How does the work done by a cord-over-a-pulley work? 

I can use the integral of power with respect to time to solve problem.
In this situation, what is the work done on the cart by the tension?
Please do not use calculus.
 A: I evaluated the work done (change in kinetic energy of the cart) using two approaches.  I thought the second approach could possibly be simpler, but it is not.  I had to use calculus for both approaches.  I cannot think of a simple way to evaluate the work.  The second approach evaluates the work done on a system of particles, a topic not addressed in many textbooks that discuss the motion of a system of particles.
Consider the system of interest as the cart. The work done on the cart is due to the net force on the cart.  The forces on the cart are $T$, the tension in the cord acting at an angle $\theta$ on the cart, $mg$ down, and $N$ the normal force up on the cart.  The only non-zero net force is in the horizontal direction due to $Tcos\theta$. This assumes no force of friction between the cart and the ground. Note that $\theta$ is not constant but increases as the cart moves left, and that needs to be accounted for in evaluating the work. The figure below shows the evaluation for work on the cart; the result is $W = T(d_2 - d_1)$ where $d_2 - d_1$ is the total distance the cord moves to the left of the pully.

To see if there is a simpler approach to evaluating the work, take the system as: the cart, the pully, and the portion of the cord from the pully to the cart.
First, the general relationships for a system of particles are summarized in the figure below. The general relationship for work done on a system of particles is not addressed in many texts, but is developed by Goldstein. [Goldstein, Classical Mechanics]  These relationships depend on the position of the particles with respect to the origin of an inertial reference frame, labeled $O$ in the figure, which can be specified as desired.

The general relationships can be applied to the system of interest here.  See the figure below.  This system is not a rigid body because the cart moves, but the only body in the system with mass is the cart, assuming a massless pully and massless cord.  The origin  $O$ is taken for convenience of calculation as the point where the cord outside the system enters the system. The only external force applied is  $\vec F_{ext} = \vec T$ , and the only body in the system to which this external force is applied is the massless pully; there is no external force applied to the cart.  The cart experiences an internal force from the cord, $\vec T$.  So the equation for work done on this system is due to the internal force on the cart.  As developed in the figure, using this approach we need to evaluate the same integral as was previously evaluated for the case where the system in the cart alone.  So, this approach does not simplify the evaluation of work.
I am not aware of any other approach that would simplify the evaluation of work.

The tension force in the cord is constant over the entire cord assuming a massless string and massless pully and no friction loss on the pully. Once the cart starts to move, the cord very quickly stretches a minor bit to provide tension, but then the cord acts as a rigid body and there is no change in the length of the cord.  The speed of each particle in the cord is the same.  Specifically, a mass element $\Delta m$ with speed $v$ in the string moves a distance $\Delta s = v \Delta t$ in the time $\Delta t$, and since the string length is constant, over $\Delta t$ the distance $\Delta s$ is constant for each element $\Delta m$ of the string therefore $v$ is constant for element of the string
