Relation between Force, Time, and Energy 
*

*Is energy involved independent of force applied over time?

*In other words, if I wish to apply a force for much more time would I require more energy?
 A: In general, it takes no energy to apply a force. Energy is described as the potential to do work, which has the same units as energy. Work is defined as a force applied during some distance. From these definitions, it is clear that the duration of the force does not directly impact how much energy is required.
For instance, Earth exerts a force of gravity over the Moon. This force persists and the Moon does not get closer (at least not significantly for this purpose). But Earth does not lose any energy by maintaining this force. If the object acted on does not move parallel to the force, no work is done and no energy is used, so Earth can hold on to the Moon indefinitely.
That said, if the force causes the object to move and a longer duration would make it move more, then it would require more energy. Also, if you were thinking about holding a heavy book or something, it does cost more energy to hold something up in the same place for longer times simply because you are using the energy to keep your muscles contracted. If you don't apply a signal (voltage) to your muscles, they don't contract and you can't hold anything. Thus, to hold up the book longer, you need to use more energy to keep your arm from going limp. But this is different from mechanical energy.
TLDR: mostly no, energy is independent of a forces duration. Sometimes yes though.
A: First, to clear something up, the amount of work done does not depend on the amount of time a force is applied, but on the distance over which the force acts. If you and your friend, Alfred, use the same force to push a block from point A to point B, but it takes you ten years and Alfred ten seconds, you both end up doing the same amount of work, and hence $\Delta E$ is the same.
Now, consider that your force acts over a longer distance...  
One way to answer the question is to consider the amount of work $dW$ done by a force $\vec{F}$ over a path $d\vec{s}$. This is given by $dW = \vec{F} \cdot d\vec{s}$. Therefore, if the force acting on the particle is perpendicular to the particle's path, no work is done ($\cos{\pi/2}=0$) and the particle's energy does not change. Central forces such as gravity can accomplish this.
Another way to answer this is to consider the type of force involved. There are two types of forces: conservative and nonconservative. If the work done by the force in moving a particle from point A to point B is independent of the path taken from A to B, then the force involved in conservative. If the work done is dependent on the path taken from A to B, then the force involved is nonconservative.
Consider a particle moving from A to B along one path, then from B back to A along a different path
$$
A \rightarrow C \rightarrow B \\
B \rightarrow D \rightarrow A
$$
For conservative forces, the work done is independent of the path, so the amount of energy gained in going from $A \rightarrow C \rightarrow B$ is equal and opposite to the amount of energy gained in returning to A along the path $B \rightarrow D \rightarrow A$. Therefore, the net energy gained is zero. For nonconservative forces, the energy gained/lost along the two paths can be different.
A: The energy you input is equal to the work you perform, that is force times distance. Thus the energy will grow if you continue to apply force and the system on which you apply force continues to move. The rate at which you bring this energy to the system per unit of time is the power, if it is constant through time, then energy equals power times duration.
Note that the energy you input in the system is not necessarily equal to the energy you spend, because there may be losses internal to your own "system" (your muscles if you're acting manually, a motor,...)
A: Exerting a force over a time interval is not always related to energy being put into the system.
The amount of energy given to a system by a force is called the work. This is computed by calculating the projection of the force onto the displacement made by the object.
$$ \text{(Change in Energy)} = W = \vec{F}\cdot d\vec{l} = F dl \cos(\theta)$$
This means that you don't necessarily need to input energy when exerting a force.
$\textbf{Example 1}$
Place a textbook on a level table. You should see that the textbook doesn't (it's velocity is $0$). 
Since it isn't moving Newton's first law tells us that the forces acting on the book must be balanced. 
One of the forces is the force of gravity on the textbook. This will be directed downward toward the floor and has a magnitude equal to the weight of the text. For the sake of argument lets say the book weighs $5 \text{ lbs}$. 
The other force is provided by the table. In order to balance against the force of gravity it must be directed upwards and also have a magnitude of $5\text{ lbs}$.
The book experiences no displacement ($d\vec{l}=0)$ therefore we conclude that the work donw is $0$. 
So we have a situation in which forces are being exerted on the book but no energy is being inputted into the system. It should also be clear that these forces will act as long as we leave the book there and therefore there is no need to input energy into the system. 
$\textbf{Example 2}$
We have established now that we only need to give energy to a system when exerting a force if our force causes a displacement. However there are even situations in which an object is displaced by no work is done.
Consider a mass hanging by a string. You pick up the free end of the string and start to twirl the mass around so that it traverses a circle at a steady rate. In this case the instantaneous displacement of the mass is tangent to the circle. 
From experience we can conclude that a string can only support a force which is directed along its length. Therefore if we are careful to keep our hand at the center of the circle we know the force will be directed along the radius of the circle.
When a radius intersects a tangent at circumference of a circle they meet at a right angle. From this we can conclude that no work is done since, 
$$ W = F dl \cos(90^\circ) = 0.$$
What this means is that although we have to do work (give energy to the system) to get the mass going in a circle we don't have to put more energy into the system to keep it going.
If you have any experience doing this you will probably believe that the analysis is wrong since in practice once you start twirling the string the object immediately falls down. This is because our experiment is taking place in the presence of gravity which I didn't take into account in the above discussion. So technically you would have to do the experiment in free fall to really see it work.
$\textbf{Example 3} $
Suppose we want to accelerate an object from rest up to some speed $v$, but only have an energy $E$ available to us.  
Suppose we are going to use a constant force $F$ to accelerate the object. Then we can suppose that it takes a time $T$ to get the object up to speed. There average speed is $v/2$ which lets us compute the distance traveled while accelerating ($L$).
$$ L = \frac{vT}{2} $$
We can then compute the work done by the constant force,
$$ W = FL = \frac{FvT}{2},$$
and we can relate this to the kinetic energy using the work energy theorem,
$$ \frac{mv^2}{2} = \frac{FvT}{2} $$
$$ v= \frac{FT}{m} $$
From this we can see that the work done in terms of the force and the duration of time is,
$$ E =  W = \frac{F^2T^2}{2m}$$
It should be clear that for a fixed energy $E$ we can exert the force over a arbitrarily long time so long as we are willing to reduce the magnitude of the force. 
$\textbf{Example 4}$
In the above example if we don't hold the energy delivered to the system constant but instead hold the force at a fixed value then it should be clear that a longer duration of force will require a larger energy input. Which I think is what you were referring to in your question.
A: You might want to use conservation of momentum for that purpose.
For example, if you apply a force to an object, you can write the conservation of momentum like this:
$$
m\cdot v_1+\int_{t_1}^{t_2}F\,\mathrm dt=m\cdot v_2
$$
I used this equation because I was given a time-dependent tension force and I needed to input time as the integral variable.
A: Energy is the capacity of ability of a system to create natural events.Work is the transformation of energy from one kind to another and the transfer of energy from one object to another.
