Work and energy problem I am trying to solve the problem (see image), my concern is about the 2 factor, can anyone please help me understand this.

 A: The factor of two is necessary because when the motor moves the rope a certain distance, the crate only rises half of that distance because of the pulley configuration.  Therefore, if the crate rises $15\,\mathrm m$, the pulley had to pull up $30\,\mathrm m$ of rope to make that happen.
We can prove this mathematically as follows.  Let $\ell$ denote the total length of the rope.  Let $x$ denote the horizontal length of rope in the diagram, and let $y$ denote the vertical distance from the pulley to the pulley holding the crate.  Then we have
\begin{align}
  x + 2y = \ell
\end{align}
The factor of two comes from the fact that there are two vertical segments of rope with length $y$.  Hence, if $\Delta x$ denotes a change in $x$, and if $\Delta y$ denotes a change in $y$, then we have
\begin{align}
  \Delta x = -2\Delta y.
\end{align}
A: Another way of looking at the answer of joshphysics;
Let $z$ be the amount of cable pulled in by the motor, so that:$$z=2x$$ which in turn implies that$$x=\frac{z}{2}$$and the force exerted by the motor is $$F=600+2x^2=600+2\left( \frac{z}{2}\right) ^2=600+\frac{z^2}{2}$$Integrating to find the work done (note upper limit!):$$W=\int_0^{30}\left( 600+\frac{z^2}{2}\right) dz$$
If we evaluate this integral, subtract the potential energy of the crate , and equate the remainer to the KE of the crate, we find the same correct answer...
