Question about Kinetic Energy and Force acting on a body Below is the graph of a body that rests on a friction-less flat surface. On this body force F is applied and it's value is being depicted in the graph below. We can see that the body moves from point O to $x_1$, $x_2$ etc.

I want to ask something that i can't understand. Why does kinetic energy increase from point O to $x_1$? If someone could explain i would be very grateful.  
 A: Because of the Work-Energy Theorem. The work done on an object by an external force $\vec{F}$ is
$$W=\vec{F}\cdot \Delta \vec{x},$$
where $\Delta \vec{x}$ is the displacement. The Work-Energy theorem says that the work is equal to the change in kinetic energy:
$$\Delta K=W.$$
So the reason that the kinetic energy is changing for the particle is that $\vec{F}\cdot \Delta \vec{x}$ is positive for the region $(0,x_1)$. Although the force might be getting smaller, it is still positive (if we read that graph as "The force in x-direction" and the object is moving along the x-direction).
A: 
How is KE accumulated and thus increasing? This is my primary concern. Also, what's happening with velocity? Is it increasing? Because if KE is increasing then velocity also should increase, right?

Between $x=0$ and $x=x_1$, a net force, say $F$, is acting on the body. With Newton's Second Law we get:
$$F=ma,$$
with $a=\frac{dv}{dt}$ the acceleration and $v$ the velocity.
So:
$$F=m\frac{dv}{dt}$$
$$dv=\frac{F}{m}dt$$
$$v=\frac1m \int_0^tFdt$$
Here $F$ is not a constant (in fact it's decreasing) in time but it's always positive and that means that up to $x=x_1$, $v$ (and kinetic energy $\frac{mv^2}{2}$) keeps increasing.
