Answer
$KE=1.6 \ J$
Work Step by Step
We call $\Delta x$ how far on the incline the block goes.
Thus, we find using conservation of energy that:
$ \frac{1}{2}kx^2 = mgh + \mu F_n \Delta x$
$ \frac{1}{2}kx^2 = mgh + \mu mgcos\theta \Delta x$
$ \frac{1}{2}(2000)(.1)^2 = (4.5)(\Delta x sin30) + (.5)(4.5)cos30 \Delta x$
$\Delta x = 2.38$
This is greater than 2, so the block reaches the top. We find its kinetic energy there:
$ \frac{1}{2}(2000)(.1)^2 = (4.5)(\Delta x sin30) + (.5)(4.5)cos30 \Delta x+\frac{1}{2}mv^2 $
$ \frac{1}{2}(2000)(.1)^2 = (4.5)(2 sin30) + (.5)(4.5)cos30(2)+KE $
$KE=1.6 \ J$
