Power required to move an object 
*

*Edited to better reflect the answer I was after *


I have a trolley with laden weight of 110kg that needs to ascend an incline of 6 degrees at a speed of 10km/h. Provision is made to reach that speed within 5sec.
For the sake of argument, ignore any friction on the wheels.
I've used this http://www.dummies.com/education/science/physics/calculating-the-force-needed-to-move-an-object-up-a-slope/
And ascertained the force required to move the trolley will be 113N (if I did it correctly), but I am not sure how now to add the speed requirement to get the required Wattage of the motor?
I understand there will be an initial requirement to get to 10km/h in say 5s, then a requirement to sustain that momentum.
I am looking for (a) the initial power required by the motor to accelerate the load to 10km/h, and then (b) the continuous power demand required to maintain the speed.
 A: I'll restate the problem that the OP wants to solve as follows: 
Given a trolley with a mass of $m=110\,\mbox{kg}$ on an $\alpha=6^\circ$ incline, what is the constant power required so that we can accelerate the trolley up the incline such that it can reach a speed of $v_0=10\,\mbox{km/h}$ within a time of $T=5\,\mbox{s}$? We assume the trolley experiences no friction, and is thus subject only to a gravitational force corresponding to a gravitational acceleration of $g=9.81\,\mbox{m/s}^2$.
To solve this problem, we write down the required power $P$ as a function of acceleration $a=v'$ as
$$P=m\,v'(t)\,v(t)+m\,g\,\sin(\alpha)\,v(t)$$
Solving for $v'$ we find the ordinary differential equation
$$v'(t)=\frac{P}{m\,v(t)} - g\,\sin(\alpha).$$
I used Mathematica to solve the above ODE with the initial condition $v(t=0)=0$. There might be more elegant solutions than this one, but figuring those out would cost me a bit of effort...
Mathematica's solution of this differential equation is somewhat ugly, and contains the Lambert W function (also known as the product logarithm, defined as the principal solution of the equation $z=W \mbox{e}^W$). We get
$$v(T)=P\,\frac{W\left(-e^{-\frac{m\,\left(g\,\sin(\alpha)\right)^2}{P}\,T-1}\right)+1}{\sin(\alpha)\,g\,m}.$$
Solving for $P$ results in
$$\hat{P}=\frac{G+1}{G+1+W\left(- \left(G+1\right)\mbox{e}^{-(G+1)}\right)},$$
where we have used the symbols $\hat{P}=\frac{P}{m\,g\,\sin(\alpha)\,v_0}$ and $G=\frac{g\,\sin(\alpha)}{v_0/T}$. 
A plot of the function $\hat{P}(G)$ is shown below:

Plugging in the data, we have $G\approx1.85$, which gives $\hat{P}\approx1.0766$, or
$$P\approx337.31\,\mbox{W}.$$
This looks about right; notice that at the beginning of the acceleration, less power is needed to push the trolley uphill (since it's slower), and that the acceleration will therefore be larger in the beginning and become less later on. Once we have reached the designated speed $v_0$, we can reduce the power to $\hat{P_e}=1$, or $P_e=m\,g\,\sin(\alpha)\,v_0\approx313.32\,W$.
A: So you are asking of the power requirement, assuming there is no friction.
The definition of power is $P=\frac{dW}{dt}$.
If we plug in the definition of work $W=\vec{F}\dot{}\vec{s}$ and simplify it, using that the force is constant and in the same direction as the movement, it becomes: $P=\frac{dW}{dt}=\frac{d(\vec{F}\dot{}\vec{s})}{dt}=\frac{d(Fs)}{dt}=F\frac{ds}{dt}+s\frac{dF}{dt}=F\frac{ds}{dt}=Fv$. So in words: the power ist simply the product of force and speed.
If you are motivated: You can add friction easily and analogous (F is still the force needed due to the slope of the curve): $P=(F+F_\mathrm{friction})v$. The friction can be divided in two parts: one due to air proportional to $v^2$, and one due to the contact with the concrete proportional to the perpendicular part of the weight force. $$F_\mathrm{friction}=F_\mathrm{air}+F_\mathrm{concrete}=c_1v^2+c_2F_{\perp}=c_1v^2+c_2mg\cdot\sin(6^\circ)$$
