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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 with1 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 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 with1 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.
CBSE | Class 11 | Laws of Motion | NCERT | Physics
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 with1 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) Stone mass = 0.1 kg

10 ms^(-2)= acceleration

 

F = mg = 0.1 x 10 = 1.0 N is the net force.

 

The force is applied vertically and downwards.

 

(b) The train maintains a steady speed. As a result, the acceleration will be zero. So there is no force acting on the stone due to the motion of the train. Therefore, the force acting on the stone will remain the same (1.0 N)

 

(c) When the train accelerates at 1m/s2, the stone experiences an additional force of F’ = ma = 0.1 x 1 = 0.1 N. The force acts in the horizontal direction.

 

But as the stone is dropped, the force F’ no longer acts and the net force acting on the stone F = mg = 0.1 x 10 = 1.0 N. (vertically downwards).

 

(d) As the stone

i

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