Answer
$v = 5.5~m/s$
Work Step by Step
We can find the speed at the bottom of the ramp:
$K_1+U_1+W = K_2+U_2$
$0+U_1+W = K_2+0$
$K_2 = U_1+W$
$\frac{1}{2}mv^2 = mgh-F_f~L$
$v^2 = \frac{(2)(mgh-F_f~L)}{m}$
$v^2 = \frac{(2)(mgL~sin~\theta-L~mg\mu_k~cos~\theta)}{m}$
$v^2 = (2)(gL~sin~\theta-L~g\mu_k~cos~\theta)$
$v = \sqrt{(2)(gL~sin~\theta-L~g\mu_k~cos~\theta)}$
$v = \sqrt{(2)[(9.8~m/s^2)(3.7~m)~(sin~39^{\circ})-(3.7~m)~(9.8~m/s^2)(0.28)~(cos~39^{\circ})]}$
$v = 5.5~m/s$
