Work energy theorem The work energy theorem states that the work done along a path from a point (1) to point (2) is the difference of the kinetic energies between the last and the first points. But if the force was moving the particle in a constant speed, then the work would become zero according to the equation since the kinetic energy is conserved, but intuitively the force did a work because it had transported the particle and changed its position.. I need an explanation for my misunderstanding 
 A: The state of constant motion along a straight path or being at rest defines a state of the system known as inertia. Force is some agency that disturbs the state of inertia. Force causes acceleration (precisely, a change in momentum). Newton's second law of motion states that  
$$\vec{F}=\frac{d\vec{p}}{dt}$$  
If the body is moving at a constant velocity, then there is no change in momentum and so  
$$\frac{d\vec{p}}{dt}=0\implies\vec{F}=0$$  
Hence constant motion means no external unbalanced force is acting on the body. Since there is no force, there is no work done (Since $W=\int\vec{F}\cdot d\vec{s}$).   
Hence no work is necessary to be done on a body to maintain it at a constant speed, because no force is required for a body to move with constant velocity. There is displacement of course. The fact is that the displacement that happens here during a constant motion here is not the consequence of the force. The displacement term appearing in the equation of work:  
$$\text{Work done}= \text{Force} \times \text{displacement along the direction of applied force}$$    
is the displacement as a consequence of the applied force. The dot product between force and displacement (picks up the component of displacement in the direction of the force) vanishes if the force and displacement are independent of each other (or orthogonal to each other in some sense). Here the displacement happens not due to any force. So, the component of force that caused the displacement is zero. Hence there is no work done.   
Then by the work energy theorem, there is no change in the kinetic energy of the body. Hence the conservation of energy. This makes sense. No work is done on the body as there is no force to do that. The body is under constant motion, which means momentum is conserved. That is, $p=\text{constant}\implies \Delta p=0$ and so $K.E=p^2/2m=\text{constant}\implies\Delta K.E=(\Delta p)^2/2m=0$.  
Conclusions:  


*

*An unbalanced force causes acceleration. It cannot make objects move at constant velocity. Constant velocity is achieved only in the absence of an external force.  

*According to Newton's second law, a force is not necessary to keep an object in motion (there is motion, a constant one, possible even in the absence of a force). However the motion will be an unaccelerated one, and takes place along a straight line.  

*So, in constant motion, the displacement is not due to some force. The force is absent and so the work done is zero.   

*Since the object is moving at a constant velocity, its momentum is not changing, which means that the kinetic energy of the body is not changing. So work done is zero implies the change in kinetic energy is zero. Hence the work- energy theorem.

A: The definition of work in a mechanical situation is $$W_{1\to 2}=\int_{\vec{r}_1}^{\vec{r}_2} \vec{F}_{net}\cdot\mathrm{d}\vec{r}.$$
Here we see that if the net force has a component along the direction of motion, then work is done on the particle.  If the force is completely perpendicular to the path of motion, zero work is done.
We also know that if a particle is moving at constant speed, then $$\vec{v}\cdot\vec{v} = v^2 =\mathrm{constant},$$ and the time derivative of this will be zero:
$$\frac{\mathrm{d}}{\mathrm{d}t}\left(\vec{v}\cdot\vec{v}\right)=0=2\left(\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}\cdot\vec{v}\right)=2\left(\vec{a}\cdot\vec{v}\right).$$
We see here that either $\vec{a}=0$, which means the net force is zero, or the net force is perpendicular to the instantaneous velocity, and hence, perpendicular to the path, so that zero work is done.
A: 
your statement  may be revised as
 If a Force is acting on the particle/body/system  and at some point of time is placed at  point number (1) - 
if it  gets displaced to point number (2) then the amount of work performed on the body/particle /system will change its total  energy -
 the amount of change is the difference between its energy at point (2) and point (1). 
so work energy theorem states that work done by the  net unbalanced force is stored in the energy of the particle/body or system.
Let  us understand  by some example-
 we have a body  kept  on an inclined plane
and somebody starts pushing the body from a point (1) to  some higher point (2) on the ramp by application of pushing  force -
 the work done by the force can be calculated by scalar product of Force vector and the displacement vector.
Now work energy theorem will lead us to an equivalence of
work done W = change in energy = Energy E(2) - Energy E(1)
The energy of the body may contain two parts - the Potential Energy and Kinetic Energy. As the body was at rest at point (1) its Kinetic Energy will be zero but its Potential Energy will be there.
At point (2)  its potential energy has changed but a kinetic energy may also be there by the action of force.
So  W= K.E.(2)+P.E.(2) - P.E. (1)
Now your question can be handled that imagine a situation when a force is acting on the body but does not produce acceleration and the body maintains a constant speed-
 then  the question arises as to how you know that force is acting on it ?
No doubt its a thought experiment - but the definition of Newtonian force is through change in momentum-....
( and if the force is  doing some work this work must be stored somewhere (conservation of Energy  must hold).
I just try to give an example-  say a body  is kept on the table and somebody picks it up by applying a force and moves it  to a height h -
all along the path  having an uniform speed so the initial speed is  say v and it is maintained by the agent applying the force and moves it to height  h


*

*as far as K.E. is concerned its the same but the agent's force did a work  equal to the change in potential energy of the body 


from point(1) at the table to height h above - so  naturally the work done by the force is not lost .
 try to think it over by taking other  physical examples.
