Is the energy formula 'force multiplied by displacement' ($E=W=Fs$) correct? Why? Is there a reason that the definition of energy is work, and work is defined as force multiplied by displacement ($E=W=Fs$)? It seems to me a strange formula.
'Momentum' is what puts a motionless mass into motion; momentum is mass multiplied by velocity ($p=mv$). Also the definition of force in physics is the change of momentum per unit time; force is the 'time derivative' of momentum ($F=Δp/Δt$). So why the definition of energy should be 'force multiplied by displacement' and not for example momentum?
For example, if the definition of energy should be momentum (mass multiplied by velocity $p=mv$), then it means if the velocity of an object becomes two times, it's energy becomes two times, but the current kinetic energy formula in physics $EK=(1/2)mv^2$ says that the energy becomes four times, which is different.
Is the energy formula 'force multiplied by displacement' ($E=W=Fs$) correct? Why?
 A: 
Is the energy formula 'force multiplied by displacement' ($==$) correct? Why?

First, $E=W$ is incorrect, it should be $\Delta E=W$.
The formula $W=Fs$ is not wrong, but it is very limited. It applies only for a constant force parallel to the displacement. It should not be considered a definition of work, but rather an example of work or a special case of work.
A better definition for mechanical work is $$W_{mech}=\int \vec F \cdot \vec v \ dt $$ where $\vec v$ is the velocity of the material at the point of application of force $\vec F$. A definition which includes non-mechanical work is $\Delta E= Q + W$ where $Q$ is the heat added to the system and $W$ is the work done on the system (the opposite sign convention of thermodynamics, but the convention that is consistent with the question).

So why the definition of energy should be 'force multiplied by displacement' and not for example momentum?

As I mentioned above, energy is not defined that way. In the most general definition, energy is defined according to Noether’s theorem as the quantity that is conserved because the laws of physics are the same yesterday, today, and tomorrow. Since it is conserved, energy is neither created nor destroyed, only transferred from one system to another. Work is that transfer of energy.
From the definition of energy you can figure out the formula for transferring energy mechanically, and thus obtain the above formula.
A: 
Is there a reason that the definition of energy is work

The definition of energy is not work. The definition of energy is the capacity for doing work. Work is one of the two mechanisms for transferring energy. The other is heat.

'Momentum' is what puts a motionless mass into motion

It is a change in momentum (a net force), not momentum, that puts a motionless mass into motion. Momentum keeps the mass moving.

So why the definition of energy should be 'force multiplied by
displacement' and not for example momentum?

Again, energy is not defined force multiplied by displacement. That defines work, which is energy transfer.

...but the current kinetic energy formula in physics $EK=(1/2)mv^2$
says that the energy becomes four times, which is different.

It's different because momentum does not equal energy.

Is the energy formula 'force multiplied by displacement' ($E=W=Fs$)
correct? Why?

No, because force multiplied by displacement equals work and work is the transfer of energy, not the energy itself. Work results in a change in energy. Per the work energy theorem the net work done on an object equals its change in kinetic energy.
Hope this helps.
A: Starting from just Newton’s Laws, you could develop the ability to work out the dynamics of some simple systems (for example, one-dimensional cases where all forces and displacements occur along a single line) without ever needing the concept of energy. You’d notice right away that momentum is conserved, and learn to apply that rule to get answers easily.
However, for even slightly more complicated cases, you would find that to determine anything about the motion, you’d need vector calculus.
The fact that the scalar we call energy is conserved (never created or destroyed, only converted from one form to another or passed from place to place) saves you from the hassle of vector calculus in some of these cases.
A: Ultimately the 'force times displacement' concept arises from the fact that acceleration is the second derivative of position.
Position, velocity, and acceleration form a remarkable trio.
I will use the following commonly used letters:
't' for time
's' for 'situ', the position coordinate
'v' for velocity
'a' for acceleration
$$ v = \frac{ds}{dt} \qquad a =\frac{dv}{dt} $$


For simplicity I wil discuss the case of uniform acceleration.
Generalization to non-uniform acceleration is straightforward, and the case of uniform acceleration is already sufficient to explain what is going on. (That is, to jump directly to non-uniform acceleration would be an instance of premature generalization.)
A uniform acceleration means that for every unit of time the amount of change of velocity is the same.
That is: uniform acceleration means that the velocity increases linear with time.
Since acceleration is the second derivative of position the velocity increases quadratic with distance.
That is: taking a second derivative introduces a quadratic relation. With uniform acceleration: velocity increases linear with time, and increases quadratic with distance.

