of the stone, m = 0.1 kg

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

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

F = ma = mg

= 0.1 × 10 = 1 N

Acceleration due to gravity always acts in the downward direction.

(b)1 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 N.

(c)1 N; vertically downward

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

Therefore, the net force acting on the stone, F' = ma = 0.1 × 1 = 0.1 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.

F = mg = 1 N

This force acts vertically downward.

(d)0.1 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 m/s2

The net 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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What is the correct answer to: 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)1 N; vertically downward

Mass of the stone, m = 0.1 kg

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

As per Newton’s second law of motion, the net force acting on the stone...

How do you solve 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. step by step?

Step-by-step explanation:
• (a)1 N; vertically downward

• Mass of the stone, m = 0
• 1 kg

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Student Discussion

ARNAB BHANJA DEO
Class · · , · Aug 20, 2020
Thanks a lot for your help
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Vakyanshsrivastav
Class · · , · Aug 24, 2018
mast hai
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Ambika
Class · · , · Jul 30, 2018
It was of great help
Thank you
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Ansif
Class · · , · Feb 26, 2018
Very helpful work
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bhh
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anami
Class · · , · Jan 29, 2018
good work
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anami
Class · · , · Jan 29, 2018
great answers
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anami
Class · · , · Jan 29, 2018
good job
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anami
Class · · , · Jan 29, 2018
good work keep it up
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Faizan
Class · · , · Sep 22, 2016
Its was of great help .keep up the goodwork
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