(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