Simple Work/Energy Problem

I think your solution is correct. Sometimes the book is wrong.

 

How can that be true if the applied force is horizontal? How is that force affecting the gravitational potential energy?

Work = [tex]\Delta[/tex]KE = [tex]\int[/tex][tex]\vec{F}[/tex] [tex]\bullet[/tex] d[tex]\vec{r}[/tex] = [tex]\int[/tex]Fdrcos(0) = [tex]\int^{v_{f}}_{v_{i}}[/tex]vdv = (1/2)m([tex]v_{f}^{2}[/tex] - [tex]v_{f}^{2}[/tex])

From the picture, it looks like the initial and final velocities are at non-horizontal angles, so wouldn't you have to find the horizontal components of those velocities to use KE?

Also, assuming state 1 is the initial position and state 2 is the final position, from the picture it looks like the change in height should be negative, since the object moves downhill.

 

I solved the problem and obtained the answer of -0.09m. To do this is assumed that the velocity at state 1 is in the x direction and the velocity at state 2 is in a random direction. I then set up my work kinetic energy equations for both the x and y directions. In the x i got Fx = 1/2(3.2kg)(vf)^2-1/2(3.2kg)(8.8)^2= (132N)(.8m). In the y direction i got (9.8m/s^2)*h = 1/2(Vf)^2. The final equation to tie it all togther is just ((Vfx)^2+(Vfy)^2)^.5 = 11.9m/s . This last equation is just setting the known final velocity magnitude of 11.9m/s equal to the square root of the sum of the squares of its components. Thus from the first two equations we can find (vfx)^2 which is just a number and (vfy)^2 which is a function of height. Hence the only unkown is the h and it will turn out negative.