More work done at a higher velocity with the same force? I'm reading mechanics at undergraduate level and I'm wondering if I understand power correctly.
Let's suppose we have mass and we push on it with the constant net force $ \mathbf{F} $ along the positive x-axis (there is no friction.) If $ \mathbf{v_{t_1}}$ is the velocity along x-axis at some time $t_1$ the power is $ \mathbf{F} \cdot \mathbf{v_{t_1}} $ 
This implies that the power increases with increasing velocity. So after some time $t_2 = t_1 + dt$ it is true that $ \mathbf{F} \cdot \mathbf{v_{t_1}} < \mathbf{F} \cdot \mathbf{v_{t_2}} $
Is this correct? 
If so it just feels strange to me as I have earlier thought, in high school, that the power is constant if we apply a constant force on a mass. One of the reasons I think it's strange is that if the velocity of the mass goes to light-speed then a very small push on the mass will change the power dramatically and will cause a big change in the kinetic energy of the mass. Where as if you push, with the same force, on the mass when it has a low velocity it will cause a small change in power and kinetic energy. 
If all this is correct then I think it implies that if we want to dramatically change the kinetic energy of an object, we should apply a force on it when the object has a very large velocity. (I don't know how this would turn out in practice though.)
 A: Power is, of course, the rate of energy transfer.   The amount of energy transferred is the product of force and distance.   In fact, $P=\mathbf{F\cdot{v}}$   is actually a rewriting of   $$P={\mathbf{F\cdot\frac{d}{\mathsf{t}}}}$$
In the absence of friction, the mass is accelerated by the constant force.   Although  the change in velocity, $a\Delta t$, at a later time interval is the same as the change in velocity, $a\Delta t$, at an earlier time interval of the same duration, the distance traveled during the later time interval, $$\mathbf{d_2}=\mathbf{v_{t_2}}\Delta t+a\Delta t$$ is greater than $$\mathbf{d_1}=\mathbf{v_{t_1}}\Delta t+a\Delta t$$
So yes, the power is increasing with increasing velocity.
At speeds growing significantly close to light speed, it can't be ignored that $a$ gets smaller and smaller with the same constant applied force, because the inertial mass gets bigger and bigger.  So the power would approach a constant value.
A: $\text{work done = force}\times \text{displacement}$  
$\text{power} = \dfrac{\text{work done}}{\text{time taken}} = \text{force} \times \text{velocity}$  
So if a force moves a greater distance in unit time the rate of working (power) will be larger.
Drop a mass of $2 \,\rm kg$ at rest and allow it to fall $5\, \rm m$.  
Instantaneously at the start the power will be zero and at the end it will be $200 \, \rm W$ the fall taking a time of one second and the loss of gravitational potential energy (and gain in kinetic energy) being $100 \, \rm J$.
Now look at the motion over the next second during which the mass falls a distance of $15\, \rm m$.  
At the start the power will be $200 \, \rm W$ and at the end it will be $400 \, \rm W$ the fall taking a time of one second and the loss of gravitational potential energy (and gain in kinetic energy) being $300 \, \rm J$.
Possible your uncertainty is due the fact that one quantity is dependent on distance and the other dependent on time?
So trelling further in a given time requires the expenditure of more energy.
