Work and kinetic energy on an incline? In my 100-level university physics course, we are just starting to touch on work and kinetic energy. While I feel I have a good understanding of the concepts, there is one problem on the homework that I am struggling with:

"A mass $m = 17~\mathrm{kg}$ is pulled along a horizontal floor with no friction for a distance $d = 5.8~\mathrm m.$ Then the mass is pulled up an incline that makes an angle $\theta$ = $35^\circ$ with the horizontal and has a coefficient of kinetic friction $\mu_k = 0.38.$ The entire time, the massless rope used to pull the block is pulled parallel to the incline at an angle of $\theta$ = 35${^\circ}$ (thus, on the incline, it is parallel to the surface) and has a tension $T = 86~\mathrm N.$
(a) What is the work done by the tension before the block goes up the incline?
(e) How far up the incline does the block travel (measured along the incline) before coming to rest?"

In my search for a solution to (e), I came across one on Yahoo! Answers which said to solve for $\Delta x$ in the following equation:
$$W_\textrm{gravity} + W_\textrm{kinetic friction} - T|\Delta x| = (a)$$
Which, using the direction of $T$ as my positive x-axis, I expanded:
$$|mg\sin\theta||\Delta x|\cos\theta_{\Delta x, g} + |\mu_kmg\cos\theta||\Delta x|\cos\theta_{\Delta x, k} - T|\Delta x| = Td\cos\theta$$
Assuming the equation I found is a solution, what I am failing to understand is why and how the work up the incline relates to the work on the horizontal surface. Any help understanding this equation would be much appreciated.
 A: Start by assuming that the 17 kg mass is the system and now consider what happens.
Initially the external force (tension) is doing work on the system and the system is gaining kinetic energy so when the system is at the start of the incline it has an amount of kinetic energy which you can evaluate.
Work is done on the system and this results in an increase in the kinetic energy of the system. 
On the slope in terms of forces acting on the system you have the tension up the slope ($86$ N) and the component of the weight of the system (~ $140$ N) down the slope and the frictional force down the slope.
So there is a net force down the slope which will result in the system slowing down and thus overall the system will do work.
As the system is doing work its kinetic energy will decrease until eventually it becomes zero.  
(Work done on system)$_{\text{before slope}}$ = change in kinetic energy of system  
Change in kinetic energy of system = (Bet work done by system)$_{\text{on slope}}$
The (Work done by system)$_{\text{on slope}}$ has three contributions:  


*

*(Work done on system)$_{\text{tension}} $

*(Work done by system)$_{\text{friction}}$

*(Work done by system)$_{\text{component of weight}} $

A: I'm not going to comment on whether that answer is correct, but to answer your question, why do you think the work up the incline relates to the work along the horizontal surface?
The question (e) doesn't ask about work, it asks how far it goes before coming to rest. (What does it do once it's at rest? Does it start heading back down, or does it just sit there? Why does it come to rest?) Does comparing the work done before it reaches the incline to the work done along the incline seem to be the way to solve it?