Demonstrating that quadratic relation mathematically:
(1) and (2) are used to get from (3) to (7)
$$ ds = v \ dt  \qquad (1)  $$
$$ dv = a \ dt  \qquad (2)   $$

The integral for acceleration from a starting point $s_0$ to a final point $s$
$$ \int_{s_0}^s a \ ds \qquad (3)  $$
Use (1) to change the differential from $ds$ to $dt$. Since the differential is changed the limits change accordingly.
$$ \int_{t_0}^t a \ v \ dt \qquad (4) $$
Change the order:
$$ \int_{t_0}^t v \ a \ dt  \qquad (5) $$
Change of differential according to (2), with corresponding change of limits.
$$ \int_{v_0}^v v \ dv  \qquad (6)  $$
So we have:
$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2  \qquad (7) $$
(7) is purely a mathematical statement. (7) follows logically from these properties:
-velocity is the time derivative of position
-acceleration is the time derivative of velocity

In order to obtain a uniform acceleration we need a force with the following somewhat rare property: the amount of acceleration that it causes must be independent from the existing velocity of the object that is being acted upon.
(If the acceleration is provided by yourself, let's say you are pushing some sort of cart, then the acceleration will level off quickly, you can run only so fast.)
Gravity is remarkable in the following sense: it seems you cannot "saturate" gravity. No matter how fast an object is already falling, gravity just keeps increasing the velocity. That is: the acceleration from gravity is independent from the existing velocity.
We multiply both sides of (7) with a factor $m$ for 'mass'.
$$ \int_{s_0}^s m a \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2  \qquad (8) $$
And then we substitute $m a$ with $F$
$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2  \qquad (9) $$
So:
By just evaluating the product of force and displacement we get an expression in terms of $\tfrac{1}{2}mv^2$, an expression that we already know from collision experiments. In perfectly elastic collisions the quantity $\tfrac{1}{2}mv^2$ is conserved.
If we choose to regard the quantity $\tfrac{1}{2}mv^2$ as a meaningful quantity, then the relation described by (9) transfers that meaning to the quantity 'force times displacement'.

It is necessary to be cautious with (9). That relation is applicable only when the force that is involved has the property that the acceleration that it causes is independent from the already existing velocity of the object that it is acting upon.
A: Is the energy formula force multiplied by displacement   correct?

which work is needed to move the mass along the path from $~s_0~$ to $~s_1$
starting with Newton’s second law
$$\frac {d}{dt}\mathbf p =\mathbf F\quad\text{or}\\
\mathbf p\cdot \frac {d}{dt}\mathbf p =\mathbf p\cdot\mathbf F\\
\int \mathbf p\cdot d\mathbf p=\int \mathbf F\cdot\mathbf p\,dt$$
hence
$$\frac 12 \mathbf p\cdot \mathbf p= \int \mathbf F\cdot\mathbf p\,dt$$
with $~\mathbf p=m\,\mathbf v~$ and if m is constant you obtain
$$\frac{m^2}{2} \mathbf v\cdot \mathbf v= m\,\int \mathbf F\cdot\mathbf v\,dt\\
\underbrace{\frac{m}{2} \mathbf v\cdot \mathbf v}_{\text{KE}}= \int \mathbf F\cdot\,\frac{d\mathbf R(s)}{ds}\,\frac{ds}{dt} \,dt=\underbrace{\int\,\mathbf F(s)\cdot \frac{d\mathbf R(s)}{ds}\, ds}_{\text{Work}}$$
hence
the kinetic energy is equal work only if the mass m is constant and is equal "force multiplied by displacement $(~W=\mathbf F\,s~)~$"   if the force $~F~$ is constant and the path is straight line $~( \frac{d\mathbf R(s)}{ds}=1~)$
A: All answers before, have stated the formula and it's relation to kinetic energy. That isn't what OP is asking. Newton assumed energy was proportional to momentum. The key experiment in understanding why kinetic energy/work it is defined to be this was émilie du châtelets sand experiment, Where she related the penetration depth of balls dropped into in sand to the velocity of balls, She noticed that the penetration depth was proportional to Mv^2 . which meant that the total energy isn't proportional to MV like Newton suggested.  this was the experiment that led to the concept of kinetic energy, and the equivalent concept of work
