Acceleration provided by static frictional force Let us assume a box which is kept inside a compartment of a train moving with constant velocity. And also that surface of compartment provides some friction(static) to box.
Let us denote static friction by $f_s$. And let that constant  velocity of train is $v$. Now due to some external force the velocity of train got increased to $v_1$. So we have some acceleration now. Using that we can calculate the force applied on the train.
My questions:-
$(1)$ When the train is moving with constant velocity $v$, what is the impended velocity of block?(i.e the velocity of the block if the static friction was absent)
$(2)$ After the train accelerated, what is the acceleration does it produce on the block?
 A: TL;DR The force is applied to the train, but this force is not directly exerted to the box. Train gets acceleration and starts changing its velocity. The box on the train floor wants to keep its current velocity. Static friction between the box and the train floor opposes relative motion between the two, and as a result the box also gets (some) acceleration. The key to understand this situation is to observe the box and the train from the inertial reference frame. The earth (ground) can serve as a good approximation for inertial reference frame.
In the absence of static/kinetic friction force, as seen by ground observer, the box keeps its (current) velocity while the train accelerates. Below I discuss a more realistic scenario in which there is some friction between the box and the train floor.

Friction forces
There are two types of friction forces, namely static friction force and kinetic friction force. Both are defined via normal force exerted on the object:
$$F_{fs} = \mu_s n, \qquad F_{fk} = \mu_k n$$
where $\mu_s$ and $\mu_k$ are coefficients of static and kinetic friction, respectively, and $n$ is magnitude of the normal force. It should be noted that the coefficient of static friction is given as its maximum value, and in general $\mu_{s,\text{max}} \geq \mu_k$ (see figure below).

Source: H. D. Young, R. A. Freedman, "University Physics with Modern Physics in SI Units", 15th ed., 2019.
The friction forces always act along the surface in the direction which opposes relative motion between the two surfaces:

*

*the static friction force acts when there is no relative movement between the two surfaces;


*the kinetic friction force acts when there is relative movement between the two surfaces, and is smaller in magnitude compared to the static friction force.
In the example of box sitting on the train floor, if we assume the floor is horizontal then the normal force magnitude equals weight of the box:
$$n = mg$$

Box does not slip
The static friction force provides horizontal force component to the box which gives it acceleration. Since the box does not move relative to the train, it means its acceleration $a_\text{box}$ equals that of the train $a_\text{train}$. As seen by a ground observer, the second Newton's law of motion gives
$$m_\text{box} a_\text{box} = \mu_s n_\text{box} \qquad \rightarrow \qquad m_\text{box} a_\text{box} = \mu_s m_\text{box} g$$
From this we can determine the coefficient of static friction that actually acts on the box
$$\mu_s^\star = \frac{a_\text{train}}{g}$$
If $\mu_s^\star$ is larger than maximum coefficient of static friction $\mu_{s,\text{max}}$, the box starts slipping. In other words, the static friction force just cannot provide that much of acceleration to the box.

Box is slipping
When the box starts slipping, its acceleration is different to that of the train $a_\text{box} \neq a_\text{train}$, and there is some relative motion between the two. In this case the kinetic friction force is exerted on the box which provides (some) acceleration
$$m_\text{box} a_\text{box} = \mu_k n_\text{box}$$
From this we can calculate the acceleration of the box as
$$a_\text{box} = \mu_k g$$

Direct answers

When the train is moving with constant velocity $v$, what is the impended velocity of block? (i.e the velocity of the block if the static friction was absent)

When the train is moving at constant velocity, there is no train acceleration. If the box is initially at rest, it remains at rest. If the box is initially moving relative to the train, it keeps moving in the absence of static/kinetic friction force, as there is nothing to provide negative acceleration.

After the train accelerated, what is the acceleration does it produce on the block?

When the train accelerates, there are two possible scenarios for the box:

*

*If the train acceleration is small enough, the static friction force will give the same acceleration to the box such that there is no relative motion between the box and the train;


*If the train acceleration is large enough, the box will start slipping in the train. The box will still get some acceleration, but smaller than the train acceleration.
In the absence of static/kinetic friction force, the box keeps its (current) velocity relative to a ground observer. Hence, it moves relative to the (accelerating) train.
