Essential University Physics: Volume 1 (3rd Edition)

$KE=1.6 \ J$
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: $\frac{1}{2}(2000)(.1)^2 = (4.5)(\Delta x sin30) + (.5)(4.5)cos30 \Delta x+\frac{1}{2}mv^2$ $\frac{1}{2}(2000)(.1)^2 = (4.5)(2 sin30) + (.5)(4.5)cos30(2)+KE$ $KE=1.6 \ J$