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

Concept: Newton’s second law — net force equals mass times acceleration ($F = ma$). In free fall, the only force is gravity; when the stone is at rest relative to an accelerating train, the net force must provide the same acceleration, and the vertical weight is balanced by the vertical normal reaction (it is NOT part of the net force in that case).

(a) Just after being dropped from a stationary train, the stone is in free fall.

Only gravity acts:

$F = mg = 0.1 \times 9.8 = 0.98\ \text{N}$, vertically downward.

(b) The train moves at constant velocity — no horizontal acceleration.

Just after dropping, the stone still has only gravity:

$F = 0.98\ \text{N}$, vertically downward.

(c) The train accelerates at $1\ \text{m/s}^2$ horizontally. Just after dropping, the stone retains the train’s horizontal velocity but no longer experiences the train’s horizontal force — so only gravity acts:

$F = 0.98\ \text{N}$, vertically downward.

(d) The stone is at rest relative to the accelerating train and lies on its floor, so it does not accelerate vertically — its weight $mg$ is exactly balanced by the floor's normal reaction $N$, and this balanced pair contributes nothing to the net force. The only unbalanced force is the horizontal one needed to give the stone the train's horizontal acceleration:

directed horizontally, in the direction of the train's motion (its acceleration).

The key is to apply Newton’s second law: net force = mass × acceleration. Once you identify what acceleration the stone actually has, the force follows directly. Air resistance is neglected, so the only real force in free fall is gravity.

Let’s go case by case.

1. Case (a) — dropped from a stationary train

   Just after release, the stone is in free fall. The only force acting is its weight, vertically downward.

   Weight = $mg = 0.1 \times 9.8 = 0.98\ \text{N}$.

   So the net force is $0.98\ \text{N}$ downward.

2. Case (b) — dropped from a train moving at constant velocity

   Constant velocity means zero acceleration. Just after release, the stone still has the same horizontal velocity as the train (inertia), but no horizontal force acts on it. The only force is still gravity.

   Net force = $0.98\ \text{N}$ downward, exactly as in (a).

   Watch out

   A common mistake is to think the stone “keeps” the train’s force after release. It doesn’t — once released, the train no longer pushes it. The stone’s horizontal motion is due to inertia, not a force.

3. Case (c) — dropped from an accelerating train

   The train accelerates at $1\ \text{m/s}^2$. Just after release, the stone is no longer in contact with the train, so no horizontal force from the train acts on it. Only gravity acts.

   Net force = $0.98\ \text{N}$ downward.

   Tip

   The stone’s horizontal velocity at the instant of release equals the train’s velocity at that instant, but since no horizontal force acts, its horizontal acceleration is zero. The net force is purely vertical.

4. Case (d) — stone lying on the floor of an accelerating train, at rest relative to the train

   Here the stone is in contact with the train and moves with it. The train accelerates at $1\ \text{m/s}^2$, so the stone must also accelerate at $1\ \text{m/s}^2$ horizontally.

   By Newton’s second law, the net horizontal force on the stone is

in the direction of the train’s acceleration.

Vertically, weight and normal reaction cancel (stone doesn’t accelerate vertically), so the net force is purely horizontal.

The stone is at rest relative to the train, so its acceleration equals the train’s acceleration.

Concept: Distinguishing Free Fall from Motion "At Rest Relative to an Accelerating Frame"

Step 1: Cases (a), (b), (c) — the stone is dropped (loses contact with the train)

The instant it is released, the train can no longer exert any force on it. The only force remaining is gravity, regardless of whether the train was stationary, moving at constant velocity, or accelerating at the moment of release (its own horizontal velocity at release is irrelevant to the force on it afterward — only contact forces matter, and contact is gone):

Step 2: Case (d) — the stone stays on the train's floor

Now the stone is in contact with the accelerating train and moves with it, so it shares the train's horizontal acceleration $a = 1\ \text{m s}^{-2}$.

Step 3: Resolve forces on the stone in case (d)

Vertically: weight $mg$ down, normal reaction $N$ up. Since the stone does not accelerate vertically, $N = mg$ — these two cancel and contribute nothing to the net force.

Horizontally: the only unbalanced force is what accelerates the stone with the train, $F_x = ma$.

Step 4: Compute

Final Answer:

(a) 0.98 N downward; (b) 0.98 N downward; (c) 0.98 N downward; (d) 0.1 N horizontal (in the direction of the train's motion) — not a diagonal combination with gravity, since the vertical forces are balanced in case (d).